MA0 1

Introduction to University Mathematics (Proofs)

Every lesson so far, in one document · 10 chapters

Lesson 2

Predicates, Quantifiers, and Proofs

Taught 12 Aug 2026

Predicates

There is a problem: So far, what we’ve built (propositional logic) has no way to express the classic (correct) argument “All men are mortal. Socrates is a man. Therefore Socrates is mortal,” because propositional logic treats each proposition as an indivisible whole. It therefore has no way to look inside a statement and relate the object (Socrates), the property (being a man), and the consequence (being mortal); without that internal structure, we cannot formally justify the conclusion.

What is missing is the ability to distinguish the object of our speech from the description we make about it, so we introduce predicate logic, which allows us to reason about objects and their properties.

Variables and Predicates

Recall that a term is a symbol that denotes an object: constants such as the natural number $5$ or $\pi$ are terms denoting specific objects, while an unquantified variable such as $x$ is a term whose value has not yet been specified. We can then form expressions describing the object denoted by a variable, such as $\operatorname{man}(x)$ to say ” $x$ is a man,” or $x>3$ to say ” $x$ is greater than $3$ .” We call these expressions predicates (also known as variable propositions), since they make a claim that depends on one or more variables.

Definition 2.1 (Predicate).

Let $x_1, \ldots, x_n$ be variable symbols. We say $\varphi(x_1, \ldots, x_n)$ is an $n$ -ary predicate if its truth value is determined once values are assigned to each of the variables $x_1, x_2, \ldots, x_n$ (or equivalently, replacing each variable by a term denoting a specific object produces a proposition).

A $1$ -ary (monadic) predicate describes a property, e.g., $\text{IsPrime}(x)$ .
A $2$ -ary (dyadic) predicate describes a relation between two terms, e.g., $\text{IsGreaterThan}(x, y)$ .
An $n$ -ary predicate describes a relation among $n$ terms.

Note that we would usually write such a declaration as: let $P(x)$ be the statement ” $x^2-1=0$ .”

Remark.

A term by itself is not a complete sentence, so it has no truth value; likewise, as hinted earlier the resulting expressions above ( $x>3$ ) are still not propositions because their truth value depends on the actual value of $x$ .

Applying a predicate to the appropriate number of terms gives an atomic formula, the simplest kind of formula in first-order logic; for example, if $P$ is the predicate “is prime,” then $P(x)$ and $P(7)$ are atomic formulas. If all the terms in an atomic formula denote specific objects, then the formula has a definite truth value and is a proposition.

Example 2.2.

Let $R(x, y, z)$ denote $x + y = z$ , where the variables range over the integers. Then $R(2, -1, 5)$ is $\bot$ since $2 + (-1) \neq 5$ ; $R(3, 4, 7)$ is $\top$ since $3 + 4 = 7$ ; and $R(x, 3, z)$ is still a predicate, since $x + 3 = z$ can be $\top$ or $\bot$ depending on the values of $x$ and $z$ .

Remark.

It is common to use capital letters such as $P, Q, R$ to denote predicates.

A predicate’s variables take values in a domain $U$ , called the universe of discourse (in the previous example, $U$ is the integers). Once values from $U$ are substituted, the predicate becomes a proposition with truth value $\top$ or $\bot$ ; hence we can now apply the logical connectives introduced earlier.

Example 2.3.

Let $P(x)$ denote $x > 0$ . Then the following are true: $P(3) \lor P(-1),$ $P(3) \to \neg P(-1);$ and the following are false: $P(3) \land P(-1),$ $P(3) \to P(-1)$ .

More generally, we can build new predicates out of old ones using connectives. Expressions constructed from predicates and logical connectives whose truth value still depends on one or more variables are called propositional functions.

Example 2.4.

Using $P(x)$ as above, the following are propositional functions:

R(x, y) \defeq P(x) \to P(y), \qquad S(y) \defeq P(3) \land P(y)

$R(x, y)$ is a predicate in two variables; $S(y)$ is a predicate in one variable (since $P(3)$ is already a proposition).

Problem 2.1.

Consider the expressions $\text{man}(x)$ , $x > 3$ , and $2 + 3 = 5$ . Which of these are propositions and which are predicates? For each predicate, give one substitution that makes it $\top$ and one that makes it $\bot$ .

Remark (History and Significance of Predicate Logic).

Aristotle developed a limited form of predicate logic through his theory of syllogisms. The modern version was independently developed by Frege and Peirce between the 19th and 20th centuries, roughly 2000 years later, mirroring the timeline of propositional logic itself. Predicate logic (also called first-order logic) is now the standard language for mathematical statements and is equally fundamental in computer science, appearing in database queries, logic programming (Prolog), automated theorem proving, software verification, and symbolic AI. Some mathematicians, notably Hilbert, hoped it would be a complete system for all of mathematics, but Gödel’s incompleteness theorem showed otherwise: no fixed collection of axioms can prove all true mathematical statements.

Introduction to Sets

We take a quick detour to study sets. Sets are probably the most important structure in mathematics right now; a precise answer as to why they are so important would fill an entire course, but for our purposes it might suffice as a rather crude approximation to say that sets are the building blocks of all mathematics.

Being so important, we introduce them now and will study them in greater detail later. For the moment, the following informal definition will suffice.

Definition 2.5 (Set (to be revised)).

A set is nothing more than a collection of objects. The objects in the collection are called its elements or members. We usually denote sets by uppercase letters $A,B,S,\dots$ , and elements by lowercase letters $a,b,x,\dots$ . If $S$ is a set, then $x \in S$ means ” $x$ is an element of $S$ ;” sometimes we also write $S \ni x$ for $x \in S$ and say ” $S$ contains $x$ .” Also worth talking about, is it’s negation: $x \notin S$ or $\neg(x \in S)$ , means ” $x$ is not an element of $S$ ,” or ” $x$ does not belong to $S$ ;” when needed (not often), some write $S \not\ni x$ which stands for ” $S$ does not contain $x$ .”

Example 2.6.

Let $S = \{2,3,5\}$ . Then $3 \in S$ , but $4 \notin S$ .

Remark.

Membership is a relation between an element and a set. That said, a set can itself be an element of another set. For example, $\{2,3\} \in \{\{2,3\},5\},$ because the collection on the right has two elements: the set $\{2,3\}$ and the number $5$ ; The problem arises when this is confused with the case where only the original objects are elements: $\{2,3\} \notin \{2,3,5\}.$ So $a \in A$ asks whether $a$ is literally a member of the collection $A$ , not merely whether its contents appear among the elements of $A$ .

Problem 2.2.

Let $S = \{2,3,5\}.$ Which of the following are true?

(a) $3 \in S$

(b) $4 \notin S$

(c) $5 \in S$

(d) $1 \in S$

There is one refinement we will need: Not every collection is useful in mathematics, for example, “a collection of all beautiful paintings” doesn’t make sense because whether a painting belongs to it depends on the judge; with a different panel of judges we might get a different result, “beauty is in the eye of the beholder” afterall.

Either way, to combat this we normally restrict ourselves to what is called a well-defined set. Informally, a well-defined set is a set such that, for every object, there is a specific objective rule that allows us to determine whether the object belongs to the set or not; one or the other must happen, but not both. Well aside from being a mouthfull, this definition runs into the same exact problems we introduced earlier; our current model cannot describe this so we finally introduce one which can.

Quantifiers

Next we make claims about all values in a domain, or assert that at least one satisfying value exists. These are the universal and existential quantifiers.

The Universal Quantifier

Definition 2.7 (Universal Quantifier).

Let $\varphi(x)$ be a predicate with variable $x$ ranging over a universe of discourse $U$ . The universal quantification of $x$ in $\varphi$ , written $\forall x\,(\varphi(x))$ , is the proposition asserting that $\varphi(a)$ is $\top$ for every element $a \in U$ . It is read “for all $x$ , $\varphi(x)$ .”

Determining the truth of $\forall x\,(\varphi(x))$ amounts to exhaustive verification: examine every object in the universe, and if even one makes $\varphi$ false, the entire statement is false. If every object passes, the statement is true; a value $a$ for which $\varphi(a) = \bot$ is called a counterexample.

If the universe is the collection $U = \{a_1, a_2, \ldots, a_n\}$ , the universal quantifier reduces to a conjunction:

\forall x\,(\varphi(x)) \equiv \varphi(a_1) \land \varphi(a_2) \land \cdots \land \varphi(a_n)

Remark (Indexed Conjunction and Disjunction).

When a conjunction or disjunction extends over a collection of terms, we write

\bigwedge_{i=1}^{n} \varphi(a_i) \;\defeq\; \varphi(a_1) \land \varphi(a_2) \land \cdots \land \varphi(a_n), \qquad \bigvee_{i=1}^{n} \varphi(a_i) \;\defeq\; \varphi(a_1) \lor \varphi(a_2) \lor \cdots \lor \varphi(a_n)

More generally, for any index set $S$ the notations $\bigwedge_{a \in S} \varphi(a)$ and $\bigvee_{a \in S} \varphi(a)$ denote the conjunction and disjunction, respectively, of $\varphi(a)$ over every $a$ in $S$ . In this notation, the finite-universe equivalences above become $\forall x\,(\varphi(x)) \equiv \bigwedge_{a \in U} \varphi(a)$ and, as we shall see shortly, $\exists x\,(\varphi(x)) \equiv \bigvee_{a \in U} \varphi(a)$ .

Specifying the Quantification Set

Earlier our description of a quantification states that it “ranges over a universe of discourse,” take for example the sentence

\forall x\, (x^2 \geqslant 0)

which may look true, but as it stands it is meaningless. What is $x$ ? “For every $x$ ” from where? If $x$ ranges over colours, or over sentences, then $x^2 \geqslant 0$ is not false, it is not even grammatical. We need to know where the object $x$ comes from before we can ask whether the claim about it is true.

If we amend the sentence so that $x$ ranges over the real numbers, we get a well-formed statement, and a true one. If instead we let $x$ range over the complex numbers, we get an equally well-formed statement that is now false: the imaginary unit $i$ is a complex number with $i^2 = -1$ , and $-1 \geqslant 0$ fails. (We will make little real use of the complex numbers in these notes, but they are a convenient source of counterexamples.)

Example 2.8.

Let $P(x)$ denote ” $x > 0$ .”

If the universe is the integers, then $\forall x\,(P(x))$ is $\bot$ : the value $x = 0$ is a counterexample, since $P(0)$ is false. If $U$ is the positive integers (the natural numbers greater than 0), then $\forall x\,(P(x))$ is $\top$ . The truth value of a universally quantified statement depends on the choice of universe.

The Existential Quantifier

Definition 2.9 (Existential Quantifier).

Let $\varphi(x)$ be a predicate with variable $x$ ranging over a universe $U$ . The existential quantification of $x$ in $\varphi$ , written $\exists x\,(\varphi(x))$ , is the proposition asserting that there is at least one element $a \in U$ for which $\varphi(a)$ is $\top$ . It is read “there exists an $x$ such that $\varphi(x)$ .”

Where the universal quantifier demands exhaustive verification, the existential quantifier demands a search: examine objects in the universe until one satisfies $\varphi$ . If such an object is found, it is called a witness and the statement is true. If the entire universe is exhausted without finding a witness, the statement is false.

If the universe is the collection $U = \{a_1, a_2, \ldots, a_n\}$ , the existential quantifier reduces to a disjunction:

\exists x\,(\varphi(x)) \equiv \varphi(a_1) \lor \varphi(a_2) \lor \cdots \lor \varphi(a_n)

If $\forall x\,(\varphi(x))$ is true and $U$ is non-empty, then $\exists x\,(\varphi(x))$ must also be true: if every element satisfies $\varphi$ , at least one does.

Example 2.10.

Let $P(x)$ denote ” $x > 0$ .”

$\exists x\,(P(x))$ is $\top$ when the universe is the integers (witness: $x = 1$ ), $\top$ when $U$ is the positive integers, and $\bot$ when $U$ is the negative integers.

The following table summarises the two quantifiers:

Statement	True when	False when
$\forall x\,(P(x))$	$P(x)$ is $\top$ for every $x \in U$	There exists a counterexample: some $a$ with $P(a) = \bot$
$\exists x\,(P(x))$	There exists a witness: some $a$ with $P(a) = \top$	$P(x)$ is $\bot$ for every $x \in U$

Remark (Quantifiers and Finite Domains).

If the domain is finite, quantifiers are technically unnecessary: $\forall x\,(P(x))$ is a conjunction and $\exists x\,(P(x))$ is a disjunction, both expressible in propositional logic. Quantifiers are needed for infinite domains, where no finite conjunction or disjunction suffices.

Scope and Variable Binding

Definition 2.11 (Scope, Bound Variables, and Free Variables).

The scope of a quantifier is the portion of the formula to which it applies, typically delimited by parentheses. A variable that falls within the scope of a quantifier is bound to that quantifier. A variable not bound by any quantifier is free. A formula with free variables is a predicate, and becomes a proposition only when all free variables are either substituted by terms or bound by quantifiers.

This is why we bracket the body of a quantifier. Quantifiers bind more tightly than all propositional connectives, so an unbracketed formula is read as narrowly as possible:

\forall x\, P(x) \lor Q(x) \quad \text{means} \quad \bigl(\forall x\,(P(x))\bigr) \lor Q(x)

This expression has a free variable $x$ in $Q(x)$ and is therefore a predicate. The two $x$ ‘s are in fact independent: the expression could equivalently be written $\bigl(\forall y\,(P(y))\bigr) \lor Q(x)$ . By contrast, $\forall x\,(P(x) \lor Q(x))$ has both occurrences of $x$ inside the brackets, so both are bound, and the result is a proposition.

Earlier it was shown that a quantifier always comes with a set attached, either named outright or fixed in advance as the universe of discourse. Writing that set into the quantifier itself is common, and has its own notation.

Definition 2.12 (Bounded Quantifiers).

Let $A$ be a set and let $P(x)$ be a predicate. The bounded quantifiers are defined by

\forall x \in A\,(P(x)) \;\defeq\; \forall x\,(x \in A \to P(x)), \qquad \exists x \in A\,(P(x)) \;\defeq\; \exists x\,(x \in A \land P(x))

read “for every $x$ in $A$ , $P(x)$ ” and “there is an $x$ in $A$ with $P(x)$ .” When several variables share a set we abbreviate further, writing $\forall x, y \in A\,(P(x,y))$ for $\forall x \in A\, \forall y \in A\,(P(x,y))$ , and likewise $\exists x, y \in A\,(P(x,y))$ for $\exists x \in A\, \exists y \in A\,(P(x,y))$ .

The restriction enters the universal quantifier through an implication and the existential one through a conjunction, and the pairing cannot be swapped. The statement $\exists x\,(x \in A \to P(x))$ is made true by any object outside $A$ whatsoever, since a false antecedent makes the implication true, so it says nothing about the elements of $A$ at all.

The Unique Existential Quantifier

It is often useful to assert that exactly one object satisfies a predicate.

Definition 2.13 (Unique Existential Quantifier).

The unique existential quantification $\exists!\, x\,(\varphi(x))$ asserts that exactly one element of $U$ satisfies $\varphi$ . It is defined in terms of the other quantifiers:

\exists!\, x\,(\varphi(x)) \;\equiv\; \exists x\,\bigl(\varphi(x) \land \forall y\,(\varphi(y) \to y = x)\bigr)

This reads: “there exists an $x$ such that $\varphi(x)$ , and any $y$ satisfying $\varphi(y)$ must equal $x$ .” It takes a bounded form like the other two: for a set $A$ ,

\exists!\, x \in A\,(P(x)) \;\defeq\; \exists!\, x\,(x \in A \land P(x))

Example 2.14.

Let $U$ be the integers and let $P(x)$ denote ” $x + 1 = 0$ .” Then $\exists!\, x \in U\,(P(x))$ is $\top$ : the unique witness is $x = -1$ . If instead $P(x)$ denotes ” $x > 0$ ,” then $\exists!\, x \in U\,(P(x))$ is $\bot$ , since every positive integer satisfies $P$ and there are infinitely many of them.

Formalising Arguments

We can now symbolise the prime-number argument that propositional logic could not handle. Let $U$ be the integers, let $P(x)$ denote ” $x$ is a prime number greater than $2$ ,” and let $O(x)$ denote ” $x$ is odd.”

“Every prime number greater than $2$ is odd” becomes $\forall x \in U\,(P(x) \to O(x))$ .
“The number $7$ is a prime number greater than $2$ ” becomes $P(7)$ .
By substituting $x = 7$ into (1) we obtain $P(7) \to O(7)$ . Since $P(7)$ is true by (2) and the conditional $P(7) \to O(7)$ is true, bivalence forces $O(7)$ to be true: ” $7$ is odd.”

More broadly, the four classical categorical propositions of Aristotelian logic can be expressed in first-order form. Given predicates $S(x)$ and $R(x)$ over a universe $U$ :

Type	Statement	First-order form
A	All $S$ are $R$	$\forall x \in U\,(S(x) \to R(x))$
E	No $S$ is $R$	$\forall x \in U\,(S(x) \to \neg R(x))$
I	Some $S$ is $R$	$\exists x \in U\,(S(x) \land R(x))$
O	Some $S$ is not $R$	$\exists x \in U\,(S(x) \land \neg R(x))$

Remark (The Connective Trap).

A common error is to write $\forall x \in U\,(S(x) \land R(x))$ , which makes the much stronger claim that everything in the universe is both $S$ and $R$ . Equally, $\exists x \in U\,(S(x) \to R(x))$ is almost always vacuously true: any element that is not $S$ makes the implication true and serves as a witness.

Example 2.15 (Translating Natural Language).

Let $U$ be all people, let $S(x)$ denote ” $x$ is a student in this class,” and let $P(x)$ denote ” $x$ has written a programme in Python.”

“Every student in this class has written a programme in Python” is formalised as $\forall x \in U\,(S(x) \to P(x))$ .

“Some student in this class has written a programme in Python” is formalised as $\exists x \in U\,(S(x) \land P(x))$ .

Quantifier Negation

The universal and existential quantifiers are duals, connected by negation in a manner analogous to De Morgan’s laws for conjunction and disjunction.

Theorem 2.16 (Quantifier Negation).

For any predicate $\varphi(x)$ with universe $U$ :

(1) $\neg\bigl(\forall x\,(\varphi(x))\bigr) \equiv \exists x\,(\neg\varphi(x))$

(2) $\neg\bigl(\exists x\,(\varphi(x))\bigr) \equiv \forall x\,(\neg\varphi(x))$

Discussion.

The propositional laws of the last chapter say nothing about quantifiers, so we read each quantifier as what it abbreviates: a universal statement is a conjunction over the whole universe, an existential one a disjunction. Written that way, each claim is an instance of the generalised De Morgan law (the negation of a long conjunction is the long disjunction of the negations, and conversely), and the last step reads the answer back as a quantifier. The second part is the first with the two connectives exchanged.

Proof.

The two quantifiers expand into an indexed conjunction and an indexed disjunction over $U$ :

\forall x\,(\varphi(x)) \equiv \bigwedge_{a \in U} \varphi(a), \qquad \exists x\,(\varphi(x)) \equiv \bigvee_{a \in U} \varphi(a)

(1)

\begin{aligned} \neg\bigl(\forall x\,(\varphi(x))\bigr) &\equiv \neg\Bigl(\bigwedge_{a \in U} \varphi(a)\Bigr) && \text{by the expansion of } \forall \\ &\equiv \bigvee_{a \in U} \neg\varphi(a) && \text{by generalised De Morgan} \\ &\equiv \exists x\,(\neg\varphi(x)) && \text{by the expansion of } \exists \end{aligned}

(2)

\begin{aligned} \neg\bigl(\exists x\,(\varphi(x))\bigr) &\equiv \neg\Bigl(\bigvee_{a \in U} \varphi(a)\Bigr) && \text{by the expansion of } \exists \\ &\equiv \bigwedge_{a \in U} \neg\varphi(a) && \text{by generalised De Morgan} \\ &\equiv \forall x\,(\neg\varphi(x)) && \text{by the expansion of } \forall \end{aligned}

Remark (De Morgan over a Universe).

The generalised De Morgan’s laws were stated for $n$ propositions, and the proof above applies them to a conjunction indexed by $U$ , which need not be finite (we explore this in a second). When $U = \{a_1, \ldots, a_n\}$ , the step is the corollary verbatim:

\neg\bigl(\varphi(a_1) \land \cdots \land \varphi(a_n)\bigr) \equiv \neg\varphi(a_1) \lor \cdots \lor \neg\varphi(a_n)

Which has an immediate consequences: $\forall x\,(\varphi(x)) \equiv \neg\bigl(\exists x\,(\neg\varphi(x))\bigr)$ and $\exists x\,(\varphi(x)) \equiv \neg\bigl(\forall x\,(\neg\varphi(x))\bigr)$ .

Example 2.17 (Negating in English).

Let $U$ be the real numbers.

“There is a real number $x$ with $x^2 = -1$ ,” in symbols $\exists x \in U\,(x^2 = -1)$ , is $\bot$ . Its negation is the universal statement “for every real number $x$ , $x^2 \neq -1$ ,” which is $\top$ .
“Every real number is positive or negative,” in symbols $\forall x \in U\,(x > 0 \lor x < 0)$ , is $\bot$ : the counterexample is $x = 0$ . Its negation is $\exists x \in U\,\neg(x > 0 \lor x < 0)$ , and De Morgan’s laws turn the negated disjunction into a conjunction, giving $\exists x \in U\,(x \leqslant 0 \land x \geqslant 0)$ : “there is a real number that is neither positive nor negative.” It is $\top$ , with $x = 0$ doing the existing.

Negation and Bounded Quantifiers

The bounded notation of Definition 2.12 is coherent with negation. Applying the theorem to a set $S$ :

\neg\bigl(\forall x \in S\,(P(x))\bigr) \equiv \exists x \in S\,(\neg P(x)) \quad \text{and} \quad \neg\bigl(\exists x \in S\,(P(x))\bigr) \equiv \forall x \in S\,(\neg P(x))

The restricted domain is preserved under negation: only the predicate is negated, not the domain condition. Unfolding the first one confirms this. Writing the bounded quantifier out in full, negating, and using the negation of a conditional:

\neg\bigl(\forall x\,(x \in S \to P(x))\bigr) \equiv \exists x\,\neg(x \in S \to P(x)) \equiv \exists x\,(x \in S \land \neg P(x))

which is the bounded existential $\exists x \in S\,(\neg P(x))$ .

Problem 2.3.

Show the negation of the existential claim to be true.

Distribution of Quantifiers over Connectives

The universal quantifier distributes over conjunction, and the existential quantifier distributes over disjunction:

\forall x\,(P(x) \land Q(x)) \equiv \forall x\,(P(x)) \land \forall x\,(Q(x))

\exists x\,(P(x) \lor Q(x)) \equiv \exists x\,(P(x)) \lor \exists x\,(Q(x))

However, the reverse pairings do not hold in general:

\exists x\,(P(x) \land Q(x)) \;\not\equiv\; \exists x\,(P(x)) \land \exists x\,(Q(x))

\forall x\,(P(x) \lor Q(x)) \;\not\equiv\; \forall x\,(P(x)) \lor \forall x\,(Q(x))

For both non-equivalences, a single counterexample suffices: Let $U = \{a, b\}$ , with $P(x)$ true exactly of $a$ and $Q(x)$ true exactly of $b$ . Then $\exists x,P(x) \land \exists x,Q(x)$ is $\top$ (witnesses $a$ and $b$ ), but $\exists x,(P(x) \land Q(x))$ is $\bot$ , since $P(a) \land Q(a)$ fails on the second conjunct and $P(b) \land Q(b)$ on the first. Likewise $\forall x,(P(x) \lor Q(x))$ is $\top$ (it holds at $a$ via $P$ and at $b$ via $Q$ ), while $\forall x,P(x) \lor \forall x,Q(x)$ is $\bot$ , since $P(b)$ and $Q(a)$ both fail.

Problem 2.4.

Come up with your own example as to why this fails, what situation does it not fail?

Validity and Satisfiability in Predicate Logic

Just as in propositional logic, a quantified statement with all variables bound can be classified by its truth behaviour across all possible interpretations. A statement is valid if it is true for every domain and every choice of predicates (the analogue of a tautology), satisfiable if there exists at least one domain and choice of predicates making it true, and unsatisfiable if no such choice exists.

Example 2.18.

The statement $\forall x\,(\neg S(x)) \leftrightarrow \neg\exists x\,(S(x))$ is valid: it is an instance of Theorem 2.16 and holds for any predicate $S$ and any domain.

The statement $\forall x\,(F(x) \leftrightarrow T(x))$ is satisfiable: taking $F \equiv T$ makes it true, while taking the domain to be the integers with $F(x) \equiv (x > 0)$ and $T(x) \equiv (x < 0)$ makes it false.

The statement $\forall x\,(F(x) \land \neg F(x))$ is unsatisfiable: it asserts that every element simultaneously satisfies and fails to satisfy $F$ , contradicting the principle of bivalence.

Problem 2.5.

Let $P$ be a predicate over an arbitrary universe $U$ . Determine whether each of the following statements is valid, satisfiable, or unsatisfiable, and justify your answer.

(a) $\exists!\, x \in U\,(P(x)) \to \exists x \in U\,(P(x))$

(b) $\forall x \in U\,(P(x)) \to \exists!\, x \in U\,(P(x))$

(c) $\exists!\, x \in U\,(\neg P(x)) \to \neg\forall x \in U\,(P(x))$

Problem 2.6.

Let $U$ be all English texts, and let $P(x)$ , $Q(x)$ and $R(x)$ denote ” $x$ is a clear explanation”, ” $x$ is satisfactory”, and ” $x$ is an excuse” respectively. Formalise the following in predicate logic:

(a) All clear explanations are satisfactory.

(b) Some excuses are unsatisfactory.

(c) Some excuses are not clear explanations.

Problem 2.7.

Let $Z$ be the integers and let $R$ be the real numbers. Using Theorem 2.16 , negate each of the following statements and simplify. State whether the original or its negation is true.

(a) $\forall x \in Z\,(x^2 \geqslant 0)$

(b) $\exists x \in R\,(x^2 < 0)$

(c) $\forall x \in R\,\exists y \in R\,(x + y = 0)$

Nested Quantifiers

The statements encountered so far have involved a single quantifier binding a single variable. Many mathematical claims, however, involve multiple variables and require several quantifiers applied in sequence. A nested quantifier is a quantifier that falls within the scope of another quantifier.

Example 2.19 (Additive Inverse).

The statement “every real number has an additive inverse” involves two variables: the number itself and its inverse. Letting $U$ be the real numbers, the formalisation is:

\forall x \in U\, \exists y \in U\,(x + y = 0)

The outer quantifier $\forall x \in U$ asserts that the claim holds for every real $x$ . The inner quantifier $\exists y \in U$ asserts that, for each such $x$ , a suitable $y$ exists. The two quantifiers are nested: $\exists y \in U$ lies within the scope of $\forall x \in U$ .

A nested quantified statement can be decomposed by treating each inner quantification as a propositional function. For instance, $\forall x \in U\, \exists y \in U\,(x + y = 0)$ can be read as $\forall x \in U\,(Q(x))$ , where $Q(x) \defeq \exists y \in U\,(P(x, y))$ and $P(x, y) \defeq (x + y = 0)$ . Note that $Q(x)$ is itself a predicate: the existential quantifier binds $y$ , but $x$ remains free until the outer $\forall x \in U$ binds it.

Order of Quantifiers

The order in which quantifiers appear is critical. Consider the same predicate $P(x, y) \defeq (x + y = 0)$ , with $U$ still the real numbers.

$\forall x \in U\, \exists y \in U\,(P(x, y))$ asserts: “for every real $x$ , there exists a real $y$ such that $x + y = 0$ .” This is $\top$ : for any given $x$ , the witness $y = -x$ works. Crucially, the witness may differ for each $x$ .

$\exists y \in U\, \forall x \in U\,(P(x, y))$ asserts: “there exists a single real $y$ such that $x + y = 0$ for every real $x$ .” This is $\bot$ : no single number is the additive inverse of every real number.

The difference shows up in how one would search for the answer. In the first statement, we walk through the values of $x$ one at a time and look for a fresh $y$ at each step. In the second, we pick a candidate $y$ first and then test it against every $x$ . The quantifier order dictates which variable is chosen first, and therefore which one the other is allowed to depend on.

Example 2.20 (Order Changes the Meaning).

Let $U$ be the real numbers and let $p(m, n)$ denote ” $m > n$ .” Then $\forall n \in U\, \exists m \in U\,(p(m, n))$ says “for every number there is a larger one,” which is $\top$ ; the witness $m = n + 1$ works for each $n$ . Flipping the two quantifiers, $\exists m \in U\, \forall n \in U\,(p(m, n))$ says “there is a number larger than every number,” which is $\bot$ ; among other embarrassments, such an $m$ would have to satisfy $m > m$ (this will be important in a second).

Example 2.21 (Captains).

Let $X$ be the professional footballers, let $Y$ be the clubs, and let $p(x, y)$ denote ” $x$ is the captain of $y$ .” Four formally similar statements have four very different meanings:

$\forall x \in X\, \exists y \in Y\,(p(x, y))$ : every footballer is some club’s captain. $\bot$ .
$\exists y \in Y\, \forall x \in X\,(p(x, y))$ : there is one club whose captain is every footballer at once. $\bot$ , and its dressing room does not bear thinking about.
$\forall y \in Y\, \exists x \in X\,(p(x, y))$ : every club has a captain. $\top$ .
$\exists x \in X\, \forall y \in Y\,(p(x, y))$ : there is one footballer who captains every club. $\bot$ .

Only the quantifier order and the choice of which set each variable is drawn from separate these.

Problem 2.8.

Negate statement (3) of the example above, “every club has a captain,” pushing the negation all the way inside. Then say the result in football English.

However, quantifiers of the same type may be freely reordered:

\forall x \in U\, \forall y \in U\,(P(x, y)) \equiv \forall y \in U\, \forall x \in U\,(P(x, y))

\exists x \in U\, \exists y \in U\,(P(x, y)) \equiv \exists y \in U\, \exists x \in U\,(P(x, y))

Both nested universal quantifiers demand that $P$ hold for every pair, and both nested existential quantifiers demand that at least one pair satisfies $P$ . In neither case does the order in which pairs are examined affect the outcome.

Quantification Fixes a Variable

Let $X$ be the even integers greater than or equal to $4$ , let $P$ be the prime numbers, and for $n \in X$ and $a, b \in P$ let $Q(n, a, b)$ denote ” $a + b = n$ .” Consider

\forall n \in X\, \exists a, b \in P\,(Q(n, a, b))

The initial quantification $\forall n \in X$ serves to fix a value of $n$ , which is then used for the rest of the statement. The assertion that follows, that there are primes $a$ and $b$ with $a + b = n$ , is made about that fixed but arbitrary value. Read as a whole, the statement says that no matter which $n$ is chosen from $X$ , we can find $a$ and $b$ satisfying $Q$ . The order of quantification is what tells us those values may depend on the chosen $n$ .

Take a particular value. We know $8 \in X$ , since $8$ is even and $8 \geqslant 4$ . Can we find $a, b \in P$ with $a + b = 8$ ? We can take $a = 3$ and $b = 5$ . Now take $n = 14$ ; our choices have to be different from before, and $a = 3$ , $b = 11$ will do (as will $a = b = 7$ ). This is what we mean when we say that $a$ and $b$ depend on $n$ .

Problem 2.9.

What is the difference between the statement above and the following one?

\exists n \in X\, \exists a, b \in P\,(Q(n, a, b))

Is it $\top$ or $\bot$ ? Why?

Remark.

The universal statement $\forall n \in X\, \exists a, b \in P\,(Q(n, a, b))$ is the Goldbach Conjecture, met as a proposition of unknown truth value in the previous chapter. Checking $n = 8$ and $n = 14$ settles two instances; it does not settle the statement, because a universal quantifier over an infinite set is not exhausted by any number of instances; but a single $n$ for which no such $a$ and $b$ exist would settle it the other way.

Example 2.22 (Translating from Mathematics).

The statement “the sum of two positive integers is always positive” contains implicit quantifiers and a hidden domain. Making these explicit step by step:

Rewrite with explicit quantifiers: “for every two integers, if both are positive, then their sum is positive.”
Introduce variables: “for all integers $x$ and $y$ , if $x > 0$ and $y > 0$ , then $x + y > 0$ .”
Formalise, taking $U$ to be the integers:

\forall x, y \in U\,(x > 0 \land y > 0 \to x + y > 0)

Example 2.23 (Carrying Restrictions Along).

The Archimedean principle says that if $b$ and $s$ are positive numbers, there is a natural number $n$ with $ns > b$ . This holds however large $b$ is and however small $s$ is.

Let $U$ be the real numbers, let $N$ be the natural numbers, and let $p(b, s, n)$ denote ” $ns > b$ .” If we agree in advance that $b$ and $s$ range over the positive reals and $n$ over the natural numbers, the principle encodes compactly. If instead all three variables range over $U$ , we have to carry the restrictions ourselves:

\forall b, s \in U\,\Bigl((b > 0 \land s > 0) \to \exists n\,\bigl(n \in N \land p(b, s, n)\bigr)\Bigr)

read “for every $b$ and $s$ , if both are positive, then there is a natural number $n$ satisfying $ns > b$ .” Writing $L$ for the positive real numbers, the bounded quantifiers of Definition 2.12 package the same statement as

\forall b, s \in L\, \exists n \in N\,(p(b, s, n))

The two say the same thing, and the second is shorter.

Example 2.24 (Translating Natural Language).

Let $U$ be all people and let $L(x, y)$ denote ” $x$ loves $y$ .”

\begin{array}{ll} \text{Everybody loves somebody.} & \forall x \in U\, \exists y \in U\,(L(x, y)) \\[4pt] \text{There is someone who is loved by everyone.} & \exists x \in U\, \forall y \in U\,(L(y, x)) \\[4pt] \text{There is someone who loves someone.} & \exists x, y \in U\,(L(x, y)) \\[4pt] \text{Everyone loves themselves.} & \forall x \in U\,(L(x, x)) \end{array}

Observe how the English phrasing obscures the quantifier order; “Everybody loves somebody” places the universal quantifier first: each person has their own someone. “There is someone who is loved by everyone” places the existential first: a single person is loved by all. Despite their superficial similarity, the two statements are logically independent.

Problem 2.10.

Let $U$ be all students in a class and let $F(x, y)$ denote ” $x$ and $y$ are friends.” Formalise the following in predicate logic, stating the set each variable is drawn from:

(a) Everyone has a friend.

(b) There is someone who is friends with everyone.

(c) No one is friends with everyone.

(d) There exists a pair of students who are not friends with each other.

Example 2.25 (Truth Values of Nested Statements).

Let $U$ be the real numbers. Determine the truth value of each statement:

(a) $\forall x \in U\, \exists y \in U\,(x^2 = y)$ is $\top$ : for any $x$ , choose $y = x^2$ .

(b) $\forall x \in U\, \exists y \in U\,(x = y^2)$ is $\bot$ : taking $x = -1$ , there is no real $y$ with $y^2 = -1$ .

(c) $\exists x \in U\, \forall y \in U\,(xy = 0)$ is $\top$ : choose $x = 0$ .

(d) $\exists x, y \in U\,(x + y \neq y + x)$ is $\bot$ : addition of real numbers is commutative, so $x + y = y + x$ for all $x, y \in U$ . This is the negation of $\forall x, y \in U\,(x + y = y + x)$ , which is $\top$ .

Problem 2.11.

Let $U$ be the integers. Determine the truth value of each statement, and justify your answer.

(a) $\forall x \in U\, \exists y \in U\,(x \cdot y = 1)$

(b) $\exists x \in U\, \forall y \in U\,(x + y = y)$

(c) $\forall x, y \in U\, \exists z \in U\,(x + z = y)$

Negating Nested Quantifiers

Theorem 2.16 extends to nested quantifiers by repeated application. Each quantifier flips ( $\forall \leftrightarrow \exists$ ) and the negation pushes inward:

\neg\bigl(\forall x\, \exists y\,(P(x, y))\bigr) \equiv \exists x\, \neg\bigl(\exists y\,(P(x, y))\bigr) \equiv \exists x\, \forall y\,(\neg P(x, y))

At each step, one quantifier is negated. The process terminates when the negation reaches the predicate.

Example 2.26.

Let $U$ be the real numbers and negate the statement $\forall x \in U\, \exists y \in U\,(x + y = 0)$ , “every real number has an additive inverse.”

Applying the negation rules from outside in:

\neg\bigl(\forall x \in U\, \exists y \in U\,(x + y = 0)\bigr) \equiv \exists x \in U\, \forall y \in U\,(x + y \neq 0)

In English: “there is a real number with no additive inverse.” This is $\bot$ , confirming that the original statement is $\top$ .

Problem 2.12.

Let $U$ be the real numbers. Negate each of the following statements and simplify. State whether the original or its negation is true.

(a) $\forall x, y \in U\,(x + y = y + x)$

(b) $\exists x \in U\, \forall y \in U\,(xy = 0)$

(c) $\forall x \in U\, \exists y, z \in U\,(y^2 + z^2 = x)$

Remark (Prenex Normal Form).

A formula is in Prenex Normal Form (PNF) if all quantifiers appear at the front, followed by a quantifier-free predicate:

Q_1 x_1\, Q_2 x_2\, \cdots\, Q_k x_k\,\bigl(\psi(x_1, x_2, \ldots, x_k)\bigr)

where each $Q_i$ is either $\forall$ or $\exists$ , each variable ranges over the universe fixed in advance, and $\psi$ contains no quantifiers. For example, the statement $\exists x\,(P(x)) \to \exists x\,(Q(x))$ is not in PNF because quantifiers appear on both sides of $\to$ . Rewriting using the defining equivalence of the conditional and renaming variables for clarity:

\neg\bigl(\exists x\,(P(x))\bigr) \lor \exists y\,(Q(y)) \equiv \forall x\,(\neg P(x)) \lor \exists y\,(Q(y)) \equiv \forall x\, \exists y\,(\neg P(x) \lor Q(y))

The last expression is in PNF. Every statement in predicate logic can be converted to PNF using quantifier negation, variable renaming, and the distribution rules established earlier.

Problem 2.13.

Let $P$ and $Q$ be predicates over a universe $U$ . Convert the following to Prenex Normal Form:

(a) $\forall x\,(P(x)) \to \exists x\,(Q(x))$

(b) $\neg\bigl(\forall x\, \exists y\,(P(x, y))\bigr) \lor \forall z\,(Q(z))$

Proofs

Valid Arguments and Inference Rules

We can use all of this to do what a mathematician (or an AI) was meant to do in the first place: write proofs. In the first chapter of the notes we built a proof system for propositional equivalences, but equivalence proofs are symmetric (carrying identical truth values under every assignment), but much of mathematics proceeds asymmetrically (we know certain facts and wish to derive new ones, in one direction only). If $p \to q$ is known and $p$ is true, we may conclude $q$ , but not the reverse. The tools for this one-way reasoning are called inference rules, and a proof is a structured chain of such inferences leading from premises to a conclusion.

Arguments and Validity

Definition 2.27 (Argument).

An argument in propositional logic is a finite sequence of propositions $p_1, p_2, \ldots, p_n, q$ . The propositions $p_1, \ldots, p_n$ are the premises and $q$ is the conclusion. The argument is valid if the truth of all premises guarantees the truth of the conclusion: $(p_1 \land p_2 \land \cdots \land p_n) \to q$ is a tautology. An argument that is not valid is invalid.

We write a valid argument with the premises above a horizontal line and the conclusion below:

\frac{p_1 \quad p_2 \quad \cdots \quad p_n}{q}

An inference rule is an argument form that remains valid regardless of which particular propositions are substituted for its variables: just as the axioms of earlier notes gave us equivalences that hold universally, inference rules give us one-directional deductions that hold universally.

Remark (The Turnstile).

The notation $p_1, p_2, \ldots, p_n \vdash q$ (read ” $p_1, \ldots, p_n$ entail $q$ ”) means that $q$ can be derived from the premises $p_1, \ldots, p_n$ using axioms, inference rules, and previously established theorems. When no premises are needed, we write $\vdash q$ , meaning $q$ is derivable from the axioms alone.

We start off with the most famous.

Modus Ponens

The basic rule of deduction is Modus Ponens (from the Latin modus ponendo ponens, “the method of affirming by affirming”):

\frac{p \to q \quad p}{q}

If we know $p \to q$ and $p$ , we may conclude $q$ . Its validity rests on the tautology

((p \to q) \land p) \to q

To see why this is a tautology, suppose both $p \to q$ and $p$ are true: by the truth table of the conditional, the only circumstance under which $p \to q$ is true and $p$ is true is when $q$ is also true. Hence $q$ must hold.

Example 2.28.

Earlier established the universal statement “every prime greater than $2$ is odd.” Let $p$ denote “7 is prime and greater than 2” and $q$ denote “7 is odd.” The universal claim gives $p \to q$ , and we can verify $p$ directly (7 is prime, and $7 > 2$ ); modus Ponens yields $q$ : the integer 7 is odd.

Problem 2.14.

Let $p$ denote “it is snowing” and $q$ denote “the lecture is cancelled.” Suppose we know $p \to q$ and $p$ . State the conclusion and identify the inference rule. Now suppose instead we know $p \to q$ and $q$ . Can we conclude $p$ ? Justify your answer using the truth table of the conditional.

Modus Tollens

Theorem 2.29 (Modus Tollens).

For any propositions $p$ and $q$ ,

\frac{p \to q \quad \neg q}{\neg p}

Discussion.

The claim is that a rule of inference is valid: granted the two premises, the conclusion follows. We already have one such rule, Modus Ponens, so rather than argue from truth values we rewrite the premises until that rule applies. The premise $p \to q$ is equivalent to its contrapositive $\neg q \to \neg p$ , and the second premise $\neg q$ is precisely the antecedent of that conditional; Modus Ponens then delivers $\neg p$ . So the proof is the contrapositive equivalence followed by one application of Modus Ponens.

Proof.

Assume $p \to q$ and $\neg q$ . By the Contrapositive equivalence, $p \to q \equiv \neg q \to \neg p$ . Since $\neg q$ is true, Modus Ponens applied to $\neg q \to \neg p$ and $\neg q$ yields $\neg p$ .

The Latin name modus tollendo tollens (“the method of denying by denying”) captures the reasoning: if the consequence of an implication fails, its premise must also fail.

Example 2.30.

If $n$ is even, then $n^2$ is even. We observe that $49$ is not even and by Modus Tollens, 7 is not even.

Hypothetical Syllogism

Theorem 2.31 (Hypothetical Syllogism).

For any propositions $p$ , $q$ , $r$ ,

\frac{p \to q \quad q \to r}{p \to r}

Discussion.

Again a rule of inference, but the conclusion is itself a conditional, and a conditional is established by assuming its antecedent and deriving its consequent. So there are three assumptions in play: the two premises, and $p$ taken on for the sake of the argument. From $p$ and the first premise Modus Ponens gives $q$ ; from $q$ and the second premise it gives $r$ . Having derived $r$ from $p$ , we discharge the assumption and conclude $p \to r$ . Modus Ponens is used twice, and nothing else.

Proof.

Assume $p \to q$ and $q \to r$ . We wish to show $p \to r$ . Assume $p$ . From $p$ and $p \to q$ , Modus Ponens gives $q$ . From $q$ and $q \to r$ , a second application of Modus Ponens gives $r$ . Since assuming $p$ led to $r$ , we conclude $p \to r$ .

This inference is also called the chain rule or the transitivity of implication. A multi-step deduction is a chain of hypothetical syllogisms. For instance, suppose we know “if $n$ is divisible by $4$ , then $n$ is even” and “if $n$ is even, then $n^2$ is even.” Hypothetical Syllogism yields: “if $n$ is divisible by $4$ , then $n^2$ is even.”

Implication Elimination

Theorem 2.32 (Implication Elimination).

For any propositions $p$ and $q$ , $(p \to q) \vdash (p \vdash q)$ .

Discussion.

The statement relates the two arrows: the connective $\to$ , which lives inside a proposition, and the turnstile $\vdash$ , which lives between propositions and asserts derivability. It says that from $p \to q$ we may derive $q$ from $p$ . So we assume $p \to q$ and must show $p \vdash q$ , that is, derive $q$ from the assumption $p$ ; Modus Ponens applied to the assumption and the hypothesis does this. The converse direction is the discharge step used in the previous theorem.

Proof.

Assume $p \to q$ . We must show $p \vdash q$ . Assume $p$ . From $p \to q$ and $p$ , Modus Ponens yields $q$ . Thus $p \vdash q$ .

This gives one direction of the link between a conditional and a derivation: if $p \to q$ is known, then from the assumption $p$ one may derive $q$ . The reverse direction will appear later in Direct Proof, where we show that if assuming $p$ leads to $q$ , then the conditional $p \to q$ is established. Together, the two directions show how the turnstile and the conditional correspond: $p \vdash q$ if and only if $\vdash p \to q$ .

Further Propositional Inference Rules

Several additional inference rules arise from tautologies:

Conjunction. From two established truths, their conjunction follows.

\frac{p \quad q}{p \land q}

The tautological justification is immediate, but Conjunction can also be derived from Modus Ponens alone via reductio.

Proof.

Assume $p$ and $q$ . Suppose for contradiction that $\neg(p \land q)$ . By De Morgan’s Law (Theorem 6, 1), $\neg(p \land q) \equiv \neg p \lor \neg q$ . By the Conditional axiom, $\neg p \lor \neg q \equiv p \to \neg q$ . Since $p$ holds, Modus Ponens yields $\neg q$ . But $q$ is assumed, a contradiction. Therefore $p \land q$ .

Simplification. From a conjunction, either conjunct may be extracted.

\frac{p \land q}{p}

Addition. A known truth may be weakened to a disjunction.

\frac{p}{p \lor q}

Disjunctive Syllogism. If one disjunct is eliminated, the other must hold.

\frac{p \lor q \quad \neg p}{q}

Resolution. Two clauses sharing a complementary literal may be combined, eliminating the shared variable.

\frac{\neg p \lor r \quad p \lor q}{q \lor r}

Constructive Dilemma. Two conditionals with a disjunction of their antecedents yield the disjunction of their consequents.

\frac{(\alpha \to \gamma) \quad (\beta \to \delta) \quad (\alpha \lor \beta)}{\gamma \lor \delta}

Problem 2.15.

Prove that all of these are tautologies.

Notice the relationships among these rules. Disjunctive Syllogism is a special case of Resolution (set $r = \bot$ and apply the Identity axiom). Addition is dual to Conjunction, and Simplification is a special case of Modus Ponens (since $(p \land q) \to p$ is a tautology).

Remark (Resolution Subsumes Other Rules).

Resolution can recover other rules. It gives Hypothetical Syllogism: rewrite $p \to q$ as $\neg p \lor q$ and $q \to r$ as $\neg q \lor r$ (by the Conditional axiom), then a single resolution step on $q$ yields $\neg p \lor r$ , which is $p \to r$ . It can also recover Modus Ponens: express $p$ as $p \lor \bot$ and $p \to q$ as $\neg p \lor q$ , and resolution gives $q \lor \bot \equiv q$ by the Identity axiom. For this reason Resolution is the basis of automated theorem proving in computer science.

The following table collects the propositional inference rules for reference.

\small \begin{array}{lll} \textbf{Name} & \textbf{Rule} & \textbf{Tautology} \\ \hline \text{Modus Ponens} & p,\; p \to q \;\vdash\; q & ((p \to q) \land p) \to q \\ \text{Modus Tollens} & \neg q,\; p \to q \;\vdash\; \neg p & ((p \to q) \land \neg q) \to \neg p \\ \text{Hyp.\ Syllogism} & p \to q,\; q \to r \;\vdash\; p \to r & ((p \to q) \land (q \to r)) \to (p \to r) \\ \text{Conjunction} & p,\; q \;\vdash\; p \land q & (p \land q) \to (p \land q) \\ \text{Simplification} & p \land q \;\vdash\; p & (p \land q) \to p \\ \text{Addition} & p \;\vdash\; p \lor q & p \to (p \lor q) \\ \text{Disj.\ Syllogism} & p \lor q,\; \neg p \;\vdash\; q & ((p \lor q) \land \neg p) \to q \\ \text{Resolution} & \neg p \lor r,\; p \lor q \;\vdash\; q \lor r & ((\neg p \lor r) \land (p \lor q)) \to (q \lor r) \\ \text{Constr.\ Dilemma} & (\alpha \to \gamma),\; (\beta \to \delta),\; \alpha \lor \beta \;\vdash\; \gamma \lor \delta & ((\alpha \to \gamma) \land (\beta \to \delta) \land (\alpha \lor \beta)) \to (\gamma \lor \delta) \\ \text{Disj.\ Elimination} & (p \to r),\; (q \to r),\; p \lor q \;\vdash\; r & ((p \to r) \land (q \to r) \land (p \lor q)) \to r \\ \text{Ex Falso} & p,\; \neg p \;\vdash\; q & (p \land \neg p) \to q \end{array}

Problem 2.16.

Identify the inference rule applied in each step of the following argument. Premises: (i) $p \to q$ , (ii) $\neg q \lor r$ , (iii) $\neg r$ .

From (ii) and (iii), conclude $\neg q$ .
From (i) and step 1, conclude $\neg p$ .

Building Complex Arguments

Inference rules are used in chains. A valid argument is a sequence of steps in which each step is either a premise or follows from earlier steps by a single inference rule.

Example 2.33.

We show the following argument is valid. Premises: (1) $p \land (p \to q)$ , (2) $q \to r$ . Conclusion: $r$ .

\begin{array}{rll} 1. & p \land (p \to q) & \text{Premise} \\ 2. & q \to r & \text{Premise} \\ 3. & p & \text{Simplification from 1} \\ 4. & p \to q & \text{Simplification from 1} \\ 5. & q & \text{Modus Ponens from 3, 4} \\ 6. & r & \text{Modus Ponens from 5, 2} \end{array}

Problem 2.17.

Show that the following argument is valid by listing each step and the inference rule used. Premises: (1) $p \to q$ , (2) $q \to (r \land s)$ , (3) $\neg r \lor u$ , (4) $p$ . Conclusion: $u$ .

Inference Rules for Quantified Statements

The rules above govern propositional logic. When predicates and quantifiers enter the picture, additional rules are needed to bridge the gap between universal or existential claims and their specific instances.

Universal Instantiation (UI). If a predicate holds for every element of the universe, it holds for any particular element $c$ :

\frac{\forall x\, P(x)}{P(c)} \quad \text{for any } c \text{ in the universe}

Universal Generalisation (UG). If $P(c)$ can be established for an arbitrary element $c$ (one about which no special assumptions are made), then $\forall x\, P(x)$ follows:

\frac{P(c) \text{ for arbitrary } c}{\forall x\, P(x)}

The word “arbitrary” is critical. If the proof of $P(c)$ exploits any property peculiar to $c$ , the generalisation is invalid.

Existential Instantiation (EI). If at least one element satisfies $P$ , we may introduce a name $c$ for such an element:

\frac{\exists x\, P(x)}{P(c) \text{ for some } c}

The name $c$ must be fresh: not previously used in the argument.

Existential Generalisation (EG). If a particular element $c$ satisfies $P$ , then at least one element does:

\frac{P(c)}{\exists x\, P(x)}

Universal Modus Ponens (UMP). Combining Universal Instantiation with Modus Ponens gives the quantifier rule used most often:

\frac{\forall x\,(P(x) \to Q(x)) \quad P(c)}{Q(c)}

Example 2.34.

Write $R$ for the real numbers, and suppose we know $\forall x\,(x > 0 \to \sqrt{x} \in R)$ , the universe being $R$ , and that $9 > 0$ . Universal Modus Ponens yields $\sqrt{9} \in R$ .

Problem 2.18.

Identify the quantifier inference rule used in each step. The universe is the positive integers.

“Every multiple of $6$ is a multiple of $3$ .” (Premise)
” $18 = 6 \cdot 3$ , so $18$ is a multiple of $6$ .” (Premise)
“Therefore, $18$ is a multiple of $3$ .”
“Therefore, there exists a positive integer that is a multiple of $3$ .”

Example 2.35 (Combining Quantifier Rules).

We show that “a student in this class has not read the book” and “every student in this class passed the first exam” together imply “someone who passed the first exam has not read the book.”

Let $C(x)$ denote ” $x$ is in this class,” $B(x)$ denote ” $x$ has read the book,” and $E(x)$ denote ” $x$ passed the exam.” The premises are $\exists x\,(C(x) \land \neg B(x))$ and $\forall x\,(C(x) \to E(x))$ . We derive $\exists x\,(E(x) \land \neg B(x))$ .

\begin{array}{rll} 1. & \exists x\,(C(x) \land \neg B(x)) & \text{Premise} \\ 2. & \forall x\,(C(x) \to E(x)) & \text{Premise} \\ 3. & C(a) \land \neg B(a) \text{ for some } a & \text{EI from 1} \\ 4. & C(a) \to E(a) & \text{UI from 2} \\ 5. & C(a) & \text{Simplification from 3} \\ 6. & E(a) & \text{Modus Ponens from 4, 5} \\ 7. & \neg B(a) & \text{Simplification from 3} \\ 8. & E(a) \land \neg B(a) & \text{Conjunction from 6, 7} \\ 9. & \exists x\,(E(x) \land \neg B(x)) & \text{EG from 8} \end{array}

Fallacies

Definition 2.36 (Fallacy).

A fallacy is an argument form that appears valid but is not: the premises do not logically guarantee the conclusion, even though the reasoning may seem persuasive.

Two fallacies are especially common. Both arise from misapplying the conditional.

Affirming the Consequent. The argument ” $p \to q$ ; $q$ ; therefore $p$ ” is invalid. A counterexample: let $p = \bot$ and $q = \top$ . Then $p \to q$ is $\top$ and $q$ is $\top$ , but $p$ is $\bot$ .

Denying the Antecedent. The argument ” $p \to q$ ; $\neg p$ ; therefore $\neg q$ ” is equally invalid. With $p = \bot$ and $q = \top$ , both premises are satisfied but $\neg q$ is $\bot$ .

Both fallacies confuse an implication with its converse or its inverse: $p \to q$ does not entail $q \to p$ , nor does it entail $\neg p \to \neg q$ .

Example 2.37.

“If it rained, the pitch is wet. The pitch is wet. Therefore it rained.” This affirms the consequent. The pitch might be wet because the sprinklers were on. The converse “if the pitch is wet, then it rained” does not follow from the original implication.

Problem 2.19.

Identify the fallacy in each argument and provide a counterexample (an assignment under which the premises are true but the conclusion is false).

“If $x > 2$ , then $x > 1$ . The number $\tfrac{3}{2}$ satisfies $x > 1$ . Therefore $\tfrac{3}{2} > 2$ .”
“If $x > 0$ , then $x^2 > 0$ . The number $-1$ does not satisfy $x > 0$ . Therefore $(-1)^2 \leqslant 0$ .”

Proof Techniques

With inference rules in hand, we turn to the practical matter of constructing proofs. A proof is a finite sequence of statements, each of which is an axiom, a premise, or a consequence of earlier statements by an inference rule. What dictates the strategy is the shape of the claim: whether it asserts a conditional, a biconditional, a conjunction, a disjunction, or a quantification. Each shape comes with a direct method that builds the claim up from its parts, and an indirect method that assumes the negation and derives a contradiction.

Since these notes are about writing proofs, not merely admiring them, we will not simply state a result and prove it. Each proof comes with a Discussion, where we plan the argument and air the thought process, followed by the Proof itself, written in full English sentences with proper syntax and grammar. The two are kept apart on purpose. A discussion names the form of the statement, says which pieces have to be proved and by what method, and points to the definition or rule each piece will use; the proof then carries that plan out.

We have not built the numbers yet (although we have been using them), and doing so is where this course is headed. But until then we help ourselves to the arithmetic and order rules of school algebra, and we name our objects in words: a whole number is one of $\ldots, -2, -1, 0, 1, 2, \ldots$ , and a real number is any point of the number line. Every rule borrowed along the way is proved from the ground up once we have the objects to prove it about, and where a proof below leans on a borrowed rule we say which one (we will collect on any debt as we continue).

Two properties of whole numbers come up often enough in the examples to be worth fixing now.

Definition 2.38 (Even and Odd).

A whole number $n$ is even if it can be written as $n = 2k$ for some whole number $k$ , and odd if it can be written as $n = 2k + 1$ for some whole number $k$ . Every whole number is one or the other and never both.

Definition 2.39 (Prime and Composite).

A whole number greater than $1$ is composite if it is a product of two whole numbers each at least $2$ , and prime otherwise.

Proving Conditional Claims

Recall that $p \to q$ asserts that whenever $p$ holds, $q$ necessarily holds. The conditional is already true in every case where $p$ fails, so no proof need consider those cases: it is enough to suppose $p$ and deduce $q$ .

The strategies are collected below. The justification column speaks about our proofs rather than inside them, and it uses $\implies$ for that outer step, keeping it apart from the connective $\to$ that lives in the statements themselves; $\impliedby$ is the same arrow read right to left.

\small \begin{array}{lll} \textbf{Strategy} & \textbf{Approach} & \textbf{Justification} \\ \hline \text{Trivial proof} & \text{Show } q \text{ is true} & q \text{ true} \implies p \to q \text{ true} \\ \text{Vacuous proof} & \text{Show } p \text{ is false} & p \text{ false} \implies p \to q \text{ true} \\ \text{Direct proof} & \text{Assume } p,\text{ derive } q & p \vdash q \;\implies\; \vdash p \to q \\ \text{Contrapositive} & \text{Assume } \neg q,\text{ derive } \neg p & p \to q \equiv \neg q \to \neg p \\ \text{Contradiction} & \text{Assume } p \land \neg q,\text{ derive } r \land \neg r & (p \land \neg q) \to \bot \;\implies\; p \to q \end{array}

The last two are called indirect proofs, since they establish the conditional without constructing a chain from $p$ to $q$ .

The first two are degenerate, and they are worth naming only so that they are recognised when they occur. A trivial proof ignores the hypothesis altogether: “if $x > 5$ , then $x \cdot 0 = 0$ ” holds because the conclusion holds for every $x$ . A vacuous proof shows instead that the hypothesis is never met: “for every real number $x$ , if $x < x$ , then $x = 7$ ” holds because no $x$ satisfies $x < x$ .

Definition 2.40 (Direct Proof).

A direct proof of a conditional statement $p \to q$ proceeds by assuming $p$ and deriving $q$ through a sequence of valid inferences. The assumption of $p$ is not asserted as fact; it is a hypothesis under which the argument operates.

This strategy reflects the Deduction Rule, a meta-logical principle: if assuming $p$ allows us to derive $q$ using valid inferences, then the conditional $p \to q$ is established. The Deduction Rule connects the turnstile ( $\vdash$ ) and the conditional ( $\to$ ): the statement $p \vdash q$ (a syntactic derivation) becomes $p \to q$ (a logical truth).

Strategy (Direct proof of a conditional).

Claim: $p \to q$

Suppose $p$ holds.
Prove that $q$ holds.

Proposition 2.41.

If $x > 1$ , then $x^2 > 1$ .

Discussion.

We will assume $x > 1$ and show $x^2 > 1$ . Our tool is the well-known algebraic rule (which we prove later on) that if $a > b$ and $c > d$ with all four numbers positive, then $ac > bd$ .

Proof.

Assume $x > 1$ . Since $x > 1 > 0$ , both $x$ and $1$ are positive, so multiplying the inequality $x > 1$ by itself yields $x^2 > 1 \cdot 1 = 1$ . Thus $x^2 > 1$ .

Our statements will grow considerably more complicated than this, but the shape stays the same. With the proof done, we may now say with full mathematical confidence that whenever a number $x$ satisfies $x > 1$ , it automatically satisfies $x^2 > 1$ .

As we know the contrapositive $\neg q \to \neg p$ is logically equivalent to $p \to q$ , so proving either one proves the other. The swap is worth making when $p$ or $q$ carries a negation inside it, since negating it turns the assumption into a positive assertion we can compute with.

Strategy (Contrapositive proof of a conditional).

Claim: $p \to q$

Suppose $\neg q$ holds.
Prove that $\neg p$ holds.

This is the direct strategy applied to $\neg q \to \neg p$ .

Proposition 2.42.

If $x^3 < 0$ , then $x < 0$ .

Discussion.

Here $p$ is " $x^3 < 0$ " and $q$ is " $x < 0$ ". Proving this directly is awkward, since we would begin knowing a fact about $x^3$ and have to dig our way down to a fact about $x$ . The contrapositive reverses this. Since $\neg q$ is " $x \geqslant 0$ " and $\neg p$ is " $x^3 \geqslant 0$ ", the contrapositive $\neg q \to \neg p$ reads “if $x \geqslant 0$ , then $x^3 \geqslant 0$ ”, where we start from $x$ and build up to $x^3$ ; that direction is straightforward. So we will assume $x \geqslant 0$ and show $x^3 \geqslant 0$ , announcing at the outset that we are proving the contrapositive.

Proof.

We prove the contrapositive: if $x \geqslant 0$ , then $x^3 \geqslant 0$ . Assume $x \geqslant 0$ . Since all terms are non-negative, multiplying $x \geqslant 0$ by itself thrice gives $x \cdot x \cdot x \geqslant 0 \cdot 0 \cdot 0$ , that is, $x^3 \geqslant 0$ . The contrapositive holds, so the original statement “if $x^3 < 0$ , then $x < 0$ ” holds as well.

Indirect proofs

Every indirect method below uses proof by contradiction.

Definition 2.43 (Proof by Contradiction).

A proof by contradiction (Latin: reductio ad absurdum) establishes a proposition $p$ by assuming $\neg p$ and deriving a contradiction, that is, a statement of the form $q \land \neg q$ for some proposition $q$ . Since a sound proof system cannot derive a false conclusion from true premises, the assumption $\neg p$ must be false, so $p$ is true.

The opening move of such a proof is written so often that i abbreviate it to AFSOC standing for “assume for the sake of contradiction.” What we assume is the logical negation of the claim, so each indirect strategy below is really the negation rules of earlier read as instructions.

So in the case of the direct proof, the indirect method rests on the negation of a conditional, $\neg(p \to q) \equiv p \land \neg q$ . Assuming the negation therefore means granting ourselves the hypothesis and the failure of the conclusion at once, which is one more fact to work with than a direct proof has.

Strategy (Indirect proof of a conditional).

Claim: $p \to q$

AFSOC that $p$ holds and $q$ fails.
Find a contradiction.

Proposition 2.44.

For all real numbers $x$ and $y$ , if $x + y > 2$ , then $x > 1$ or $y > 1$ .

Discussion.

Our statement is the conditional $p \to q$ , where $p$ is " $x + y > 2$ " and $q$ is the disjunction ” $x > 1$ or $y > 1$ ”. A direct proof would have to settle which half of $q$ to aim at, and the hypothesis does not decide that for us, so we go indirectly. The negation of the whole statement is $p \land \neg q$ , and De Morgan turns $\neg q$ into ” $x \leqslant 1$ and $y \leqslant 1$ ”. That leaves us with three inequalities: we add the two coming from $\neg q$ and set the result against $p$ .

Proof.

AFSOC that $x + y > 2$ and that the conclusion fails, so that $x \leqslant 1$ and $y \leqslant 1$ . Adding these two inequalities gives $x + y \leqslant 2$ . But we assumed $x + y > 2$ , and no number is both greater than $2$ and at most $2$ . Therefore, whenever $x + y > 2$ , at least one of $x > 1$ , $y > 1$ holds.

Remark (Contradiction Against Contrapositive).

The two indirect routes are easy to confuse. A proof by contradiction assumes the negation of the entire statement, $p \land \neg q$ , and derives an absurdity from it. A contrapositive proof assumes only $\neg q$ and derives $\neg p$ , which is a direct proof of an equivalent statement and never mentions absurdity at all. In practice both are often labelled “proof by contradiction,” but the contrapositive route is usually the cleaner of the two, and it is the one to reach for first.

Problem 2.20.

Prove that if $0 < x < 1$ , then $x^2 < x$ .

Problem 2.21.

Prove that for all whole numbers $n$ , if $3n + 2$ is odd, then $n$ is odd.

Problem 2.22.

Prove that for all real numbers $x$ , if $x^5 < 0$ , then $x < 0$ .

Proving Biconditional Claims

The connective $\leftrightarrow$ is defined entirely in terms of $\to$ : asserting $p \leftrightarrow q$ is asserting the conjunction $(p \to q) \land (q \to p)$ . That gives the obvious strategy, prove one conditional and then the other, and it exposes the most common mistake we see, which is proving one of them and stopping. The two directions are labelled ( $\implies$ ) and ( $\impliedby$ ), and each may be proved by whichever of the conditional methods suits it; there is no requirement that both use the same one.

Strategy (Direct proof of a biconditional).

Claim: $p \leftrightarrow q$

Prove that $p \to q$ , by any of the conditional methods.
Prove that $q \to p$ , by any of the conditional methods.

Consider ” $n$ is even if and only if $n^2$ is even”. Before the proof, recall the two forms an even and an odd whole number take, $n = 2k$ and $n = 2k + 1$ .

Proposition 2.45.

$n$ is even if and only if $n^2$ is even.

Discussion.

The two conditionals are $p \to q$ , “if $n$ is even, then $n^2$ is even”, and $q \to p$ , “if $n^2$ is even, then $n$ is even”. The first is straightforward: write $n = 2k$ and show that $n^2 = 2(\text{whole number})$ . The second is trickier, since its hypothesis tells us about $n^2$ when we want a conclusion about $n$ ; so we take its contrapositive $\neg p \to \neg q$ , “if $n$ is odd, then $n^2$ is odd”, assume $n = 2k + 1$ , and show that $n^2 = 2(\text{whole number}) + 1$ .

Proof.

( $\implies$ ) First, suppose $n$ is even, so $n = 2k$ for some whole number $k$ . Then $n^2 = (2k)^2 = 4k^2 = 2(2k^2)$ , and since $2k^2$ is a whole number, $n^2$ is even.

( $\impliedby$ ) The remaining conditional is “if $n^2$ is even, then $n$ is even”, and we prove its contrapositive: if $n$ is odd, then $n^2$ is odd. Suppose $n = 2k + 1$ for some whole number $k$ . Squaring, $n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1,$ and since $2k^2 + 2k$ is a whole number, $n^2$ is odd. The contrapositive holds, so the second conditional does too. Both conditionals are proven, and with them the biconditional.

Strategy (Indirect proof of a biconditional).

Claim: $p \leftrightarrow q$

AFSOC that $\neg(p \to q) \lor \neg(q \to p)$ .
In the first case, $p \land \neg q$ holds. Find a contradiction.
In the second case, $q \land \neg p$ holds. Find a contradiction.

Whether this route is worth taking depends entirely on what $p$ and $q$ are, and the direct one usually wins; but it helps when both directions are hard to prove directly: the negations $p \land \neg q$ and $q \land \neg p$ each give two facts to work with, where the direct proof has one.

Proposition 2.46.

Let $x$ be a real number. Then $x = 0$ if and only if no real number $y$ satisfies $xy = 1$ .

Discussion.

Our statement is the biconditional $p \leftrightarrow q$ , where $p$ is " $x = 0$ " and $q$ is “no real number $y$ satisfies $xy = 1$ ”. Both directions are hard to prove directly for the same reason, which is that $q$ is itself a denial: proving it outright would mean ruling out every candidate $y$ one at a time, thus negating gets round both problems. In the first case we hold $p$ along with $\neg q$ , and $\neg q$ produces an actual $y$ with $xy = 1$ , into which we may substitute $x = 0$ . In the second we hold $q$ along with $\neg p$ , and $x \neq 0$ is exactly the condition under which we may divide by $x$ , which gives a $y$ that $q$ says does not exist.

Proof.

AFSOC that one of the two conditionals fails.

Suppose first that $p$ holds and $q$ fails, so that $x = 0$ and some real number $y$ satisfies $xy = 1$ . Then $1 = xy = 0 \cdot y = 0$ , and $1 \neq 0$ .

Suppose instead that $q$ holds and $p$ fails, so that no real number $y$ satisfies $xy = 1$ , and $x \neq 0$ . Since $x \neq 0$ we may divide by it, and $y = \frac{1}{x}$ is then a real number with $xy = 1$ , which $q$ denies.

Neither case can stand, so both conditionals hold and the biconditional with them.

A third route is neither direct nor indirect. Rather than linking $p$ and $q$ to each other, we find an intermediary statement $r$ and prove $p \leftrightarrow r$ and $r \leftrightarrow q$ ; the chain then forces $p$ and $q$ to share a truth value. The acronym TFAE, “the following are equivalent,” names the theorems built this way: a list of conditions, all asserted to imply one another, proved by running a cycle of implications through the list.

Strategy (Intermediary proof of a biconditional).

Claim: $p \leftrightarrow q$

Find a statement $r$ .
Prove that $p \leftrightarrow r$ , by any of the methods above.
Prove that $r \leftrightarrow q$ , by any of the methods above.

Proposition 2.47.

For a real number $x$ , the following are equivalent: $x^2 = x$ ; the product $x(x - 1)$ is zero; $x = 0$ or $x = 1$ .

Discussion.

Our statement puts three conditions on a real number $x$ and asserts that any one of them forces the other two, so we take $r$ to be " $x(x-1) = 0$ " and prove the two biconditionals $x^2 = x \leftrightarrow r$ and $r \leftrightarrow$ ” $x = 0$ or $x = 1$ ”, each of them directly. The first is a rearrangement, and it runs in both directions because subtracting $x$ from both sides is undone by adding it back. The second is the zero product rule, which we borrow here and prove below.

Proof.

Take $r$ to be the statement " $x(x - 1) = 0$ ".

( $x^2 = x \leftrightarrow r$ ) Subtracting $x$ from both sides of $x^2 = x$ gives $x^2 - x = 0$ , and $x^2 - x = x(x - 1)$ , so $r$ holds. Adding $x$ to both sides of $x(x-1) = 0$ reverses the step, so the two conditions imply one another.

( $r \leftrightarrow x = 0$ or $x = 1$ ) We borrow the zero product rule: a product of two real numbers is zero exactly when one of the factors is zero. (It is proved below, in the section on disjunctions.) So $x(x - 1) = 0$ holds exactly when $x = 0$ or $x - 1 = 0$ , and the latter says $x = 1$ . The chain of equivalences carries the truth value from the first condition to the third, so the three are either all true or all false.

Remark (Always Check the Converse).

A common error is to prove one direction of a biconditional and assume the other follows. Consider the equation $\sqrt{x - 3} + \sqrt{x + 4} = 7$ . Squaring both sides and simplifying produces $x = 12$ as the sole candidate. Squaring can introduce extraneous solutions, so we must substitute back: $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$ , and here the candidate is genuine. Contrast this with $x + \sqrt{x} = 0$ : squaring leads to candidates $x = 0$ and $x = 1$ , but $1 + \sqrt{1} = 2 \neq 0$ , so only $x = 0$ is a true solution.

Problem 2.23.

Prove that a whole number $n$ is even if and only if $n + 1$ is odd. Then prove it a second time by the indirect method.

Problem 2.24.

Prove that a real number $x$ satisfies $x^3 = x$ if and only if $x = 0$ , $x = 1$ , or $x = -1$ .

Proving Conjunctions

A conjunction asks for two proofs; The only thing worth saying about the direct method is that both halves must actually appear.

Strategy (Direct proof of a conjunction).

Claim: $p \land q$

Prove that $p$ holds.
Prove that $q$ holds.

Proposition 2.48.

If $x > 1$ , then $x^2 > x$ and $x^3 > x^2$ .

Discussion.

Let $p$ be " $x^2 > x$ " and $q$ be " $x^3 > x^2$ ", with the hypothesis $x > 1$ available to both. For $p$ we borrow a second rule of order, that if $a > b$ and $c > 0$ , then $ac > bc$ ; since $x > 1 > 0$ , multiplying $x > 1$ by $x$ gives $x^2 > x$ . For $q$ we multiply $x > 1$ by $x^2$ instead, which the same rule permits once we know $x^2$ is positive, and $p$ has just told us that $x^2 > x > 1$ .

Proof.

Assume $x > 1$ . Since $x > 1 > 0$ , the number $x$ is positive, so multiplying both sides of $x > 1$ by $x$ gives $x^2 > x$ , which is the first half.

For the second half, note that $x^2 > x > 1 > 0$ , so $x^2$ is positive as well. Multiplying both sides of $x > 1$ by $x^2$ gives $x^3 > x^2$ . Both halves hold, so their conjunction does.

By De Morgan’s law the negation of $p \land q$ is a disjunction, so the indirect method splits into two cases, each of which must be run to an absurdity.

Strategy (Indirect proof of a conjunction).

Claim: $p \land q$

AFSOC that $\neg p \lor \neg q$ holds.
In the first case, $\neg p$ holds. Find a contradiction.
In the second case, $\neg q$ holds. Find a contradiction.

Problem 2.25.

Prove the proposition above a second time by the indirect method, assuming that $x > 1$ and that $x^2 \leqslant x$ or $x^3 \leqslant x^2$ . Compare the two proofs and say which you would rather read.

Proving Disjunctions

A disjunction asserts that at least one of two statements holds. If we can tell in advance which one it is, there is nothing to discuss: prove that one and ignore the other. That is the direct method.

Strategy (Direct proof of a disjunction).

Claim: $p \lor q$

Prove that $p$ holds, or else prove that $q$ holds.

The interesting case is when neither half holds on its own, because $p$ and $q$ are predicates and which of them is true varies with the object. Rather than sort out exactly which objects satisfy which, we argue that if $p$ holds the claim is already proved and $q$ need not be looked at, which leaves only the cases where $p$ fails, and there we must guarantee $q$ . Since $p \lor q \equiv q \lor p$ , the roles may be swapped, and we assume whichever negation is more useful.

Strategy (Proof of a disjunction by otherwise).

Claim: $p \lor q$

Suppose $\neg p$ holds.
Prove that $q$ holds.

Proposition 2.49.

For every real number $x$ , either $x \leqslant 1$ or $x^2 > x$ .

Discussion.

Our statement is the disjunction $p \lor q$ , where $p$ is " $x \leqslant 1$ " and $q$ is " $x^2 > x$ ", and neither half holds for every $x$ : the number $\tfrac{1}{2}$ satisfies $p$ and fails $q$ , while $2$ does the reverse. Thus, we take the “otherwise” route: here $\neg p$ reads $x > 1$ , and multiplying that inequality by the positive number $x$ gives $q$ .

Proof.

Suppose $x \leqslant 1$ fails, so $x > 1$ . Then $x$ is positive, and multiplying both sides of $x > 1$ by $x$ gives $x^2 > x$ . Hence in every case at least one of the two holds.

The negation of a disjunction is a conjunction of negations, so the indirect method grants us both failures at once.

Strategy (Indirect proof of a disjunction).

Claim: $p \lor q$

AFSOC that $\neg p \land \neg q$ holds.
Find a contradiction.

Proposition 2.50.

For all real numbers $x$ and $y$ , if $xy = 0$ , then $x = 0$ or $y = 0$ .

Discussion.

This is the zero product rule we borrowed for the three equivalent conditions above: the conclusion is a disjunction, so we assume the hypothesis $xy = 0$ and negate the conclusion, which by gives us $x \neq 0$ and $y \neq 0$ together. A non-zero real number may be divided by, so we divide $xy = 0$ through by $x$ and read off a statement about $y$ that the second half of the negation denies.

Proof.

Assume $xy = 0$ , and AFSOC that both halves of the conclusion fail, so that $x \neq 0$ and $y \neq 0$ . Since $x \neq 0$ we may divide both sides of $xy = 0$ by $x$ , which gives $y = 0$ . But we assumed $y \neq 0$ , and no number is both zero and non-zero. Therefore at least one of $x = 0$ , $y = 0$ holds.

Problem 2.26.

Prove that for every real number $x$ , either $x = 0$ or $x^2 > 0$ .

Problem 2.27.

Prove the proposition about $x \leqslant 1$ and $x^2 > x$ a second time, by the indirect method.

Proving Universal Claims

To prove $\forall x \in S,\, p(x)$ we take an arbitrary element of $S$ and show that $p$ holds of it. Nothing may be assumed about that element beyond its membership in $S$ , and Universal Generalisation then gives the claim for every element of $S$ .

Strategy (Direct proof of a universal claim).

Claim: $\forall x \in S,\; p(x)$

Let $y \in S$ be arbitrary and fixed.
Prove that $p(y)$ holds.

Proposition 2.51.

For all whole numbers $n \geqslant 3$ , the number $n^2 - 1$ is composite.

Discussion.

Our statement is universally quantified over all integers $n \geq 3$ . To prove that $n^2 - 1$ is composite, we must, by definition, produce two integers each strictly greater than $1$ whose product is $n^2 - 1$ . The identity $n^2 - 1 = (n+1)(n-1)$ gives us exactly such a factorisation. The only possible concern is whether the smaller factor, $n-1$ , is at least $2$ . The hypothesis $n \geq 3$ settles this: subtracting $1$ from both sides of $n \geq 3$ yields $n - 1 \geq 2$ , which is precisely the lower bound we need. (Had we allowed $n = 2$ , the factorisation would read $3 \cdot 1$ , which is prime and therefore would fail.)

Proof.

Let $n \geqslant 3$ be any whole number. Notice that $n^2 - 1 = (n + 1)(n - 1)$ , a product of two whole numbers. Since $n \geqslant 3$ , the smaller factor satisfies $n - 1 \geqslant 2$ , so $n^2 - 1$ is a product of two whole numbers each at least $2$ . Thus $n^2 - 1$ is composite.

Remark (Arbitrary Is Not Specific).

Consider the statement “for every whole number $n$ , $n^2 \geqslant 0$ ”. The following is not a valid proof: “let $n$ be an arbitrary whole number, say $n = 17$ . Then $17^2 = 289 \geqslant 0$ .” The writer has chosen a specific value, not an arbitrary one, and what has been proved is the existential claim, that some whole number has a non-negative square. A correct proof never names $n$ , and every step in it must hold for all the values $n$ might have taken.

The negation of a universal claim is an existential one, and the work is to show that no such object can exist. This is the route to take when the claim says that something does not happen, since “there is no object with this property” is a universal claim under a negation; it says that every object fails to have the property.

Strategy (Indirect proof of a universal claim).

Claim: $\forall x \in S,\; p(x)$

AFSOC that there is some $y \in S$ for which $\neg p(y)$ holds.
Find a contradiction.

Proposition 2.52.

There is no least positive real number. That is, no positive real number $a$ satisfies $a \leqslant b$ for every positive real number $b$ .

Discussion.

The statement says that every positive real number fails to be least, which is a universal claim although the negation in front of it hides the fact. Its negation is existential and supplies a positive $a$ with $a \leqslant b$ for every positive $b$ , so we assume such an $a$ and look for a positive number smaller than it. Halving serves: $\tfrac{a}{2}$ is positive whenever $a$ is, and it is smaller than $a$ , so it both must and cannot satisfy the inequality $a$ was assumed to satisfy.

Proof.

AFSOC that a least positive real number $a$ exists, so that $a > 0$ and $a \leqslant b$ for every positive real number $b$ . Consider $\tfrac{a}{2}$ . Since $a > 0$ we have $\tfrac{a}{2} > 0$ , so $\tfrac{a}{2}$ is itself a positive real number, and being one it must satisfy $a \leqslant \tfrac{a}{2}$ . But $\tfrac{a}{2} < a$ , and no number is both smaller than $a$ and at least $a$ . Therefore no least positive real number exists.

Problem 2.28.

Prove that for every real number $x$ there is a real number $y$ with $y > x$ .

Problem 2.29.

Let $x$ be a real number with $x \geqslant 0$ , and suppose $x \leqslant y$ for every positive real number $y$ . Prove that $x = 0$ .

Proving Existential Claims

An existential claim is a claim of existence: it asserts that some object belongs to a set and has a certain property. To prove one we exhibit such an object and verify, for our reader, both halves of the assertion, that the object is an element of the correct set and that it has the correct property. Existential Generalisation then yields the claim. A proof of this kind is called constructive, since the witness is produced.

Strategy (Direct proof of an existential claim).

Claim: $\exists x \in S,\; p(x)$

Define a specific object $y$ .
Prove that $y \in S$ .
Prove that $p(y)$ holds.

Proposition 2.53.

Fix real numbers $a, b, c, d, e, f$ with $ad - bc \neq 0$ . Then there are real numbers $x$ and $y$ satisfying both

ax + by = e \qquad \text{and} \qquad cx + dy = f.

Discussion.

The claim is existential, so the proof must produce a specific pair $x$ , $y$ and check that it satisfies both equations, which means we have to find the pair before we can write anything down. Multiplying the first equation by $d$ and the second by $-b$ makes the two $y$ terms cancel when we add:

\begin{aligned} adx + bdy &= de \\ -bcx - bdy &= -bf \\ \hline (ad - bc)x &= de - bf \end{aligned}

so $x = \frac{de - bf}{ad - bc}$ , which is a real number precisely because $ad - bc \neq 0$ . Cancelling the $x$ terms the same way, by multiplying the first equation by $c$ and the second by $-a$ , gives $(bc - ad)y = ce - af$ and so $y = \frac{af - ce}{ad - bc}$ .

None of this scratch work belongs in the proof. A reader does not care how we came upon the pair; they care what the pair is and why it works, and the proof is shorter and easier to check for leaving the search out.

Proof.

Since $ad - bc \neq 0$ , we may define

x = \frac{de - bf}{ad - bc} \qquad \text{and} \qquad y = \frac{af - ce}{ad - bc},

and both are real numbers. Substituting them into the two equations,

\begin{aligned} ax + by &= \frac{(ade - abf) + (abf - bce)}{ad - bc} = \frac{ade - bce}{ad - bc} = \frac{e(ad - bc)}{ad - bc} = e, \\ cx + dy &= \frac{(cde - bcf) + (adf - cde)}{ad - bc} = \frac{adf - bcf}{ad - bc} = \frac{f(ad - bc)}{ad - bc} = f. \end{aligned}

So the pair $x, y$ satisfies both equations, and such a pair exists.

In a non-constructive existence proof we establish that a witness must exist without identifying it. The indirect method is of this kind: it never produces the object; it only shows that its absence leads to a contradiction.

Strategy (Indirect proof of an existential claim).

Claim: $\exists x \in S,\; p(x)$

AFSOC that for every $y \in S$ , $\neg p(y)$ holds.
Find a contradiction.

Proposition 2.54.

Let $a$ , $b$ , $c$ be real numbers. Then two of them have a non-negative product.

Discussion.

The claim is again existential, but this time there is no pair we could name in advance, since which two work depends on the signs of $a$ , $b$ , $c$ and the statement gives us no signs. So we prove it indirectly. The negation says that every pair fails, and it hands us the three inequalities $ab < 0$ , $bc < 0$ and $ac < 0$ at once. Multiplying all three together produces $(abc)^2$ , and a square cannot be negative.

Proof.

AFSOC that no two of $a$ , $b$ , $c$ have a non-negative product, so that $ab < 0$ , $bc < 0$ , and $ac < 0$ . Multiplying the first two of these negative numbers gives a positive one, and multiplying that by the third gives a negative one, so

(ab)(bc)(ac) < 0.

But $(ab)(bc)(ac) = a^2b^2c^2 = (abc)^2$ , and we borrow the rule that the square of a real number is never negative. (It is proved below, in the section on disproving claims.) So $(abc)^2$ is both negative and non-negative, which is impossible. Therefore some two of $a$ , $b$ , $c$ have a non-negative product.

Notice what the proof does not tell us: which pair it is. It shows only that every pair cannot fail at once, which is what makes the argument non-constructive.

Unique existence asks for a second step. We find the object that works, then show it is the only one, which we do by assuming that $x$ and $y$ both work and concluding that $x = y$ .

Strategy (Proof of a unique existential claim).

Claim: $\exists!\, x \in S,\; p(x)$

Prove $\exists x \in S,\; p(x)$ , by either method above.
Suppose $p(x)$ and $p(y)$ both hold, and prove that $x = y$ .

Proposition 2.55.

There exists a unique real number $x$ such that $3x - 1 = 0$ .

Discussion.

Unique existence is two claims, so the proof comes in two halves. Existence asks for one real number satisfying $3x - 1 = 0$ , and we verify $x = \tfrac{1}{3}$ by substitution rather than display the rearrangement that found it. Uniqueness is the half that is often forgotten: producing a solution says nothing about whether there is a second one, so we suppose that $x$ and $y$ both satisfy the equation and show that the supposition forces $x = y$ .

Proof.

Consider $x = \tfrac{1}{3}$ : then $3 \cdot \tfrac{1}{3} - 1 = 1 - 1 = 0$ , so at least one real number works. For uniqueness, suppose $x$ and $y$ both satisfy the equation. Then $3x - 1 = 0 = 3y - 1$ , so $3x = 3y$ , and dividing by $3$ gives $x = y$ . Thus exactly one real number satisfies $3x - 1 = 0$ .

Problem 2.30.

Fix a real number $a$ . Prove that the equation $x^2 + 2ax + a^2 = 0$ has exactly one real solution.

Problem 2.31.

Fix a real number $a$ . Prove that there is a real number $x$ satisfying $x^3 + (1 - a^2)x - a = 0$ .

Disproving Claims

To disprove a statement is to show that its truth value is $\bot$ , and by the definition of negation that is the same as showing its negation is $\top$ . So there is nothing new to learn: write down the logical negation, then prove that statement by whichever of the strategies above fits its shape. The negation rules are what turn the disproof into a proof, and they say which shape we end up with.

Disproving a universal claim therefore means proving an existential one, and the object we produce is the counterexample we met when the quantifier was introduced. Exhibiting a single one settles the matter, however many objects satisfy the claim.

Example 2.56 (A Universal Claim).

Disprove: for all prime numbers $n$ , the number $n$ is odd.

Discussion.

Disproving a universal claim means proving its negation, which here is the existential claim that some prime fails to be odd, so one prime is all we have to produce. An even number has the form $2k$ , and any $k \geqslant 2$ makes $2k$ a product of two whole numbers each at least $2$ , hence composite; that leaves $k = 1$ as the only candidate, and it works.

Disproof.

The number $2$ is prime, and $2$ is even. Thus it is not true that all primes are odd.

Disproving an existential claim is the harder direction, since its negation is universal: every single object must fail, and there is no shortcut through exhibiting one of them.

Example 2.57 (An Existential Claim).

Disprove: there exists a real number $x$ such that $x^2 < 0$ .

Discussion.

Disproving an existential claim means proving its negation, and that negation is universal: every real number must satisfy $x^2 \geqslant 0$ , so no single well-chosen $x$ will settle it and the argument has to cover all of them at once. What stands in the way is the sign of $x$ , which we are not told and which decides how the inequality behaves under multiplication, so we split on it and run the two cases separately.

Disproof.

We show that for all real numbers $x$ , $x^2 \geqslant 0$ . Every real number falls into one of two cases, $x \geqslant 0$ or $x < 0$ . If $x \geqslant 0$ , multiplying the inequality by itself preserves the sign, so $x \cdot x \geqslant 0 \cdot 0$ , that is, $x^2 \geqslant 0$ . If $x < 0$ , multiplying through by $-1$ flips the inequality to $-x > 0$ ; multiplying $-x > 0$ by itself then gives $(-x)(-x) > 0 \cdot 0$ , so $x^2 > 0$ and in particular $x^2 \geqslant 0$ . Either way $x^2 \geqslant 0$ , so the claim is false.

Problem 2.32.

Disprove: for every real number $x$ , $x^2 > x$ .

Problem 2.33.

Disprove: there is a real number $x$ with $x^2 + x + 1 = 0$ .

The Law of Excluded Middle

Theorem 2.58 (Law of Excluded Middle).

For any proposition $p$ , $p \lor \neg p$ is a tautology.

Discussion.

To call something a tautology is to say it is equivalent to $\top$ , so the claim is the single equivalence $p \lor \neg p \equiv \top$ . The Complement axiom asserts almost exactly that, but with the disjunction written in the other order, so the statement follows from the axiom by Commutativity. In this system the law of excluded middle is an axiom with its arguments swapped.

Proof.

By the Complement axiom, $\neg p \lor p \equiv \top$ . By the Commutativity axiom, $\neg p \lor p \equiv p \lor \neg p$ . Therefore $p \lor \neg p \equiv \top$ .

The Law of Excluded Middle (LEM) asserts that every proposition is either true or false; no third possibility exists. In proofs, LEM is what allows us to split into the exhaustive cases $p$ and $\neg p$ without knowing which of them actually holds, and a conclusion reached in both cases is established without our ever finding out.

Proposition 2.59.

Let $m$ and $n$ be whole numbers. If $mn$ is even, then $m$ is even or $n$ is even.

Discussion.

Our statement is the conditional whose hypothesis is ” $mn$ is even” and whose conclusion is the disjunction ” $m$ is even or $n$ is even”. The hypothesis says nothing about $m$ on its own and no computation extracts it, so we appeal to LEM and split on whether $m$ is even. One case gives the conclusion immediately. In the other $m$ is odd, and we show that an odd $n$ would make $mn$ odd against the hypothesis, which leaves $n$ even.

Proof.

Suppose $mn$ is even. By LEM, either $m$ is even or $m$ is odd.

Case 1. If $m$ is even, the conclusion holds immediately.

Case 2. Suppose $m$ is odd, so $m = 2j + 1$ for some whole number $j$ . If $n$ were also odd, say $n = 2k + 1$ , then $mn = (2j + 1)(2k + 1) = 4jk + 2j + 2k + 1 = 2(2jk + j + k) + 1,$ so $mn$ would be odd, contradicting the hypothesis that $mn$ is even. Therefore $n$ is even.

In both cases at least one of $m$ , $n$ is even.

Problem 2.34.

Using the Law of Excluded Middle, prove that for all whole numbers $n$ , the number $n(n + 1)$ is even.

Problem 2.35.

Look again at the proof above. Where was the Law of Excluded Middle used, where was contradiction used, and what was the contradiction? Prove the same result twice more: once using contradiction without LEM, and once using LEM without contradiction.

Ex Falso Quodlibet

The principle ex falso sequitur quodlibet (“from falsity follows whatever you like”) states that from a contradiction, any proposition may be derived:

\frac{p \quad \neg p}{q} \quad \text{for every } q

Its validity rests on the tautology $(p \land \neg p) \to q$ : since $p \land \neg p$ is identically false, the conditional holds for every $q$ by vacuous truth.

Example 2.60 (Consequences of a Contradiction).

Suppose we accept the false premise $-1 = 1$ . Adding $1$ to both sides gives $0 = 2$ , a false consequence. Squaring both sides gives $1 = 1$ , a true consequence. From a single contradiction, both true and false statements follow. Once a contradiction enters, the proof system can no longer distinguish truth from falsehood, and every proposition becomes derivable.

We rarely invoke Ex Falso directly, but every indirect proof above rests on it: if assuming the negation of our claim produces a contradiction with our true premises, then that negation cannot coexist with those premises, and the claim must hold.

Using Assumptions in Proofs

When we state a theorem it usually has hypotheses and a conclusion, and those hypotheses are ours to use: we add them to our toolkit for the duration of the proof, along with whatever facts we develop along the way.

A disjunction means cases. Suppose that at some point we have assumed or deduced $p \lor q$ . Knowing the disjunction holds means at least one of $p$ , $q$ holds, so we may consider each separately and argue as follows.

Strategy (Using a disjunction).

Because $p \lor q$ , we have two cases.

Case 1. Suppose $p$ holds. Then …

Case 2. Suppose $q$ holds. Then …

As long as the goal is reached in both cases, the deduction stands; this is the disjunction elimination rule. There is no need for a third case in which both hold, since it might not arise at all, and reaching the goal from one or the other on its own shows that assuming both was never necessary.

Remark (Without Loss of Generality).

It sometimes happens that two cases are identical up to a relabelling of variables. We then handle one case and dismiss the other with the phrase “without loss of generality” (abbreviated WLOG). For instance, to prove by contrapositive that “if $x$ and $y$ are whole numbers and both $xy$ and $x + y$ are even, then $x$ and $y$ are both even,” we must show that if $x$ or $y$ is odd then $xy$ or $x + y$ is odd. Since the hypotheses are symmetric in $x$ and $y$ , we may assume WLOG that $x$ is odd; the remaining case follows by swapping $x$ and $y$ .

“There exists” is not “let such an $x$ be given”. Writing $\exists x \in S,\, p(x)$ in the middle of a proof asserts only that the line is true: that some element of $S$ with the property is out there. It does not introduce one. If we go on to refer to $x$ afterwards, the step is not valid, since nothing has been named; several such elements might exist, and a reader cannot be left to guess whether we mean all of them or one in particular. To bring one into play, say so: let such an $x$ be given. Thereafter $x$ names an element with that property and may be used like any other object. Several at once are introduced the same way: “and so we deduce that there exist whole numbers $x$ , $y$ , $z$ such that $p(x, y, z)$ holds. Let such $x$ , $y$ , $z$ be given.”

" $p \to q$ " is not ” $p$ , therefore $q$ ”. Writing $p \to q$ on the page asserts that the implication holds; it does not assert $q$ . To deduce $q$ we must make it clear that we also have $p$ , and that we are using Modus Ponens on the two.

Strategy (Using a conditional).

$p \to q$ because …

Also, $p$ holds because …

Therefore $q$ holds.

Remark (Mistakes in Proofs).

A chain of equalities is valid only if every step is. Consider the “proof” that $-1 = 1$ :

\begin{aligned} -1 &= (-1)^1 \\ &= (-1)^{2/2} \\ &= \bigl((-1)^2\bigr)^{1/2} \\ &= 1^{1/2} \\ &= 1 \end{aligned}

The error is in the third step: the law $(x^a)^b = x^{ab}$ requires $x \geqslant 0$ . Each step in a proof must be a valid inference rather than a plausible manipulation.

Finally, Quantifiers can also conceal errors. The formula $(p \to q) \lor (q \to p)$ is a propositional tautology (verify this by truth table). Substituting predicates $p(n)$ = ” $n$ is odd” and $q(n)$ = ” $n$ is prime,” one might reason: “for every $n$ , either oddness implies primality or primality implies oddness.” Yet neither $\forall n\,(p(n) \to q(n))$ nor $\forall n\,(q(n) \to p(n))$ is true. The resolution is that $\forall n\,[(p(n) \to q(n)) \lor (q(n) \to p(n))]$ is not the same as $[\forall n\,(p(n) \to q(n))] \lor [\forall n\,(q(n) \to p(n))]$ . The universal quantifier does not distribute over disjunction. The first formula is indeed true (for each fixed $n$ , one of the two conditionals holds), but the second is false.

Exercises

Answers are checked in your browser, as often as you like. Nothing is sent anywhere and nothing is kept but your own work. A formula may be written with the symbols themselves or with ~ & | -> <-> ^, and \and, \or, \to expand as you type.

Exercises in Lean

The proofs below are checked in your browser. Nothing is sent anywhere, and nothing is stored but your own work. Type \to for →, \and for ∧, \< and \> for ⟨ ⟩.

The last sheet gave us three tactics, intro, exact and apply, and with them the implication, $\top$ , $\bot$ and negation. What is missing is the rest of the connectives. Each of $\land$ , $\lor$ and $\leftrightarrow$ comes with two questions: how to prove one, and what to do with one we have been handed. The quantifiers wait for the next sheet.

Conjunction

`constructor`

A conjunction is proved by proving both halves. If the goal is ⊢ p ∧ q, then constructor replaces it with the two goals ⊢ p and ⊢ q. A tactic acts on the first goal only, so from here on we write each goal’s proof under a focus dot ·, indented; the dot takes the first goal and the block under it must finish that goal off.

Example.

Put the caret on the constructor line and then on the line below it to watch one goal become two.

A conjunction may also be built in one term: ⟨hp, hq⟩ is the proof of p ∧ q made from a proof of each half. Type \< for ⟨ and \> for ⟩.

Taking one apart

If h : p ∧ q then h.1 is a proof of p and h.2 is a proof of q; they may also be written h.left and h.right.

Example.

The projection is a term, so it goes where any other proof would.

`obtain`

Where both halves are wanted, it reads better to take the hypothesis apart once and for all. obtain ⟨hp, hq⟩ := h replaces h : p ∧ q with hp : p and hq : q.

Example.

The same brackets take a conjunction apart on the left and build one on the right.

Exercise 2.13.

Two assumptions make their conjunction.

Exercise 2.14.

The order of a conjunction carries no information.

Exercise 2.15.

Nor does the bracketing.

Exercise 2.16.

Two hypotheses taken in turn, against the two taken together.

Exercise 2.17.

No statement holds alongside its own negation.

Disjunction

`left` and `right`

A disjunction is proved by proving one half and saying which. With the goal ⊢ p ∨ q, left changes it to ⊢ p and right changes it to ⊢ q. The choice is ours to make and cannot be deferred, so making the wrong one leaves a goal that cannot be closed. In term form the two are Or.inl and Or.inr, each applied to a proof of the half it names.

Example.

Nothing is known about $p$ , so the right half is the one to go for.

`rcases`

A disjunction we have been handed is weaker than a conjunction: it says that one of the two holds without saying which, so anything we conclude from it has to be concluded twice. rcases h with hp | hq splits h : p ∨ q into two goals, the first with hp : p and the second with hq : q, and the bar is what marks the split.

Example.

The goal is the same in both branches; only the hypothesis differs.

Exercise 2.18.

The order of a disjunction carries no information either, but here we must first find out which half we were handed.

Exercise 2.19.

Nor does the bracketing of a disjunction.

Exercise 2.20.

Conjunction distributes over disjunction.

Exercise 2.21.

One of De Morgan’s laws.

Exercise 2.22.

Half of the defining equivalence of the conditional.

Exercise 2.23.

And the other half.

Equivalence

Both directions

p ↔ q is the conjunction of p → q and q → p, and it behaves like one. On a goal ⊢ p ↔ q, constructor leaves the two implications as separate goals; ⟨h, k⟩ builds the equivalence from a proof of each direction.

Example.

The two goals arrive in the order the equivalence is written.

If h : p ↔ q then h.mp is the forward implication and h.mpr the backward one, so h.mp hp is a proof of q and h.mpr hq a proof of p. obtain ⟨f, g⟩ := h takes the two apart as it does for a conjunction.

Example.

An equivalence is used in whichever direction the goal asks for.

Exercise 2.24.

An equivalence may be read from either end.

Exercise 2.25.

Equivalence is transitive.

Exercise 2.26.

A statement conjoined with itself says no more than the statement.

Exercise 2.27.

Equivalent statements may be swapped for one another inside a disjunction.

Exercise 2.28.

No statement is equivalent to its own negation.

What the checker understands

Tactics

intro h	assume the hypothesis of an implication, naming it h
exact e	give the proof outright
apply f	reduce the goal to the hypotheses of f
assumption	close the goal with a hypothesis already present
trivial	close the goal True
exfalso	replace the goal with False
by_contra h	assume the negation of the goal
constructor	split ∧ into both halves, or ↔ into both directions
left / right	choose which half of a ∨ to prove
rcases h with a \| b	argue by cases on a disjunction
obtain ⟨a, b⟩ := h	take a conjunction or an existential apart
cases h	as above, keeping the name

Results you may cite

Classical.em	∀ (a : Prop), a ∨ ¬a — the law of excluded middle
Classical.byContradiction	∀ {a : Prop}, (¬a → False) → a — proof by contradiction; the tactic by_contra does this for you
Classical.byCases	∀ {a b : Prop}, (a → b) → (¬a → b) → b — split on whether a holds
not_not	∀ {a : Prop}, ¬¬a ↔ a — double negation
not_and_or	∀ {a b : Prop}, ¬(a ∧ b) ↔ ¬a ∨ ¬b — De Morgan
not_or	∀ {a b : Prop}, ¬(a ∨ b) ↔ ¬a ∧ ¬b — De Morgan
not_imp	∀ {a b : Prop}, ¬(a → b) ↔ a ∧ ¬b
and_comm	∀ {a b : Prop}, a ∧ b ↔ b ∧ a
or_comm	∀ {a b : Prop}, a ∨ b ↔ b ∨ a

From the logical core

And.intro	∀ {a b : Prop}, a → b → a ∧ b
And.left	∀ {a b : Prop}, a ∧ b → a
And.right	∀ {a b : Prop}, a ∧ b → b
And.symm	∀ {a b : Prop}, a ∧ b → b ∧ a
Or.inl	∀ {a b : Prop}, a → a ∨ b
Or.inr	∀ {a b : Prop}, b → a ∨ b
Or.elim	∀ {a b c : Prop}, a ∨ b → (a → c) → (b → c) → c
Or.symm	∀ {a b : Prop}, a ∨ b → b ∨ a
Iff.intro	∀ {a b : Prop}, (a → b) → (b → a) → (a ↔ b)
Iff.mp	∀ {a b : Prop}, (a ↔ b) → a → b
Iff.mpr	∀ {a b : Prop}, (a ↔ b) → b → a
Iff.symm	∀ {a b : Prop}, (a ↔ b) → (b ↔ a)
Iff.rfl	∀ {a : Prop}, a ↔ a
Iff.trans	∀ {a b c : Prop}, (a ↔ b) → (b ↔ c) → (a ↔ c)
True.intro	True
False.elim	∀ {a : Prop}, False → a
absurd	∀ {a b : Prop}, a → ¬a → b
id	∀ {a : Prop}, a → a
mt	∀ {a b : Prop}, (a → b) → ¬b → ¬a
Eq.refl	∀ {α : Type} (a : α), a = a
Eq.symm	∀ {α : Type} {a b : α}, a = b → b = a
Eq.trans	∀ {α : Type} {a b c : α}, a = b → b = c → a = c
congrArg	∀ {α : Type} {β : Type} {a b : α} (f : α → β), a = b → f a = f b
Exists.intro	∀ {α : Type} {p : α → Prop} (w : α), p w → ∃ x : α, p x
Exists.elim	∀ {α : Type} {p : α → Prop} {b : Prop}, (∃ x : α, p x) → (∀ y : α, p y → b) → b

Lesson 4

Functions

Taught 14 Aug 2026

More Sets

Once the axioms of set theory are in place, we can review some elementary albeit very useful constructions that these axioms enable.

Ordered Pairs

We often need to pair two elements while retaining their positions, and an ordinary two-element set cannot do this, since $\{a, b\} = \{b, a\}$ ; something more is needed if one element is to be marked as coming before the other. We could simply declare that the notation $(a, b)$ is ordered, but what does ordered mean in terms of sets?

Before choosing a set to represent $(a, b)$ , we state the property any such choice must have: whenever $(a, b)$ and $(c, d)$ are ordered pairs,

(a, b) = (c, d) \iff (a = c) \land (b = d). \tag{$*$}

In particular, the two coordinates must be recoverable from the pair.

Definition 4.1 (Ordered pair).

Let $a$ and $b$ be objects. The ordered pair $(a, b)$ is the set

(a, b) \;\defeq\; \bigl\{\{a\}, \{a, b\}\bigr\},

where $a$ is its first coordinate and $b$ its second coordinate.

The sets used to encode the pair, $\{a\}$ and $\{a, b\}$ , are supplied by the pairing axiom of the last chapter, and the set on the right exists by that same axiom applied once more, to those two. The construction is due to Kuratowski. Does it satisfy $(*)$ ?

Proposition 4.2 (Equality of ordered pairs).

For objects $a$ , $b$ , $c$ and $d$ , $(a, b) = (c, d)$ if and only if $a = c$ and $b = d$ .

Discussion.

Our statement is a biconditional. The reverse implication, $a = c$ and $b = d$ implies $(a, b) = (c, d)$ , is a direct substitution into the definition of the ordered pair. For the forward implication we assume $\bigl\{\{a\}, \{a, b\}\bigr\} = \bigl\{\{c\}, \{c, d\}\bigr\}$ and must recover the coordinates from the set; the argument splits into the cases $a = b$ and $a \neq b$ , because in the first case the pair collapses to the singleton $\bigl\{\{a\}\bigr\}$ while in the second its two elements are distinct and can be chased through the equality one at a time. The logical proof avoids the split: it expands each ordered pair, rewrites set equality as the two inclusions by mutual inclusion, and lets the distributive and absorption laws of the first chapter reduce the resulting disjunctions to the conjunction $a = c \land b = d$ .

Proof (element style).

Let $a$ , $b$ , $c$ and $d$ be arbitrary objects. For the reverse implication, suppose $a = c$ and $b = d$ . Then $\{a\} = \{c\}$ and $\{a, b\} = \{c, d\}$ ; hence $(a, b) = \bigl\{\{a\}, \{a, b\}\bigr\} = \bigl\{\{c\}, \{c, d\}\bigr\} = (c, d)$ .

For the forward implication, suppose $(a, b) = (c, d)$ . By the definition of set equality, every element of either ordered pair belongs to the other; we divide the argument into two cases, $a = b$ and $a \neq b$ .

If $a = b$ , then $(a, b) = \bigl\{\{a\}, \{a, a\}\bigr\} = \bigl\{\{a\}\bigr\}$ . Both $\{c\}$ and $\{c, d\}$ belong to $(c, d) = (a, b) = \bigl\{\{a\}\bigr\}$ , so each must equal $\{a\}$ ; hence $c = a$ and $d = a$ . Since $b = a$ as well, we obtain $a = c$ and $b = d$ .

If $a \neq b$ , then $\{a\} \neq \{a, b\}$ ; consequently $(a, b)$ has two distinct elements. Since $\{a\} \in (a, b) = (c, d)$ , either $\{a\} = \{c\}$ or $\{a\} = \{c, d\}$ . The latter equality would give $c = d = a$ , so $(c, d) = \bigl\{\{a\}\bigr\}$ would have only one element; this contradicts $(a, b) = (c, d)$ , since $(a, b)$ has two. Therefore $\{a\} = \{c\}$ , and hence $a = c$ .

Similarly, since $\{a, b\} \in (a, b) = (c, d)$ , either $\{a, b\} = \{c\}$ or $\{a, b\} = \{c, d\}$ . The former equality would give $a = b = c$ , contradicting $a \neq b$ ; therefore $\{a, b\} = \{c, d\}$ . It follows that $b = c$ or $b = d$ ; but $c = a \neq b$ , so $b \neq c$ and hence $b = d$ . Thus $a = c$ and $b = d$ .

Remark.

This logical proof is dumb and covoluted lol but why not.

Proof (logical style).

Let $a$ , $b$ , $c$ and $d$ be arbitrary objects. By the definition of set equality,

\begin{aligned} (a, b) = (c, d) &\equiv \Bigl(\bigl\{\{a\}, \{a, b\}\bigr\} \subset \bigl\{\{c\}, \{c, d\}\bigr\}\Bigr) \\ &\qquad {}\land \Bigl(\bigl\{\{c\}, \{c, d\}\bigr\} \subset \bigl\{\{a\}, \{a, b\}\bigr\}\Bigr) \\[2pt] &\equiv \Bigl(\bigl(\{a\} = \{c\} \lor \{a\} = \{c, d\}\bigr) \\ &\qquad\quad {}\land \bigl(\{a, b\} = \{c\} \lor \{a, b\} = \{c, d\}\bigr)\Bigr) \\ &\qquad {}\land \Bigl(\bigl(\{c\} = \{a, b\} \lor \{c\} = \{a\}\bigr) \\ &\qquad\qquad {}\land \bigl(\{c, d\} = \{a\} \lor \{c, d\} = \{a, b\}\bigr)\Bigr) \\[2pt] &\equiv \Bigl(\bigl(\{a\} = \{c\} \lor \{a\} = \{c, d\}\bigr) \\ &\qquad\quad {}\land \bigl(\{c\} = \{a, b\} \lor \{c\} = \{a\}\bigr)\Bigr) \\ &\qquad {}\land \Bigl(\bigl(\{a, b\} = \{c\} \lor \{a, b\} = \{c, d\}\bigr) \\ &\qquad\qquad {}\land \bigl(\{c, d\} = \{a\} \lor \{c, d\} = \{a, b\}\bigr)\Bigr) \\[2pt] &\equiv \Bigl(\{a\} = \{c\} \lor \bigl(\{a\} = \{c, d\} \land \{a, b\} = \{c\}\bigr)\Bigr) \\ &\qquad {}\land \Bigl(\{a, b\} = \{c, d\} \lor \bigl(\{a, b\} = \{c\} \land \{c, d\} = \{a\}\bigr)\Bigr) \\[2pt] &\equiv \bigl(\{a\} = \{c, d\} \land \{a, b\} = \{c\}\bigr) \\ &\qquad {}\lor \bigl(\{a\} = \{c\} \land \{a, b\} = \{c, d\}\bigr) \\[2pt] &\equiv \Bigl(\bigl(a = c \land a = d\bigr) \land \bigl(a = c \land b = c\bigr)\Bigr) \\ &\qquad {}\lor \Bigl(a = c \land \bigl((a = c \land b = d) \lor (a = d \land b = c)\bigr)\Bigr) \\[2pt] &\equiv (a = c \land a = d \land b = c) \lor (a = c \land b = d) \\ &\qquad {}\lor (a = c \land a = d \land b = c) \\[2pt] &\equiv (a = c \land a = d \land b = c) \lor (a = c \land b = d) \\[2pt] &\equiv (a = c \land a = d \land b = c \land b = d) \lor (a = c \land b = d) \\[2pt] &\equiv a = c \land b = d. \end{aligned}

Corollary 4.3 (Swapping coordinates).

For objects $a$ and $b$ , $(a, b) = (b, a)$ if and only if $a = b$ .

Proof.

If $(a, b) = (b, a)$ , then Proposition 4.2 gives $a = b$ by comparing first coordinates. Conversely, if $a = b$ , then $(a, b) = (a, a) = (b, a)$ .

Kuratowski’s is not the only possible choice, and the next problem looks at a shorter one.

Problem 4.1.

Some authors define the ordered pair by the shorter set

\langle a, b \rangle \;\defeq\; \bigl\{a, \{a, b\}\bigr\}.

Prove that this also satisfies $(*)$ , so that $\langle a, b \rangle = \langle c, d \rangle$ if and only if $a = c$ and $b = d$ . You may use the axiom of regularity from the last chapter, and in particular the conclusion of its second problem.
Show that regularity is genuinely needed, by identifying the step of your argument that fails without it, and say why the same step does not arise for Definition 4.1 .

Nesting ordered pairs builds longer ordered tuples.

Definition 4.4 (Ordered triples and longer).

Let $a$ , $b$ , $c$ and $d$ be objects. The ordered triple and ordered quadruple are

(a, b, c) \;\defeq\; \bigl((a, b), c\bigr), \qquad (a, b, c, d) \;\defeq\; \bigl((a, b, c), d\bigr),

and further tuples are assembled the same way.

Remark.

We have now overloaded the parenthesis symbols $(\ )$ once again: they are used not only to group operators and arguments, but also to enclose ordered pairs. This is usually not a problem in practice, as one can still determine from context which usage is intended.

Cartesian Products

With ordered pairs in hand we can collect all of them at once. The collection has to come from somewhere, and the power set supplies it.

Proposition 4.5 (Where ordered pairs live).

Let $A$ and $B$ be sets, let $a \in A$ and $b \in B$ . Then $(a, b) \in \mathcal{P}\bigl(\mathcal{P}(A \cup B)\bigr)$ .

Discussion.

An ordered pair is a set of two sets, so it sits two power sets up from whatever holds its coordinates, and the set holding both coordinates is $A \cup B$ . We go up one power set at a time: show each of $\{a\}$ and $\{a, b\}$ is a subset of $A \cup B$ , which puts both in $\mathcal{P}(A \cup B)$ , and then that having both as elements makes $(a, b)$ a subset of $\mathcal{P}(A \cup B)$ , which is what membership in the next power set asks for.

Proof.

Since $a \in A$ we have $a \in A \cup B$ , and since $b \in B$ we have $b \in A \cup B$ . Every element of $\{a\}$ is $a$ , and every element of $\{a, b\}$ is $a$ or $b$ , so both $\{a\} \subset A \cup B$ and $\{a, b\} \subset A \cup B$ , that is, both belong to $\mathcal{P}(A \cup B)$ .

The elements of $(a, b) = \bigl\{\{a\}, \{a, b\}\bigr\}$ are exactly those two sets, so $(a, b) \subset \mathcal{P}(A \cup B)$ , which is $(a, b) \in \mathcal{P}\bigl(\mathcal{P}(A \cup B)\bigr)$ .

Definition 4.6 (Cartesian product).

Let $A$ and $B$ be sets. The Cartesian product $A \times B$ is the set of all ordered pairs with first coordinate in $A$ and second in $B$ ,

A \times B \;\defeq\; \Bigl\{p \in \mathcal{P}\bigl(\mathcal{P}(A \cup B)\bigr) \;\Big|\; \exists a \in A\ \exists b \in B\,\bigl(p = (a, b)\bigr)\Bigr\},

so that for every object $p$ , $p \in A \times B$ exactly when $p = (a, b)$ for some $a \in A$ and some $b \in B$ .

The product needs no axiom of its own. The proposition above puts every candidate pair inside $\mathcal{P}(\mathcal{P}(A \cup B))$ , a set we are already holding, and comprehension carves the product out of it.

Example 4.7.

Let $A = \{1, 2\}$ and $B = \{r, s\}$ . Then $A \times B = \bigl\{(1, r), (1, s), (2, r), (2, s)\bigr\}$ . Order matters here in a way it did not for unordered pairs: $(1, r) \in A \times B$ while $(r, 1) \in B \times A$ , and the two products are different sets.

Problem 4.2.

Let $A$ and $B$ be sets. Show that $A \times B = \emptyset$ if and only if $A = \emptyset$ or $B = \emptyset$ .

Membership in a product is a conjunction, which adds another line to the dictionary:

\bigl((s, t) \in S \times T\bigr) \equiv (s \in S) \land (t \in T).

Example 4.8.

Order matters, and in general $S \times T \neq T \times S$ . With $S = \{1, 2\}$ and $T = \{a, b, c\}$ we have $T \times S = \{(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)\}$ , which is not the earlier set: the element $(1, a)$ lies in $S \times T$ but not in $T \times S$ , since $1 \notin T$ .

Example 4.9.

For any set $S$ , both $S \times \emptyset$ and $\emptyset \times S$ are empty. A pair in $S \times \emptyset$ would need its second coordinate from $\emptyset$ , which has nothing to give; through the dictionary, $\bigl((s, t) \in S \times \emptyset\bigr) \equiv (s \in S) \land \bot \equiv \bot$ .

Products of three or more sets work the same way, $S \times T \times R$ collecting the triples $(s, t, r)$ with $s \in S$ , $t \in T$ and $r \in R$ , and so on upwards. There is a subtlety here.

Remark (Order of operations).

The definition takes two sets at a time, so for three sets there are two readings, and they are not the same set. Applying it twice, the elements of $A \times (B \times C)$ are the pairs $\bigl(x, (y, z)\bigr)$ , which unfold to

\bigl\{\{x\},\ \bigl\{x, \{\{y\}, \{y, z\}\}\bigr\}\bigr\},

while the elements of $(A \times B) \times C$ are the pairs $\bigl((x, y), z\bigr)$ , which unfold to

\Bigl\{\bigl\{\{\{x\}, \{x, y\}\}\bigr\},\ \bigl\{\{\{x\}, \{x, y\}\}, z\bigr\}\Bigr\}.

Both should intuitively be the set of all triples $(x, y, z)$ , and they do describe the same object once we identify them. Using a term we introduce below, the identification amounts to checking that

\bigl(x, (y, z)\bigr) \longmapsto \bigl((x, y), z\bigr)

is a bijection from the first set onto the second, so the product is associative up to that correspondence. Once bijections are in hand this is a short piece of work, and the problems below ask you to do it; from then on we drop the brackets and write $A \times B \times C$ .

Problem 4.3.

We said that $S \times T \neq T \times S$ in general. Find every case in which they are equal.

Indexed Families

It is often convenient to give each set in a collection a label.

Definition 4.10 (Indexed family).

Let $I$ be a set. An indexed family consists of one object $x_i$ for each $i \in I$ , and is written $\{x_i\}_{i \in I}$ . The set $I$ is the index set and $i$ is an index. Different indices may label the same object.

Example 4.11.

Let $A$ , $C$ and $D$ be sets, take $I = \{1, 2, 3, 4\}$ , and declare $B_1 \defeq A$ , $B_2 \defeq C$ , $B_3 \defeq A$ and $B_4 \defeq D$ . The family $\{B_i\}_{i \in I}$ has four entries, and the equality $B_1 = B_3$ removes neither of them: the labels $1$ and $3$ are different even though the sets they label are equal. The braces in the notation are a convention, since what we are holding is an assignment of an object to each index, and that assignment records more than the plain set of members $\{A, C, D\}$ , which does not record that $A$ occurs twice.

Definition 4.12 (Indexed unions and intersections).

Let $\{A_i\}_{i \in I}$ be an indexed family of sets with $I$ non-empty. Its union and intersection are

\bigcup_{i \in I} A_i \defeq \bigcup \{A_i : i \in I\}, \qquad \bigcap_{i \in I} A_i \defeq \Bigl\{x \in \bigcup_{i \in I} A_i \;\Big|\; \forall i \in I\,(x \in A_i)\Bigr\},

so that $x \in \bigcup_{i \in I} A_i$ exactly when $x \in A_i$ for some $i \in I$ , and $x \in \bigcap_{i \in I} A_i$ exactly when $x \in A_i$ for every $i \in I$ .

Neither needs an axiom beyond those we have. Replacement turns the index set into the set $\{A_i : i \in I\}$ of members, since each $i$ has exactly one partner $A_i$ , and the union axiom pools what those members hold. The intersection is then carved out of that union by comprehension, exactly as the intersection of a system of sets was in the last chapter, and for the same reason: an object lying in every member lies in some member, so nothing is lost by looking only inside the union.

By the bounded quantifier convention of the last chapter, the two membership tests unabbreviate to $\exists i\,\bigl((i \in I) \land (x \in A_i)\bigr)$ for the union and $\forall i\,\bigl((i \in I) \implies (x \in A_i)\bigr)$ for the intersection.

For $I = \{1, 2\}$ the new symbols hand back the old ones, $\bigcup_{i \in I} A_i = A_1 \cup A_2$ and $\bigcap_{i \in I} A_i = A_1 \cap A_2$ . With quantifiers doing the work, the negation rules settle what it takes not to belong:

x \notin \bigcap_{i \in I} A_i \iff \exists i \in I\,(x \notin A_i), \qquad x \notin \bigcup_{i \in I} A_i \iff \forall i \in I\,(x \notin A_i).

So $x$ is outside the intersection if it misses a single $A_i$ , but outside the union only if it misses every $A_i$ .

Problem 4.4 (Informal).

Let $I = \{1, 2, 3, 4\}$ and let $A_i$ be the set of whole numbers $m$ with $i \leqslant m \leqslant 2i$ . Determine $\bigcup_{i \in I} A_i$ and $\bigcap_{i \in I} A_i$ .

Problem 4.5.

Let $I$ be a non-empty set, let $P$ be a statement not involving $i$ , and let $Q(i)$ be a predicate on $I$ . Prove the two exchange laws

P \land \exists i \in I\,Q(i) \equiv \exists i \in I\,\bigl(P \land Q(i)\bigr), \qquad P \land \forall i \in I\,Q(i) \equiv \forall i \in I\,\bigl(P \land Q(i)\bigr).

Where does the non-emptiness of $I$ enter? Show that if $I = \emptyset$ one of the two equivalences survives and the other fails.

Proposition 4.13 (Set difference over indexed families).

Let $A$ be a set and let $\{B_i\}_{i \in I}$ be an indexed family of sets with $I$ non-empty. Then

$A \setminus \bigcap_{i \in I} B_i = \bigcup_{i \in I} (A \setminus B_i)$ ;
$A \setminus \bigcup_{i \in I} B_i = \bigcap_{i \in I} (A \setminus B_i)$ .

Discussion.

Both are equalities of sets, so we compare membership on the two sides, and each comparison admits two readings. Elementwise, an element of $A \setminus \bigcap_{i \in I} B_i$ sits in $A$ and misses some $B_j$ , so it sits in $A \setminus B_j$ and hence in the union of the differences; conversely, belonging to one difference $A \setminus B_j$ already keeps the element out of the intersection. The second identity is the same argument with “some” replaced by “every”. The logical route says this through the dictionary: the definition turns membership in the intersection into a universal quantifier and membership in the union into an existential one, the negation rules swap $\forall$ for $\exists$ and back, and the exchange laws of the problem above push the fixed condition $x \in A$ through the quantifier.

Proof (element style).

For the first, suppose $x \in A \setminus \bigcap_{i \in I} B_i$ . Then $x \in A$ , but $x$ does not belong to every $B_i$ , so $x \notin B_j$ for some $j \in I$ . Hence $x \in A \setminus B_j$ , and therefore $x \in \bigcup_{i \in I} (A \setminus B_i)$ . Conversely, suppose $x$ belongs to that union. Then for some $j \in I$ we have $x \in A \setminus B_j$ , so $x \in A$ and $x \notin B_j$ . It follows that $x \notin \bigcap_{i \in I} B_i$ , whence $x \in A \setminus \bigcap_{i \in I} B_i$ .

For the second, suppose $x \in A \setminus \bigcup_{i \in I} B_i$ . Then $x \in A$ and $x \notin B_i$ for every $i \in I$ , so $x \in A \setminus B_i$ for every $i \in I$ , that is, $x \in \bigcap_{i \in I} (A \setminus B_i)$ . Conversely, membership in that intersection gives $x \in A \setminus B_i$ for every $i \in I$ , so $x \in A$ and no $B_i$ holds $x$ , which puts $x$ in $A \setminus \bigcup_{i \in I} B_i$ .

Proof (logical style).

For any $x$ ,

\begin{aligned} x \in A \setminus \bigcap_{i \in I} B_i &\equiv (x \in A) \land \neg\,\forall i \in I\,(x \in B_i) && \text{by the definitions} \\ &\equiv (x \in A) \land \exists i \in I\,(x \notin B_i) && \text{by quantifier negation} \\ &\equiv \exists i \in I\,\bigl((x \in A) \land (x \notin B_i)\bigr) && \text{by the exchange law} \\ &\equiv x \in \bigcup_{i \in I} (A \setminus B_i) && \text{by the definitions} \end{aligned}

and likewise

\begin{aligned} x \in A \setminus \bigcup_{i \in I} B_i &\equiv (x \in A) \land \neg\,\exists i \in I\,(x \in B_i) && \text{by the definitions} \\ &\equiv (x \in A) \land \forall i \in I\,(x \notin B_i) && \text{by quantifier negation} \\ &\equiv \forall i \in I\,\bigl((x \in A) \land (x \notin B_i)\bigr) && \text{by the exchange law} \\ &\equiv x \in \bigcap_{i \in I} (A \setminus B_i) && \text{by the definitions.} \end{aligned}

Corollary 4.14 (De Morgan's laws for indexed families).

Let $\{B_i\}_{i \in I}$ be an indexed family of subsets of an ambient set $U$ , with $I$ non-empty. With every complement taken in $U$ ,

\overline{\bigcap_{i \in I} B_i} = \bigcup_{i \in I} \overline{B_i}, \qquad \overline{\bigcup_{i \in I} B_i} = \bigcap_{i \in I} \overline{B_i}.

Proof.

Take $A = U$ in the two parts of the proposition. Each difference $U \setminus B_i$ is the complement $\overline{B_i}$ , and the identities follow.

Problem 4.6.

Let $A$ be a set and $\{B_i\}_{i \in I}$ an indexed family of sets with $I$ non-empty. Prove that $A \subset \bigcap_{i \in I} B_i$ if and only if $A \subset B_i$ for every $i \in I$ , and that $\bigcup_{i \in I} B_i \subset A$ if and only if $B_i \subset A$ for every $i \in I$ .

Problem 4.7.

Let $\{A_{i,j}\}_{(i,j) \in I \times J}$ be an indexed family of sets with $I$ and $J$ non-empty. Decide which inclusion between

\bigcup_{i \in I} \bigcap_{j \in J} A_{i,j} \qquad\text{and}\qquad \bigcap_{j \in J} \bigcup_{i \in I} A_{i,j}

holds for every such family, and whether equality is forced. (Harder.)

Functions

In order to do analysis it is not particularly useful to have only the notion of a set; we also need the notion of a function from one set to another. Informally, a function $f : X \to Y$ is an operation which assigns to each element, or input, $x$ in $X$ a single element, or output, $f(x)$ in $Y$ . Formulas, rules and graphs are convenient ways to describe the pairing of inputs with outputs, but the set of pairs itself is the function.

Definition 4.15 (Function).

Let $X$ and $Y$ be sets. A function from $X$ to $Y$ is a subset $f \subset X \times Y$ such that for every $x \in X$ there is exactly one $y \in Y$ with $(x, y) \in f$ . We write $f : X \to Y$ , call $X$ the domain of $f$ , written $\operatorname{dom} f$ , and $Y$ the codomain, and write $f(x)$ for the unique $y$ paired with $x$ , so that for any $x \in X$ and $y \in Y$ ,

y = f(x) \iff (x, y) \in f.

The condition on $f$ is sometimes called the vertical line test: exactly one pair of $f$ stands above each point of the domain. It is two demands at once, existence and uniqueness, and either can fail on its own.

We also write $x \mapsto f(x)$ for the pairing. The object $f(x)$ is the image of $x$ , and $x$ is a preimage of $f(x)$ , one of possibly several. Functions are also called maps, mappings or transformations, depending on the context.

Remark.

A function is often handed to us as a property rather than a set of pairs. Let $P(x, y)$ pertain to $x \in X$ and $y \in Y$ , and suppose that for every $x \in X$ there is exactly one $y \in Y$ making $P(x, y)$ true. Then

f \;\defeq\; \bigl\{p \in X \times Y \;\big|\; \exists x \in X\ \exists y \in Y\,\bigl(p = (x, y) \land P(x, y)\bigr)\bigr\}

is a set by comprehension, and the hypothesis on $P$ is exactly the vertical line test, so $f$ is a function from $X$ to $Y$ with $y = f(x)$ precisely when $P(x, y)$ holds. No further axiom is needed to turn a property into a function; the ones we already have build the set of pairs for us.

Example 4.16 (Informal).

Assuming you know what the real numbers are, take both domain and codomain to be them. The declaration $g(x) = 1/x$ assigns nothing at $x = 0$ , so existence fails there and $g$ is not a function on that domain. The condition $h(x)^2 = x^2$ offers two candidates whenever $x \neq 0$ , namely $x$ and $-x$ , so uniqueness fails and $h$ is not a function either.

Remark (Informal examples).

Several examples below assume you know the real numbers and the whole numbers, which we have not built yet. In those we write $R$ for the real numbers and $W$ for the whole numbers, and we borrow the rules of school algebra openly. Nothing in the theory rests on them; they are there because familiar objects make the definitions easier to read.

A rule that meets both requirements, assigning a unique value in the codomain to every element of the domain, is said to be well defined. One common way to present a function is to specify its domain, its codomain, and how the output $f(x)$ is generated from each input; this is an explicit definition.

Remark.

Functions obey substitution: if $x = x'$ then $f(x) = f(x')$ , since $f(x)$ was defined as the unique $y$ paired with $x$ , and $x$ and $x'$ name the same object. Equal inputs give equal outputs. Unequal inputs need not give unequal outputs, as a constant map shows.

Three functions occur often enough to deserve names.

Definition 4.17 (Inclusion, identity and constant maps).

Let $S \subset T$ . The inclusion map $\iota : S \to T$ is defined by $\iota(x) \defeq x$ . When $S = T$ this is the identity map on $S$ , written $\mathrm{id}_S$ . For sets $A$ and $B$ and a fixed $b_0 \in B$ , the constant map with value $b_0$ is $c : A \to B$ given by $c(a) \defeq b_0$ for every $a \in A$ .

Example 4.18.

There is a function from $\emptyset$ to any set $B$ , namely the empty set of pairs. Neither requirement can find an element of $\emptyset$ on which to fail, so both hold vacuously. We shall see below that it is the only one from $\emptyset$ to $B$ .

When the domain and codomain are sets of real numbers we can plot the pairs of $f \subset A \times B$ in the Cartesian plane to draw its graph; for $q(x) = x^2$ this yields the familiar parabola. An arbitrary domain might admit no geometric picture, but the underlying set of pairs remains. Drawn this way, the identity on $S$ is the diagonal, since it pairs each $x$ with itself, and a constant map is a horizontal line.

Proposition 4.19 (Equality of functions).

Functions $f$ and $g$ are equal if and only if they have the same domain and $f(x) = g(x)$ for every $x$ in it.

Discussion.

We prove the two implications of the biconditional. Suppose first that $f = g$ . Equality of the sets of ordered pairs gives equality of their first coordinates, hence equality of the domains; for an arbitrary element of the common domain the pair $(x, f(x))$ belongs to $g$ , and the uniqueness clause of Definition 4.15 gives $f(x) = g(x)$ . Conversely, assume the domains agree and the two functions have the same value at every point of the common domain. By mutual inclusion it is enough to prove the two inclusions. An arbitrary $(x, y) \in f$ has $x$ in the domain and $y = f(x)$ , so the hypotheses put $(x, y)$ in $g$ ; the reverse inclusion follows symmetrically.

Proof (element style).

Suppose first that $f = g$ . The domain of a function consists of the first coordinates of its elements, so $f$ and $g$ have the same domain. For any $x$ in it the pair $(x, f(x))$ belongs to $f$ , and since $f = g$ we have $(x, f(x)) \in g$ , which gives $f(x) = g(x)$ by the uniqueness of images in $g$ .

Conversely, assume the domains agree and $f(x) = g(x)$ for every $x$ in the common domain. If $(x, y) \in f$ , then $x$ lies in that domain and $y = f(x)$ ; since $f(x) = g(x)$ we have $y = g(x)$ , so $(x, y) \in g$ . Hence $f \subset g$ . A symmetric argument gives $g \subset f$ , so $f = g$ .

Proof (logical style).

If $f = g$ , then for every $x$ ,

\begin{aligned} x \in \operatorname{dom} f &\equiv \exists y\,\bigl((x, y) \in f\bigr) \\ &\equiv \exists y\,\bigl((x, y) \in g\bigr) \equiv x \in \operatorname{dom} g, \end{aligned}

and for $x$ in this common domain $(x, f(x)) \in f = g$ , so uniqueness gives $f(x) = g(x)$ . Conversely, for every ordered pair $(x, y)$ ,

\begin{aligned} (x, y) \in f &\equiv (x \in \operatorname{dom} f) \land \bigl(y = f(x)\bigr) \\ &\equiv (x \in \operatorname{dom} g) \land \bigl(y = g(x)\bigr) \equiv (x, y) \in g. \end{aligned}

Remark.

The codomain does not enter the criterion, because under Definition 4.15 a function is simply its set of ordered pairs. Consequently the set $\{(x, x) \mid x \in S\}$ defines the identity map $\mathrm{id}_S : S \to S$ , but for any $T$ with $S \subset T$ it equally defines the inclusion map $\iota : S \to T$ . The set of pairs is identical; the declared codomain matters when we ask whether every element of the target is reached.

Example 4.20.

There is only one function from $\emptyset$ to a given set $X$ . Any two have the same domain, namely $\emptyset$ , and agree at every point of it, since there are none, so the proposition makes them equal.

Definition 4.21 (Restriction and extension).

Let $g : B \to C$ be a function and let $A \subset B$ . The restriction of $g$ to $A$ is the function $g|_A : A \to C$ given by $g|_A(x) \defeq g(x)$ . The function $g$ is called an extension of $g|_A$ to $B$ .

Example 4.22 (Informal).

Consider

\begin{aligned} f &: W \to R, & f(x) &\defeq x^2 - 1, \\ g &: R \to R, & g(x) &\defeq x^2 - 1, \\ h &: R \to R, & h(x) &\defeq (x - 1)(x + 1). \end{aligned}

By the proposition, $g = h$ : the domains agree, and the rules of school algebra give $(x-1)(x+1) = x^2 - 1$ for every real number. The function $f$ is a different object entirely. Its domain is $W$ , so it cannot equal $g$ or $h$ , even though its values agree with theirs at every whole number. It is their common restriction, $f = g|_W = h|_W$ . Equality of functions depends on domains, not on formulas.

Remark.

Definition 4.15 builds a function as a set of pairs, and the domain and codomain are read off the notation $f : X \to Y$ rather than carried by the set. A tidier alternative is to package all three, taking a function to be the ordered triple $(X, Y, G)$ of a domain, a codomain and a set $G \subset X \times Y$ obeying the vertical line test. Nothing in what follows depends on the choice, and the problems below ask you to check that the two accounts agree.

Problem 4.8.

Let $A = \{1, 2, 3\}$ and $B = \{r, s\}$ . Which of the following are functions from $A$ to $B$ ? Justify each answer.

$\{(1, r), (2, s), (3, r)\}$ ;
$\{(1, r), (2, s)\}$ ;
$\{(1, r), (1, s), (2, r), (3, s)\}$ .

Problem 4.9.

Give sets $A$ and $B$ for which $f = \{(3, 2), (1, 1), (8, 5), (9, -4), (\pi, 1)\}$ is a function from $A$ to $B$ . What is $f(8)$ , and what are the preimages of $1$ ? Is your choice of $A$ and $B$ the only one?

Problem 4.10.

Let $X$ and $Y$ be non-empty sets. Prove that $X \times Y$ is a function from $X$ to $Y$ if and only if $Y$ has exactly one element.

Problem 4.11.

Let $f : X \to Y$ be a function and define its graph to be the subset $\{(x, f(x)) \mid x \in X\}$ of $X \times Y$ .

Show that two functions $f, \tilde{f} : X \to Y$ are equal if and only if they have the same graph.
Conversely, let $G \subset X \times Y$ be such that for each $x \in X$ the set $\{y \in Y \mid (x, y) \in G\}$ has exactly one element. Show that there is exactly one function $f : X \to Y$ whose graph is $G$ .
Suppose we define a function instead to be an ordered triple $(X, Y, G)$ with $G \subset X \times Y$ obeying the vertical line test, taking the domain to be $X$ , the codomain $Y$ , and $f(x)$ the unique $y$ with $(x, y) \in G$ . Show that this definition agrees with Definition 4.15 , in the sense that every choice of domain, codomain and property obeying the vertical line test produces a function in this sense with all the properties the earlier definition requires.

Images and Preimages

Definition 4.23 (Image of a set).

Let $f : A \to B$ be a function and let $S \subset A$ . The image of $S$ under $f$ is $f(S) \defeq \{f(x) \mid x \in S\}$ . The image of the whole domain is the range of $f$ , written $\operatorname{im} f \defeq f(A)$ .

The range is a subset of the codomain, and it may be a proper one. In terms of the ordered pairs of Definition 4.15 ,

\begin{aligned} \operatorname{im} f &= \bigl\{y \in B \;\big|\; \exists x \in A\,\bigl((x, y) \in f\bigr)\bigr\} \\ &= \bigl\{y \in B \;\big|\; \exists x \in A\,\bigl(f(x) = y\bigr)\bigr\}. \end{aligned}

Definition 4.24 (Preimage of a set).

Let $f : A \to B$ be a function and let $U \subset B$ . The preimage of $U$ under $f$ is $f^{-1}(U) \defeq \{x \in A \mid f(x) \in U\}$ . In particular $f^{-1}(\{y\}) = \{x \in A \mid f(x) = y\}$ is the set of all preimages of $y \in B$ .

The notation $f^{-1}(U)$ does not assume that $f$ has an inverse function; the definition applies to every function. The set $f^{-1}(\{y\})$ may be empty, may hold one element, or may hold several.

Example 4.25.

Let $A = \{1, 2, 3, 4\}$ and $B = \{r, s, t, u\}$ , and define $f : A \to B$ by $f(1) \defeq r$ , $f(2) \defeq s$ , $f(3) \defeq s$ , $f(4) \defeq t$ . Then $f(\{1, 3\}) = \{r, s\}$ , $f(\{2, 3, 4\}) = \{s, t\}$ and $\operatorname{im} f = \{r, s, t\}$ . The element $u$ belongs to the codomain but not to the range, and its point preimage is empty. In general $f^{-1}(\{y\})$ is non-empty exactly when $y \in \operatorname{im} f$ , and how many elements it holds records how many inputs are sent to $y$ . Here $f^{-1}(\{s\}) = \{2, 3\}$ , $f^{-1}(\{r, t\}) = \{1, 4\}$ and $f^{-1}(\{u\}) = \emptyset$ .

Example 4.26 (Informal).

Take $g : R \to R$ with $g(x) \defeq x + 1$ , and let $S$ be the set of real numbers $x$ with $0 \leqslant x \leqslant 2$ . Then $g(S)$ is the set of real numbers $y$ with $1 \leqslant y \leqslant 3$ : on the one hand $0 \leqslant x \leqslant 2$ gives $1 \leqslant x + 1 \leqslant 3$ , which proves one inclusion, and on the other, any such $y$ has $y - 1$ in $S$ with $g(y - 1) = y$ , which proves the reverse. The same calculation read backwards gives $g^{-1}\bigl(g(S)\bigr) = S$ .

Proposition 4.27 (Set laws for preimages).

Let $f : A \to B$ be a function and let $U, V \subset B$ . Then

$f^{-1}(B \setminus U) = A \setminus f^{-1}(U)$ ;
$f^{-1}(U \cup V) = f^{-1}(U) \cup f^{-1}(V)$ ;
$f^{-1}(U \cap V) = f^{-1}(U) \cap f^{-1}(V)$ .

Discussion.

We prove each equality by comparing membership at an arbitrary $x$ . By the definition of the preimage, membership says that $x \in A$ and that its image has the required property. So $f(x) \in B \setminus U$ says precisely that $f(x) \notin U$ ; membership in $U \cup V$ means membership in at least one of $U$ and $V$ ; and membership in $U \cap V$ means membership in both. Translating those three conditions back through the same definition yields the complement, the union and the intersection we want.

Proof (element style).

If $x \in f^{-1}(B \setminus U)$ , then $x \in A$ and $f(x) \notin U$ , so $x \in A \setminus f^{-1}(U)$ . Conversely, if $x \in A \setminus f^{-1}(U)$ , then $f(x) \notin U$ ; since $f(x) \in B$ this puts $f(x) \in B \setminus U$ , whence $x \in f^{-1}(B \setminus U)$ .

If $x \in f^{-1}(U \cup V)$ , then $f(x)$ belongs to $U$ or to $V$ , so $x$ belongs to $f^{-1}(U) \cup f^{-1}(V)$ . Conversely, membership in that union puts $x$ in one of the two preimages, so $f(x) \in U \cup V$ and $x \in f^{-1}(U \cup V)$ .

Finally $x \in f^{-1}(U \cap V)$ gives both $f(x) \in U$ and $f(x) \in V$ , so $x \in f^{-1}(U) \cap f^{-1}(V)$ ; and membership in both preimages gives $f(x) \in U \cap V$ , which is the reverse inclusion.

Proof (logical style).

For every $x$ ,

\begin{aligned} x \in f^{-1}(B \setminus U) &\equiv (x \in A) \land \bigl(f(x) \notin U\bigr) \equiv x \in A \setminus f^{-1}(U), \\[2pt] x \in f^{-1}(U \cup V) &\equiv (x \in A) \land \bigl((f(x) \in U) \lor (f(x) \in V)\bigr) \\ &\equiv \bigl(x \in f^{-1}(U)\bigr) \lor \bigl(x \in f^{-1}(V)\bigr) \equiv x \in f^{-1}(U) \cup f^{-1}(V), \\[2pt] x \in f^{-1}(U \cap V) &\equiv (x \in A) \land \bigl((f(x) \in U) \land (f(x) \in V)\bigr) \\ &\equiv \bigl(x \in f^{-1}(U)\bigr) \land \bigl(x \in f^{-1}(V)\bigr) \equiv x \in f^{-1}(U) \cap f^{-1}(V). \end{aligned}

The middle steps distribute the conjunct $x \in A$ across the disjunction and the conjunction, which is Distributivity and, for the third line, Idempotence. The membership conditions agree in each pair, so the corresponding sets are equal.

Preimages preserve complements, unions and intersections. Images preserve unions, but need not preserve intersections, as the problems below ask you to show.

Read as the assignments $S \mapsto f(S)$ and $U \mapsto f^{-1}(U)$ , a function $f : A \to B$ induces two maps between power sets, one from $\mathcal{P}(A)$ to $\mathcal{P}(B)$ and one from $\mathcal{P}(B)$ to $\mathcal{P}(A)$ , written $f$ and $f^{-1}$ again. The two behave differently, and the proposition is the reason: the preimage map preserves all three operations, while the image map preserves only unions.

Remark.

This asymmetry is why continuity is later stated in terms of preimages rather than images: a function is continuous exactly when the preimage of every open set is open.

Problem 4.12.

For the finite function above, determine $f(\emptyset)$ , $f(\{1, 4\})$ , $f^{-1}(\{r, s\})$ and $f^{-1}(\{s, u\})$ .

Problem 4.13.

Let $f : A \to B$ be a function and let $S, T \subset A$ . Prove that $f(S \cup T) = f(S) \cup f(T)$ . Must $f(S \cap T) = f(S) \cap f(T)$ always hold?

Surjections, Injections, Bijections

Definition 4.28 (Surjection).

A function $f : A \to B$ is surjective, or onto $B$ , if $f(A) = B$ ; equivalently, if for every $y \in B$ there is some $x \in A$ with $f(x) = y$ .

Every function is surjective onto its range.

Example 4.29 (Informal).

The successor map $s : W \to W$ given by $s(n) \defeq n + 1$ is not surjective, since $0$ has no preimage among the whole numbers. The same ordered pairs define a surjection onto $W \setminus \{0\}$ . So surjectivity depends on the declared codomain, not on the pairs alone.

Definition 4.30 (Injection).

A function $f : A \to B$ is injective, or one-to-one, if for all $x_1, x_2 \in A$ , $f(x_1) = f(x_2) \implies x_1 = x_2$ .

The contrapositive of that implication says distinct inputs have distinct images. To show a function is not injective it is enough to give two distinct inputs with the same image; to show it is not surjective it is enough to give an element of the codomain with no preimage.

Proposition 4.31 (Point preimages).

Let $f : A \to B$ be a function.

$f$ is injective if and only if $f^{-1}(\{y\})$ holds at most one element for every $y \in B$ ;
$f$ is surjective if and only if $f^{-1}(\{y\})$ is non-empty for every $y \in B$ .

Discussion.

For the first part we compare the definition of injectivity with the assertion that two members of $f^{-1}(\{y\})$ coincide. In one direction two such members have equal images; in the other, equal images make the two inputs members of one point preimage. For the second part, the definition of the preimage says $f^{-1}(\{y\})$ is non-empty exactly when some $x \in A$ satisfies $f(x) = y$ , which is what surjectivity asks.

Proof.

Suppose $f$ is injective. If $x_1, x_2 \in f^{-1}(\{y\})$ , then $f(x_1) = y = f(x_2)$ , so $x_1 = x_2$ . Conversely, suppose every point preimage holds at most one element. Whenever $f(x_1) = f(x_2) = y$ , both $x_1$ and $x_2$ belong to $f^{-1}(\{y\})$ , and hence they are equal. This proves the first part.

The second is the surjectivity condition rewritten: $f^{-1}(\{y\})$ is non-empty exactly when some $x \in A$ satisfies $f(x) = y$ .

Example 4.32 (Informal).

For a real number $c$ , the translation $\tau_c : R \to R$ given by $\tau_c(x) \defeq x + c$ is injective and surjective. If $\tau_c(x_1) = \tau_c(x_2)$ , then $x_1 + c = x_2 + c$ , and adding $-c$ to both sides gives $x_1 = x_2$ . For surjectivity, take any $y$ and put $x = y - c$ ; then $\tau_c(x) = (y - c) + c = y$ , so every element of the codomain has a preimage.

Example 4.33 (Informal).

The squaring map $q : R \to R$ with $q(x) \defeq x^2$ is neither injective nor surjective: $q(-1) = q(1)$ , and $-1$ lies outside its range because every square is non-negative.

Example 4.34 (Informal).

The product map $m : W \times W \to W$ with $m(a, b) \defeq ab$ is surjective, since $m(n, 1) = n$ for every whole number $n$ . It is not injective, since $m(1, 6) = m(2, 3)$ while $(1, 6) \neq (2, 3)$ by Proposition 4.2 .

Problem 4.14.

For the finite function of the previous section, decide whether it is injective and whether it is surjective, justifying each answer from the definitions.

Problem 4.15.

Let $A$ be non-empty and let $B$ have exactly two elements. How many functions from $A$ to $B$ are not surjective?

Problem 4.16.

A function $f : A \to B$ is a set of ordered pairs, so its pairs may be reversed to form the set $\rho(f) \defeq \{(y, x) \in B \times A \mid (x, y) \in f\}$ . Call the set of first coordinates of a set of pairs its domain.

Determine $\rho$ of the set $\{(1, r), (2, s), (3, r)\}$ of Problem 4.8 and of the function $f$ of Problem 4.9 , and say which of the two is a function.
Prove that if $f$ is not injective, then $\rho(f)$ is not a function.
Prove that if $f$ is injective but not surjective, then $\rho(f)$ is a function whose domain $D$ satisfies $D \subsetneq B$ .

Bijections

Definition 4.35 (Bijection).

A function $f : A \to B$ is bijective, or a bijection, if it is both injective and surjective. A bijection from $A$ to $B$ is also called a one-to-one correspondence between them.

Surjectivity asks for at least one preimage of each point of the codomain and injectivity permits at most one, so a bijection has exactly one.

Proposition 4.36 (Unique preimages).

A function $f : A \to B$ is bijective if and only if for every $y \in B$ there is a unique $x \in A$ with $f(x) = y$ .

Discussion.

We use the two parts of Proposition 4.31 . If $f$ is bijective, surjectivity gives a preimage of each $y \in B$ and injectivity shows no second preimage is possible. Conversely, existence of a preimage for every $y$ gives surjectivity, and its uniqueness gives injectivity. The same criterion is what will let us reverse the ordered pairs of a bijection later.

Proof.

Suppose first that $f$ is bijective. Surjectivity supplies, for each $y \in B$ , an $x \in A$ with $f(x) = y$ . If $x_1$ and $x_2$ both have this property, then $f(x_1) = f(x_2)$ , so injectivity gives $x_1 = x_2$ .

Conversely, suppose every $y \in B$ has exactly one preimage. Existence makes $f$ surjective. If $f(x_1) = f(x_2)$ , both $x_1$ and $x_2$ are preimages of the same element of $B$ , so uniqueness gives $x_1 = x_2$ and $f$ is injective as well.

For finite sets, an injection from $A$ to $B$ needs at least as many elements in $B$ as in $A$ , and a surjection needs at least as many in $A$ as in $B$ ; a bijection therefore forces the two to have the same number of elements. Comparing sets by functions rather than by counting is what will extend this to infinite sets, where counting is no longer available.

Example 4.37.

With $A = \{1, 2, 3, 4\}$ and $B = \{r, s, t, u\}$ as above, take $\beta \defeq \{(1, s), (2, u), (3, r), (4, t)\}$ . Each member of $A$ appears once as a first coordinate and each member of $B$ once as a second, so $\beta : A \to B$ is a bijection. The function $f$ of the previous section is neither injective nor surjective: $s$ is hit twice and $u$ is not hit at all.

Proposition 4.38 (An injection onto its range).

If $f : A \to B$ is injective, then the function from $A$ to $f(A)$ with the same ordered pairs is bijective.

Discussion.

The new function has the same ordered pairs as $f$ , so injectivity carries over unchanged, the definition mentioning only inputs and their images. For surjectivity, take $y$ in the new codomain $f(A)$ ; the definition of the image provides an $x \in A$ with $f(x) = y$ , which is the preimage required.

Proof.

Changing the codomain from $B$ to $f(A)$ does not alter the ordered pairs, and injectivity is a condition on those alone, so the new function is injective. Every $y \in f(A)$ is $f(x)$ for some $x \in A$ by the definition of the image, so it is also surjective.

Every translation is a bijection, by the example above, and so is the identity map $\mathrm{id}_A$ , since each $y \in A$ is its own unique preimage. The squaring map and the product map are not, each having already failed one of the two conditions.

Problem 4.17.

Let $f : A \to B$ be bijective and let $S \subset A$ . Prove that the restriction $f|_S : S \to f(S)$ is bijective.

Problem 4.18.

For each of the following, exhibit sets $A$ and $B$ , a subset $C \subset A$ and a function $f : A \to B$ meeting the condition, or show that none exists.

$f$ is surjective and $f|_C$ is surjective.
$f$ is surjective and $f|_C$ is not.
$f$ is injective and $f|_C$ is injective.
$f$ is injective and $f|_C$ is not.

Problem 4.19.

Let $f : A \to B$ be injective and let $S, T \subset A$ . Prove that $f(S \cap T) = f(S) \cap f(T)$ .

Problem 4.20.

Let $A$ , $B$ and $C$ be sets. Prove that

\bigl(x, (y, z)\bigr) \longmapsto \bigl((x, y), z\bigr)

is a bijection from $A \times (B \times C)$ onto $(A \times B) \times C$ , so that the two readings of a triple product agree up to that correspondence. Show also that the two sets are not in general equal, by exhibiting an element of one which is not an element of the other.

Composition

Definition 4.39 (Composition).

Let $f : A \to B$ and $g : B \to C$ be functions. Their composition is the function $g \circ f : A \to C$ given by $(g \circ f)(x) \defeq g(f(x))$ .

The notation is read from right to left: $f$ acts first and $g$ second. The codomain of $f$ must be the domain of $g$ , so that every value of $f$ can serve as an input of $g$ .

Example 4.40 (Informal).

For the squaring map $q$ and the translation $\tau_1$ above, $(q \circ \tau_1)(x) = (x + 1)^2$ while $(\tau_1 \circ q)(x) = x^2 + 1$ . At $x = 1$ these take the values $4$ and $2$ , so $q \circ \tau_1 \neq \tau_1 \circ q$ by Proposition 4.19 . Composition is not commutative.

Theorem 4.41 (Associativity of composition).

Let $f : A \to B$ , $g : B \to C$ and $h : C \to D$ be functions. Then $h \circ (g \circ f) = (h \circ g) \circ f$ .

Discussion.

By Proposition 4.19 we must check that the two functions have the same domain and agree at every input. Both are functions from $A$ to $D$ , so only the values are in question, and at an arbitrary $x \in A$ the definition of composition evaluates either side by applying $f$ , then $g$ , then $h$ .

Proof.

Both sides have domain $A$ . For every $x \in A$ , $\bigl(h \circ (g \circ f)\bigr)(x) = h\bigl(g(f(x))\bigr) = \bigl((h \circ g) \circ f\bigr)(x),$ so Proposition 4.19 gives the equality.

Associativity lets us write $h \circ g \circ f$ without brackets. The identity maps satisfy the identity laws one expects.

Proposition 4.42 (Identity laws).

If $f : A \to B$ , then $\mathrm{id}_B \circ f = f$ and $f \circ \mathrm{id}_A = f$ .

Discussion.

Again Proposition 4.19 reduces each identity to a comparison of domains and values. Both sides of each have domain $A$ , and at an arbitrary $x \in A$ the definitions of composition and of the identity map reduce both composites to $f(x)$ .

Proof.

For every $x \in A$ we have $(\mathrm{id}_B \circ f)(x) = \mathrm{id}_B(f(x)) = f(x)$ and $(f \circ \mathrm{id}_A)(x) = f(\mathrm{id}_A(x)) = f(x)$ . Each pair of functions has the same domain, so Proposition 4.19 applies.

Theorem 4.43 (Composition and bijections).

Let $f : A \to B$ and $g : B \to C$ be functions.

If $f$ and $g$ are injective, then $g \circ f$ is injective;
If $f$ and $g$ are surjective, then $g \circ f$ is surjective;
If $f$ and $g$ are bijective, then $g \circ f$ is bijective.

Discussion.

For the first part the definition of injectivity asks us to start from an equality of composite values; expanding the composition, injectivity of $g$ gives equality of the $f$ -values, and injectivity of $f$ then gives equality of the inputs. For the second, take an arbitrary $z \in C$ , choose $y \in B$ with $g(y) = z$ , then choose $x \in A$ with $f(x) = y$ , and evaluate the composite at $x$ . The third is the first two together with the definition of a bijection.

Proof.

Suppose $(g \circ f)(x_1) = (g \circ f)(x_2)$ , that is, $g(f(x_1)) = g(f(x_2))$ . Injectivity of $g$ gives $f(x_1) = f(x_2)$ , and injectivity of $f$ then gives $x_1 = x_2$ , which is the first part.

For the second, take $z \in C$ . Since $g$ is surjective, some $y \in B$ has $g(y) = z$ ; since $f$ is surjective, some $x \in A$ has $f(x) = y$ . Then $(g \circ f)(x) = g(f(x)) = g(y) = z$ . The third part follows from the other two.

Problem 4.21.

Let $f : A \to B$ and $g : B \to C$ . Prove that if $g \circ f$ is injective then $f$ is injective, and that if $g \circ f$ is surjective then $g$ is surjective.

Problem 4.22.

Construct finite sets $A$ , $B$ , $C$ and functions $f : A \to B$ , $g : B \to C$ for which $g \circ f$ is bijective although $f$ is not surjective and $g$ is not injective.

Inverse Functions

Definition 4.44 (Inverse function).

Let $f : A \to B$ and $g : B \to A$ be functions. The function $g$ is an inverse of $f$ if $g \circ f = \mathrm{id}_A$ and $f \circ g = \mathrm{id}_B$ . A function which has an inverse is called invertible.

The first equation returns each element of $A$ after applying $f$ and then $g$ ; the second does the same for $B$ . The problems below examine what each equation achieves on its own.

Theorem 4.45 (Invertibility and bijections).

A function is invertible if and only if it is bijective, and when an inverse exists it is unique.

Discussion.

Suppose first that $g$ is an inverse of $f$ . To get injectivity, apply $g$ to an equality $f(x_1) = f(x_2)$ and use $g \circ f = \mathrm{id}_A$ ; to get surjectivity, use $g(y)$ as a preimage of an arbitrary $y \in B$ and use $f \circ g = \mathrm{id}_B$ . Conversely, Proposition 4.36 gives exactly one reversed pair $(y, x)$ for each $y \in B$ , so the set of reversed pairs passes the vertical line test and is a function from $B$ to $A$ ; the two inverse identities are then those pairs read in the two directions. For uniqueness, two inverses are compared by sandwiching $f$ between them, where associativity and the identity laws collapse the composite in two ways.

Proof.

Let $f : A \to B$ have an inverse $g : B \to A$ . If $f(x_1) = f(x_2)$ , then $x_1 = g(f(x_1)) = g(f(x_2)) = x_2$ , so $f$ is injective. Given $y \in B$ , put $x = g(y)$ ; the second identity gives $f(x) = f(g(y)) = y$ , so $f$ is surjective.

Conversely, suppose $f$ is bijective and set $g \defeq \{(y, x) \in B \times A \mid (x, y) \in f\}$ . For each $y \in B$ , Proposition 4.36 gives a unique $x \in A$ with $f(x) = y$ , so $g$ is a function from $B$ to $A$ . Reading the pairs in each direction gives $g(f(x)) = x$ for $x \in A$ and $f(g(y)) = y$ for $y \in B$ , that is, $g \circ f = \mathrm{id}_A$ and $f \circ g = \mathrm{id}_B$ .

For uniqueness, let $g$ and $h$ both be inverses of $f$ . Associativity and the identity laws give

g = \mathrm{id}_A \circ g = (h \circ f) \circ g = h \circ (f \circ g) = h \circ \mathrm{id}_B = h.

The unique inverse of a bijection $f$ is written $f^{-1}$ . The construction in the proof gives

f^{-1} = \{(y, x) \in B \times A \mid (x, y) \in f\}, \qquad\text{so}\qquad f^{-1}(y) = x \iff f(x) = y.

Remark.

The symbol $f^{-1}$ also denotes the preimage of a set. The two uses are distinct: $f^{-1}(y)$ is an element of $A$ produced by the inverse function, and exists only for a bijection, while $f^{-1}(U)$ is a subset of $A$ and is defined for every function. For a bijection they are related by $f^{-1}(\{y\}) = \{f^{-1}(y)\}$ , the preimage of a point being the singleton of its inverse image.

Example 4.46.

Reversing the pairs of the bijection $\beta = \{(1, s), (2, u), (3, r), (4, t)\}$ gives $\beta^{-1} = \{(s, 1), (u, 2), (r, 3), (t, 4)\}$ . Also $\mathrm{id}_A^{-1} = \mathrm{id}_A$ for any set $A$ .

Example 4.47 (Informal).

The translations invert one another: $\tau_c^{-1} = \tau_{-c}$ , since $\tau_{-c}(\tau_c(x)) = x = \tau_c(\tau_{-c}(x))$ for every real number $x$ .

Proposition 4.48 (The inverse is a bijection).

If $f : A \to B$ is bijective, then $f^{-1} : B \to A$ is bijective and $(f^{-1})^{-1} = f$ .

Discussion.

The two equations defining an inverse are symmetric in $f$ and $f^{-1}$ , so they say equally that $f$ is an inverse of $f^{-1}$ . That makes $f^{-1}$ invertible, and the theorem above turns invertibility into bijectivity; its uniqueness clause then names $f$ as the inverse of $f^{-1}$ .

Proof.

The identities $f^{-1} \circ f = \mathrm{id}_A$ and $f \circ f^{-1} = \mathrm{id}_B$ also say that $f$ is an inverse of $f^{-1}$ . So $f^{-1}$ is invertible and hence bijective by Theorem 4.45 , and uniqueness of its inverse gives $(f^{-1})^{-1} = f$ .

Theorem 4.49 (Inverse of a composition).

If $f : A \to B$ and $g : B \to C$ are bijections, then $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$ .

Discussion.

Put $k \defeq f^{-1} \circ g^{-1}$ . Rather than compute $(g \circ f)^{-1}$ we show directly that $k$ meets the definition of an inverse of $g \circ f$ : evaluate $k \circ (g \circ f)$ at an arbitrary $x \in A$ and $(g \circ f) \circ k$ at an arbitrary $z \in C$ , where associativity and the inverse identities cancel the adjacent pairs. Since $g \circ f$ is bijective, uniqueness of inverses then identifies $k$ as the one.

Proof.

Put $k \defeq f^{-1} \circ g^{-1} : C \to A$ . For $x \in A$ and $z \in C$ ,

\bigl(k \circ (g \circ f)\bigr)(x) = f^{-1}\bigl(g^{-1}(g(f(x)))\bigr) = x, \qquad \bigl((g \circ f) \circ k\bigr)(z) = g\bigl(f(f^{-1}(g^{-1}(z)))\bigr) = z.

So $k$ is an inverse of $g \circ f$ , which is bijective by Theorem 4.43 , and uniqueness in Theorem 4.45 gives the formula.

Corollary 4.50 (Cancellation by a bijection).

Let $f : B \to C$ be bijective and let $g, h : A \to B$ . Then $f \circ g = f \circ h$ implies $g = h$ . Likewise, if $f : A \to B$ is bijective and $r, s : B \to C$ , then $r \circ f = s \circ f$ implies $r = s$ .

Proof.

For the first, compose on the left with $f^{-1}$ and use associativity:

\begin{aligned} g &= (f^{-1} \circ f) \circ g = f^{-1} \circ (f \circ g) \\ &= f^{-1} \circ (f \circ h) = (f^{-1} \circ f) \circ h = h. \end{aligned}

For the second, compose on the right with $f^{-1}$ :

\begin{aligned} r &= r \circ (f \circ f^{-1}) = (r \circ f) \circ f^{-1} \\ &= (s \circ f) \circ f^{-1} = s \circ (f \circ f^{-1}) = s. \end{aligned}

Problem 4.23 (Informal).

Let $a$ and $b$ be real numbers with $a \neq 0$ . Prove that $F : R \to R$ given by $F(x) \defeq ax + b$ is bijective, and determine $F^{-1}$ .

Problem 4.24.

Let $f : A \to B$ be bijective and let $S \subset A$ . Determine the inverse of $f|_S : S \to f(S)$ .

Problem 4.25.

Let $f : A \to B$ and $g : B \to A$ . Prove that $g \circ f = \mathrm{id}_A$ forces $f$ to be injective and $g$ to be surjective, while $f \circ g = \mathrm{id}_B$ forces $f$ to be surjective and $g$ to be injective.

Left and Right Inverses

Definition 4.51 (Left and right inverses).

Let $f : A \to B$ and $g : B \to A$ be functions. The function $g$ is a left inverse of $f$ if $g \circ f = \mathrm{id}_A$ , and a right inverse of $f$ if $f \circ g = \mathrm{id}_B$ . When it is both, it is an inverse in the earlier sense.

The names record the side on which $g$ sits in the composite. The last problem above gave two implications: a left inverse makes $f$ injective, and a right inverse makes $f$ surjective. The next theorem combines them.

Theorem 4.52 (Bijections via one-sided inverses).

Let $f : A \to B$ be a function. Then $f$ is bijective if and only if it has both a left inverse and a right inverse; when both exist they coincide, and their common value is $f^{-1}$ .

Discussion.

One direction is immediate: a bijection has $f^{-1}$ , which is a left and a right inverse by definition. For the other, the left inverse gives injectivity and the right inverse gives surjectivity, so $f$ is bijective and $f^{-1}$ exists. To identify the one-sided inverses with each other we evaluate the triple composite $g_l \circ f \circ g_r$ in its two groupings: associativity and the identity laws collapse the middle pair either way, leaving $g_l = g_r$ . The same calculation with $f^{-1}$ in place of either names their common value.

Proof.

Suppose $f$ has a left inverse $g_l$ and a right inverse $g_r$ . If $f(x_1) = f(x_2)$ , then $x_1 = g_l(f(x_1)) = g_l(f(x_2)) = x_2$ , so $f$ is injective. For $y \in B$ , the element $g_r(y) \in A$ satisfies $f(g_r(y)) = y$ , so $f$ is surjective. Hence $f$ is bijective and $f^{-1}$ exists by Theorem 4.45 . Associativity and the identity laws give

\begin{aligned} g_l &= g_l \circ \mathrm{id}_B = g_l \circ (f \circ g_r) \\ &= (g_l \circ f) \circ g_r = \mathrm{id}_A \circ g_r = g_r, \end{aligned}

so the two agree; and since $f^{-1}$ is itself both a left and a right inverse of $f$ , the same computation identifies their common value with $f^{-1}$ .

Conversely, if $f$ is bijective then $f^{-1}$ is a left and a right inverse by definition.

Theorem 4.53 (Injections and left inverses).

Let $A$ be non-empty. A function $f : A \to B$ is injective if and only if it has a left inverse.

Discussion.

If a left inverse exists, injectivity is the implication already noted. For the converse, an injective $f$ is a bijection onto its range by Proposition 4.38 , so it has an inverse there. On the range a left inverse is forced to be that one, since $g(f(x)) = x$ determines $g$ at every point of $f(A)$ ; on the rest of $B$ we are free, and non-emptiness of $A$ lets us send all of $B \setminus f(A)$ to one fixed element. That freedom is why left inverses are rarely unique. The hypothesis on $A$ cannot be dropped: the function from $\emptyset$ to $B$ is injective, but a function from $B$ to $\emptyset$ exists only when $B$ is empty.

Proof.

Suppose first that $g \circ f = \mathrm{id}_A$ . If $f(x_1) = f(x_2)$ , then $x_1 = g(f(x_1)) = g(f(x_2)) = x_2$ , so $f$ is injective.

Conversely, assume $f$ is injective. By Proposition 4.38 the function from $A$ to $f(A)$ with the same pairs is bijective; let $h : f(A) \to A$ be its inverse, which exists by Theorem 4.45 . Fix $x_0 \in A$ , which we may do since $A$ is non-empty, and define $g : B \to A$ by $g(y) \defeq h(y)$ for $y \in f(A)$ and $g(y) \defeq x_0$ for $y \in B \setminus f(A)$ . For every $x \in A$ the point $f(x)$ lies in $f(A)$ and $h(f(x)) = x$ , so $g(f(x)) = x$ and $g \circ f = \mathrm{id}_A$ .

For surjections the situation is different. A right inverse forces surjectivity, as we have seen; the converse, that every surjection has a right inverse, is a claim of a different kind. A right inverse must choose, for every $y \in B$ at once, one element of $f^{-1}(\{y\})$ , and when $B$ is infinite nothing among our axioms says such a simultaneous choice can be made. That is what the axiom of choice provides.

Problem 4.26.

Let $A$ be non-empty and $f : A \to B$ injective. Show that any two left inverses of $f$ agree on $f(A)$ . For the inclusion map from $\{1, 2\}$ to $\{1, 2, 3\}$ , exhibit two left inverses and say where they differ.

General Cartesian Products

With functions in hand we can take the product of an arbitrary collection of sets, not just two. First, the informal indexed family of the earlier section can now be said properly.

Remark.

An indexed family $\{A_\alpha\}_{\alpha \in I}$ of subsets of a set $A$ is formally a function $\varphi : I \to \mathcal{P}(A)$ , with $A_\alpha \defeq \varphi(\alpha)$ . It is the function itself, not its range: the range is the plain set of members, which forgets the labels, as in the earlier example, while $\varphi$ remembers which member sits at which index. Since $\varphi$ is a subset of $I \times \mathcal{P}(A)$ , it is a set by the axioms of the last chapter, and that is why we ask that all the $A_\alpha$ be subsets of one set $A$ .

Definition 4.54 (General Cartesian product).

Let $I$ be a set and let $\{A_\alpha\}_{\alpha \in I}$ be an indexed family of sets, all of them subsets of a given set. The Cartesian product of the family is

\prod_{\alpha \in I} A_\alpha \defeq \Bigl\{ f \in \mathcal{P}\Bigl(I \times \bigcup_{\alpha \in I} A_\alpha\Bigr) \;\Big|\; f \text{ is a function},\ \operatorname{dom} f = I,\ \forall \alpha \in I\,\bigl(f(\alpha) \in A_\alpha\bigr) \Bigr\}.

The notation is easier to read once we recall that a function from $I$ to $\bigcup_{\alpha \in I} A_\alpha$ is by Definition 4.15 a subset of $I \times \bigcup_{\alpha \in I} A_\alpha$ . So every candidate lies in the power set displayed above, a set we are already holding, and comprehension carves the product out of it. An element of the product picks one member from each $A_\alpha$ , all at once.

Example 4.55.

The definition subsumes the earlier one. A function $f$ defined on the two-element set $\{0, 1\}$ is determined by the pair of values $\bigl(f(0), f(1)\bigr)$ , so the functions with $f(0) \in A$ and $f(1) \in B$ correspond to the pairs $(a, b)$ with $a \in A$ and $b \in B$ . This identifies $A \times B$ with the general product for $I = \{0, 1\}$ , $A_0 = A$ and $A_1 = B$ , and we will not distinguish between the two readings from here on.

Definition 4.56 (Cartesian power).

Let $I$ and $A$ be sets. The Cartesian power $A^I$ is the product of the constant family,

A^I \defeq \prod_{\alpha \in I} A = \{f \in \mathcal{P}(I \times A) \mid f \text{ is a function with } \operatorname{dom} f = I\}.

Problem 4.27.

Let $A$ , $B$ and $C$ be sets, with $B^A$ the Cartesian power of Definition 4.56 .

Prove that if $A \subset B$ , then $A^C \subset B^C$ .
Prove that $\emptyset^A = \emptyset$ for every non-empty $A$ , and that $B^{\emptyset} = \{\emptyset\}$ for every $B$ .
Prove that $A^B \cap A^C = \emptyset$ whenever $B \neq C$ , and deduce that $\{1, 2\}^{\{1,2\}}$ and $\{1, 2\}^{\{1,2,3\}}$ are disjoint.

Problem 4.28.

Let $S$ be a set. For each $A \subset S$ , the characteristic function of $A$ is the function $\chi_A : S \to \{0, 1\}$ with $\chi_A(x) \defeq 1$ for $x \in A$ and $\chi_A(x) \defeq 0$ for $x \notin A$ ; so $\chi_A(1) = \chi_A(2) = \chi_A(4) = 1$ and $\chi_A(3) = 0$ when $S = \{1, 2, 3, 4\}$ and $A = \{1, 2, 4\}$ .

Give the formulas for $\chi_{\emptyset}$ and $\chi_S$ .
Express $\chi_{\overline{A}}$ in terms of $\chi_A$ , where $\overline{A}$ is the complement of $A$ in $S$ .
Prove that $A = B$ if and only if $\chi_A = \chi_B$ , for all $A, B \subset S$ .
Prove that $\chi_{A \cap B}(x) = \chi_A(x)\,\chi_B(x)$ and $\chi_{A \cup B}(x) = \chi_A(x) + \chi_B(x) - \chi_A(x)\,\chi_B(x)$ for every $x \in S$ .

Problem 4.29.

Keep the characteristic functions of the previous problem.

Prove that every function $u : S \to \{0, 1\}$ is $\chi_A$ for exactly one $A \subset S$ .
Suppose $S$ has $n$ elements. Determine the number of elements of $\{0, 1\}^S$ , and use the first part to count $\mathcal{P}(S)$ .

Problem 4.30.

Let $S$ be a set.

Prove that no function $g : S \to \mathcal{P}(S)$ is surjective.
Deduce that there is no bijection from $S$ to $\mathcal{P}(S)$ , and decide whether there can be a surjection from $\mathcal{P}(S)$ to $S$ .

The product of two non-empty sets is non-empty, witnessed by any pair $(x, y)$ with $x \in A$ and $y \in B$ , and the same argument covers three sets, four, and any collection we can write out in full. Once $I$ is infinite the argument fails for the reason we met in the last chapter: we have no way to string an unspecified number of such choices together. Closing that gap takes another axiom.

The Axiom of Choice

Axiom 4.57 (Choice).

Let $I$ be a non-empty set and let $\{A_\alpha\}_{\alpha \in I}$ be an indexed family of sets with $A_\alpha \neq \emptyset$ for every $\alpha \in I$ . Then there exists a function $f : I \to \bigcup_{\alpha \in I} A_\alpha$ with $\operatorname{dom} f = I$ and $f(\alpha) \in A_\alpha$ for every $\alpha \in I$ . Equivalently, a non-empty family of non-empty sets has non-empty product.

Such an $f$ is a choice function for the family. The point is the order of the quantifiers: the hypothesis grants each set an element of its own, while the conclusion assembles one selection from every set into a single function.

Remark.

The axiom is independent of the others, in the sense that neither it nor its negation follows from them. In sets carrying some structure a distinguished element can often be picked out constructively, and no axiom is needed; the axiom supplies one in full generality, where no rule for choosing is available. Mathematicians find it rather less comfortable than the rest, so it is good practice, which we will follow, to say plainly whenever it is used.

Theorem 4.58 (Every surjection has a right inverse).

Let $f : A \to B$ be surjective. Then $f$ has a right inverse $g : B \to A$ .

Discussion.

Surjectivity says, through Proposition 4.31 , that every point preimage $f^{-1}(\{y\})$ is non-empty. That is exactly the hypothesis of the axiom of choice applied to the family of point preimages indexed by $B$ , and what the axiom returns is a function $\varphi$ picking one element out of each. The union of the point preimages is $A$ , and $\varphi(y) \in f^{-1}(\{y\})$ says precisely that $f(\varphi(y)) = y$ , which is the right-inverse identity. The axiom is needed here: each preimage alone has an element, and the theorem needs one choice per preimage assembled into a single map.

Proof.

If $B$ is empty then so is $A$ , since every element of $A$ would have an image in $B$ , and the empty function is a right inverse; so assume $B$ is non-empty. For every $y \in B$ the set $f^{-1}(\{y\})$ is non-empty by Proposition 4.31 . Applying the axiom of choice to the family $\{f^{-1}(\{y\})\}_{y \in B}$ gives a function $\varphi$ from $B$ to $\bigcup_{y \in B} f^{-1}(\{y\})$ with $\varphi(y) \in f^{-1}(\{y\})$ for every $y \in B$ . Each point preimage is a subset of $A$ , so that union is $A$ and $\varphi : B \to A$ . Finally $\varphi(y) \in f^{-1}(\{y\})$ is equivalent to $f(\varphi(y)) = y$ , so $f \circ \varphi = \mathrm{id}_B$ and $\varphi$ is a right inverse of $f$ .

Corollary 4.59 (Surjections and right inverses).

A function $f : A \to B$ is surjective if and only if it has a right inverse.

Proof.

One direction is the theorem. For the other, a right inverse $g$ gives $f(g(y)) = y$ for every $y \in B$ , so $g(y)$ is a preimage of $y$ and $f$ is surjective.

Problem 4.31.

Let $I = \{1, 2, 3\}$ and let $\{M_i\}_{i \in I}$ be a family of non-empty sets. Show that $\prod_{i \in I} M_i$ is non-empty without appealing to the axiom of choice. Where would the argument break down for an index set that cannot be written out in full?

Exercises

Exercises in Lean

The proofs below are checked in your browser. Nothing is sent anywhere, and nothing is stored but your own work. Type \to for →, \and for ∧, \< and \> for ⟨ ⟩.

The last sheet gave the checker a universe of objects and the algebra of sets. This one adds the constructions of this chapter: the ordered pair, the product, the indexed family and the map.

A pair is typed as it is written, (a, b), and longer tuples nest to the left exactly as Definition 4.4 says, so (a, b, c) is ((a, b), c). The product is ×, typed \x, and it binds like ∩. A map is written f : Obj → Obj and applied by juxtaposition, f x; the domain and codomain of Definition 4.15 are carried by the statement being proved rather than by the arrow.

One piece of the last chapter is written out here for the first time as well: a set listed by its members, {x} and {x, y}. Its membership criterion is the equation the pairing axiom gives, so y ∈ {x} is y = x, and the later exercises use it to name a set with a single point in it.

Ordered pairs

The pair itself is opaque: the checker knows nothing about $\bigl\{\{a\}, \{a, b\}\bigr\}$ , only what Proposition 4.2 established about it. That proposition is Set.pair_eq, a biconditional, so .mp reads coordinates off an equality of pairs and .mpr builds one.

Example.

The forward direction turns the equality into a conjunction; the first half of it is the first coordinate.

Exercise 4.12.

Corollary 4.3 , in one direction.

Exercise 4.13.

Coordinates read off one pair and put back into another.

Exercise 4.14.

A triple is a pair of a pair, so its middle coordinate takes two steps to reach.

Products

Membership in a product is the dictionary line of the chapter: (x, y) ∈ A × B is x ∈ A ∧ y ∈ B, in the way that x ∈ A ∩ B was a conjunction on the last sheet, so ⟨_, _⟩ builds one and .left and .right take one apart.

Example.

Two memberships, one pair.

An arbitrary object is not written as a pair, and for it the criterion says the rest of Definition 4.6 : p ∈ A × B is

∃ a, ∃ b, p = (a, b) ∧ a ∈ A ∧ b ∈ B

so a membership hypothesis about an unknown $p$ hands over two coordinates, an equation and two memberships. obtain takes all five at once, and the equation is what rw then uses to turn $p$ into a pair everywhere it is needed.

Example.

Once $p$ has been rewritten, the goal is about a pair and the criterion applies to it.

Exercise 4.15.

The coordinates change places and so do the factors.

Exercise 4.16.

Nothing can serve as a second coordinate.

Exercise 4.17.

The product respects inclusion in each factor.

Exercise 4.18.

Products distribute over intersections.

Exercise 4.19.

And over unions.

Exercise 4.20.

An intersection of products is a product of intersections.

Exercise 4.21.

The inclusion that never fails.

Exercise 4.22.

The equality the last exercise fell short of, with the two mixed products restored.

Indexed families

A family $\{B_i\}_{i \in I}$ is a map from indices to sets, written B : Obj → Obj, and its union and intersection are ⋃ i ∈ I, B i and ⋂ i ∈ I, B i, typed \bigcup and \bigcap. Their criteria are the two the chapter gave: membership in the union is ∃ i, i ∈ I ∧ x ∈ B i, and membership in the intersection is ∀ i, i ∈ I → x ∈ B i. So an intersection is used by applying it to an index and a proof that the index belongs to $I$ , and a union is built with use.

Example.

The intersection applied at one index.

Example.

use supplies the index; what remains is that the index is one of ours and that $x$ lies in its set.

An object outside an intersection misses some set, but we are not given its index: all we hold is that no index can be a witness. Getting the index out is the classical step, and by_contra twice is what does it.

Example.

The first by_contra denies the index we want; the second turns that denial into membership at every index, which contradicts the hypothesis.

Exercise 4.23.

A set lies inside the intersection exactly when it lies inside every member.

Exercise 4.24.

And the union lies inside a set exactly when every member does.

Exercise 4.25.

The first part of Proposition 4.13 .

Exercise 4.26.

The second part. The hypothesis hj is the non-emptiness of $I$ , and one of the two inclusions cannot be had without it.

Exercise 4.27.

The inclusion between the two mixed families, in the direction that holds.

Maps

For a map f : Obj → Obj, the image and preimage of a set are f '' S and f ⁻¹' U, typed as two apostrophes and as \preimage. Their criteria are those of Definition 4.23 and Definition 4.24 : y ∈ f '' S is ∃ x, x ∈ S ∧ f x = y, and x ∈ f ⁻¹' U is f x ∈ U. Composition is g ∘ f, typed \o, and (g ∘ f) x is g (f x). Finally Injective f and Surjective f are the two conditions of Definition 4.30 and Definition 4.28 , ∀ x y, f x = f y → x = y and ∀ y, ∃ x, f x = y, so both are opened with intro.

Example.

The witness is the point itself, and the equation it has to satisfy is an identity.

The preimage passes membership straight through to $f x$ , which is why it preserves the operations: both sides of each law are the same statement about $f x$ .

Example.

Part three of Proposition 4.27 , which follows from the criterion.

Exercise 4.28.

Images preserve unions.

Exercise 4.29.

Every point of $S$ is sent into the image of $S$ .

Exercise 4.30.

And the image of a preimage is no bigger than the set it came from.

Exercise 4.31.

The first part of Theorem 4.43 .

Exercise 4.32.

Its second part.

Exercise 4.33.

Only the first map need be injective for the composite to be.

Exercise 4.34.

And only the second need be surjective.

Exercise 4.35.

A left inverse, written out pointwise, makes $f$ injective.

Exercise 4.36.

A right inverse makes it surjective.

Exercise 4.37.

Images preserve intersections exactly when the map is injective.

Exercise 4.38.

The inclusion two exercises above becomes an equality for an injective map.

Exercise 4.39.

And the other inclusion becomes an equality for a surjective one.

Exercise 4.40.

An image loses no more than the part removed.

Exercise 4.41.

Injectivity closes the gap.

Exercise 4.42.

The converse of the exercise three above: only an injection returns every set unchanged.

Exercise 4.43.

And nothing but a surjection.

A map $f : A \to B$ carries two maps between the power sets with it, $F : \mathcal{P}(A) \to \mathcal{P}(B)$ sending $S$ to $f(S)$ and $G : \mathcal{P}(B) \to \mathcal{P}(A)$ sending $U$ to $f^{-1}(U)$ . The checker has no power set to quantify over, so each of the four statements below says what injectivity or surjectivity of $F$ or of $G$ amounts to at the level of the sets themselves.

Exercise 4.44.

$f$ is injective exactly when $F$ is.

Exercise 4.45.

And surjective exactly when $F$ is.

Exercise 4.46.

The other pairing: $f$ is injective exactly when $G$ is surjective.

Exercise 4.47.

And surjective exactly when $G$ is injective.

What the checker understands

Tactics

intro h	assume the hypothesis of an implication, naming it h
exact e	give the proof outright
apply f	reduce the goal to the hypotheses of f
assumption	close the goal with a hypothesis already present
trivial	close the goal True
exfalso	replace the goal with False
by_contra h	assume the negation of the goal
constructor	split ∧ into both halves, or ↔ into both directions
left / right	choose which half of a ∨ to prove
rcases h with a \| b	argue by cases on a disjunction
obtain ⟨a, b⟩ := h	take a conjunction or an existential apart
cases h	as above, keeping the name
refine e	give the proof with holes left in it
have h : p := …	record an intermediate result
show p	restate the goal in an equal form
use w	give a witness for ∃
specialize h a	instantiate a ∀ hypothesis
rw [h]	rewrite with an equation, ← to go backwards
rfl	both sides compute to the same thing
decide / norm_num	settle a closed computation
tauto	close a goal that is true by pure logic

Results you may cite

Classical.em	∀ (a : Prop), a ∨ ¬a — the law of excluded middle
Classical.byContradiction	∀ {a : Prop}, (¬a → False) → a — proof by contradiction; the tactic by_contra does this for you
Classical.byCases	∀ {a b : Prop}, (a → b) → (¬a → b) → b — split on whether a holds
not_not	∀ {a : Prop}, ¬¬a ↔ a — double negation
not_and_or	∀ {a b : Prop}, ¬(a ∧ b) ↔ ¬a ∨ ¬b — De Morgan
not_or	∀ {a b : Prop}, ¬(a ∨ b) ↔ ¬a ∧ ¬b — De Morgan
not_imp	∀ {a b : Prop}, ¬(a → b) ↔ a ∧ ¬b
and_comm	∀ {a b : Prop}, a ∧ b ↔ b ∧ a
or_comm	∀ {a b : Prop}, a ∨ b ↔ b ∨ a
Set.ext	∀ {A B : Obj}, (∀ x : Obj, x ∈ A ↔ x ∈ B) → A = B — extensionality: sets with the same elements are equal
Set.ext_iff	∀ {A B : Obj}, A = B ↔ (∀ x : Obj, x ∈ A ↔ x ∈ B) — extensionality and substitution, in one biconditional
Set.subset_antisymm	∀ {A B : Obj}, A ⊆ B → B ⊆ A → A = B — mutual inclusion is equality
Set.empty_subset	∀ {A : Obj}, ∅ ⊆ A — the empty set is a subset of every set
Set.pair_eq	∀ {a b c d : Obj}, ((a, b) = (c, d)) ↔ (a = c ∧ b = d) — two ordered pairs are equal exactly when their coordinates are

From the logical core

And.intro	∀ {a b : Prop}, a → b → a ∧ b
And.left	∀ {a b : Prop}, a ∧ b → a
And.right	∀ {a b : Prop}, a ∧ b → b
And.symm	∀ {a b : Prop}, a ∧ b → b ∧ a
Or.inl	∀ {a b : Prop}, a → a ∨ b
Or.inr	∀ {a b : Prop}, b → a ∨ b
Or.elim	∀ {a b c : Prop}, a ∨ b → (a → c) → (b → c) → c
Or.symm	∀ {a b : Prop}, a ∨ b → b ∨ a
Iff.intro	∀ {a b : Prop}, (a → b) → (b → a) → (a ↔ b)
Iff.mp	∀ {a b : Prop}, (a ↔ b) → a → b
Iff.mpr	∀ {a b : Prop}, (a ↔ b) → b → a
Iff.symm	∀ {a b : Prop}, (a ↔ b) → (b ↔ a)
Iff.rfl	∀ {a : Prop}, a ↔ a
Iff.trans	∀ {a b c : Prop}, (a ↔ b) → (b ↔ c) → (a ↔ c)
True.intro	True
False.elim	∀ {a : Prop}, False → a
absurd	∀ {a b : Prop}, a → ¬a → b
id	∀ {a : Prop}, a → a
mt	∀ {a b : Prop}, (a → b) → ¬b → ¬a
Eq.refl	∀ {α : Type} (a : α), a = a
Eq.symm	∀ {α : Type} {a b : α}, a = b → b = a
Eq.trans	∀ {α : Type} {a b c : α}, a = b → b = c → a = c
congrArg	∀ {α : Type} {β : Type} {a b : α} (f : α → β), a = b → f a = f b
Exists.intro	∀ {α : Type} {p : α → Prop} (w : α), p w → ∃ x : α, p x
Exists.elim	∀ {α : Type} {p : α → Prop} {b : Prop}, (∃ x : α, p x) → (∀ y : α, p y → b) → b

Lesson 6

Relations and Order

Taught 16 Aug 2026

Relations and Order

People classify: we decide which things belong together, and mathematics does this on every page. This chapter covers relations in general, then the two most important kinds, orders and equivalences, and finally the way an equivalence cuts a set into pieces.

Relations

A relation records, for each pair of elements, whether the two stand in a given relationship. Since a pair with an order to it is an ordered pair, and the set of all of them is a Cartesian product, no new definitions are needed.

Definition 6.1 (Binary relation).

A relation $R$ on a set $S$ is a set of ordered pairs of elements of $S$ , that is, a subset $R \subset S \times S$ . We abbreviate $(x, y) \in R$ to $x \mathbin{R} y$ , and say $x$ is related to $y$ by $R$ ; if $(x, y) \notin R$ we write $x \not\mathbin{R} y$ .

Putting the symbol between the two elements is infix notation; writing $R(x, y)$ instead is prefix notation. For equivalence relations we usually take a symbol such as $\equiv$ or $\sim$ , and for strict orders the symbol $<$ .

Two relations are distinguished exactly by the pairs belonging to one and not the other, since a relation is its set of pairs. At the two extremes, $S \times S$ is the universal relation on $S$ and $\emptyset$ is the empty relation. Neither is of much interest: if everything is related to everything then being related is no mark of distinction, and if nothing is related to anything there is nothing to say.

Definition 6.2 (Reflexive, symmetric, transitive).

Let $R$ be a relation on a set $S$ . Then $R$ is

reflexive if $x \mathbin{R} x$ for every $x \in S$ ;
symmetric if $x \mathbin{R} y \implies y \mathbin{R} x$ for all $x, y \in S$ ;
transitive if $(x \mathbin{R} y \land y \mathbin{R} z) \implies x \mathbin{R} z$ for all $x, y, z \in S$ .

Example 6.3.

From a familiar children’s game, let $S = \{\text{rock}, \text{scissors}, \text{paper}\}$ and let $B$ , read “beats”, be

B = \bigl\{(\text{rock}, \text{scissors}), (\text{scissors}, \text{paper}), (\text{paper}, \text{rock})\bigr\}.

This $B$ is none of the three: rock does not beat itself, rock beats scissors while scissors does not beat rock, and rock beats scissors and scissors beats paper while rock does not beat paper.

Definition 6.4 (Equivalence relation).

A relation is an equivalence relation if it is reflexive, symmetric and transitive. When $\sim$ is an equivalence relation and $x \sim y$ , we say $x$ and $y$ are equivalent with respect to $\sim$ .

Example 6.5.

On $S = \{1, 2, 3, 4\}$ the relation $R = \{(1,2), (2,1), (1,1), (2,2)\}$ is symmetric and transitive but not reflexive, since $3 \not\mathbin{R} 3$ . Adjoining exactly what is missing gives $R_1 \defeq R \cup \{(3,3), (4,4)\}$ , which is reflexive, inherits symmetry and transitivity, and is therefore an equivalence relation. We call $R_1$ the reflexive closure of $R$ : it adds to $R$ only the pairs that reflexivity requires.

Example 6.6.

Transitivity is the condition that most often fails. Let $S$ be the partially exposed stones in a stream you wish to cross, and for $x, y \in S$ let $x \sim y$ mean that you can step directly from $x$ to $y$ without getting wet. This is reflexive, since once on a stone you can hop on the spot, and symmetric, since a step you can take one way you can take back. It is not generally transitive: that you can reach $y$ from $x$ and $z$ from $y$ says nothing about reaching $z$ from $x$ in one step.

Problem 6.1.

Check that the relation $R = \{(1,2), (2,1), (1,1), (2,2)\}$ above is symmetric and transitive, working from the definitions rather than from the picture.

Problem 6.2.

Let $A$ and $B$ be sets. Many texts call a subset $R \subset A \times B$ a relation from $A$ to $B$ . Show that every such $R$ is a relation on $A \cup B$ in the sense of Definition 6.1 .

Problem 6.3.

Let $R$ be a symmetric and transitive relation on a set $S$ , and put $T \defeq \{x \in S \mid x \mathbin{R} x\}$ . Show that $x \mathbin{R} y$ implies $x, y \in T$ , and that $R$ is an equivalence relation on $T$ .

Order Relations

Orders are the second of the two kinds of relation we shall need. We isolate the properties an order has, so that when we later build one on the natural numbers out of addition alone we know exactly what has to be checked.

Definition 6.7 (Strict linear order).

A strict linear order on a set $S$ is a relation, commonly written $<$ , satisfying

the transitivity law: for all $x, y, z \in S$ , if $x < y$ and $y < z$ then $x < z$ ;
the trichotomy law: for all $x, y \in S$ , exactly one of $x < y$ , $x = y$ , $y < x$ holds.

When $x < y$ holds we say $x$ is strictly less than $y$ . Strict linear orders are also called strict total orders, in contrast with the partial orders of the next section. The word exactly in the trichotomy law says two things: any two elements can be compared, and no two of the alternatives hold at once. Proofs use the second far more often than the first, so we isolate it.

Proposition 6.8 (Anti-reflexivity).

Let $<$ be a strict linear order on a set $S$ . Then $\neg(x < x)$ for every $x \in S$ .

Discussion.

The statement is a universally quantified negation, so we fix an arbitrary $x$ and rule out $x < x$ . Only the trichotomy law says anything about a single pair, so we apply it to the pair $x$ and $x$ . It offers three alternatives, $x < x$ , $x = x$ and $x < x$ , and exactly one of them holds. The middle one holds, since equality is reflexive, which rules out the other two; and each of those is the statement we wish to deny.

Proof.

Let $x \in S$ . Trichotomy applied to $x$ and $x$ says that exactly one of $x < x$ , $x = x$ , $x < x$ holds. Since $x = x$ holds, neither occurrence of $x < x$ can, so $\neg(x < x)$ .

Trichotomy can be traded for the weaker demand that at least one alternative hold, provided we also assume anti-reflexivity.

Proposition 6.9 (Characterising strict linear orders).

Let $<$ be a relation on a set $S$ such that

for all $x, y, z \in S$ , if $x < y$ and $y < z$ then $x < z$ ;
for all $x, y \in S$ , $(x < y) \lor (x = y) \lor (y < x)$ ;
for all $x \in S$ , $\neg(x < x)$ .

Then $<$ is a strict linear order on $S$ .

Discussion.

The definition asks for two laws, and the first is the first hypothesis verbatim, so only trichotomy needs proof. It splits in two: that at least one alternative holds is the second hypothesis, again verbatim, and it remains to show that no two hold together. There are three possible pairs, the two strict comparisons together or either strict comparison alongside $x = y$ , and we take them in turn. Two opposite strict comparisons feed into transitivity and produce $x < x$ ; a strict comparison alongside an equality lets us substitute one letter for the other and produce $x < x$ again. Each case gives $x < x$ , which the third hypothesis forbids.

Proof.

The first hypothesis is the transitivity law. For trichotomy, let $x, y \in S$ . The second hypothesis gives at least one of $x < y$ , $x = y$ , $y < x$ .

If both $x < y$ and $y < x$ held, transitivity would give $x < x$ , against the third hypothesis. If $x < y$ and $x = y$ held, substitution gives $x < x$ ; and $y < x$ with $x = y$ gives $x < x$ likewise. So no two of the alternatives hold together, exactly one holds, and that is the trichotomy law.

A strict order says which of two elements comes first. For bounds, minima and induction it is more convenient to use the relation that also allows equality.

Definition 6.10 (Associated non-strict order).

Let $<$ be a strict linear order on a set $S$ . The associated non-strict order $\leqslant$ is defined by: $x \leqslant y$ holds if and only if $(x < y) \lor (x = y)$ holds. When $x \leqslant y$ we say $x$ is less than or equal to $y$ .

Proposition 6.11 (Laws of the associated order).

Let $<$ be a strict linear order on $S$ with associated $\leqslant$ . For all $x, y \in S$ :

$(x \leqslant y) \lor (y \leqslant x)$ ;
if $x \leqslant y$ and $y \leqslant x$ then $x = y$ ;
$\neg(x \leqslant y)$ if and only if $y < x$ ;
$\neg(x < y)$ if and only if $y \leqslant x$ .

Discussion.

All four come from the list of three alternatives in trichotomy, so in each we go through the list and see which entries remain. The first is a disjunction and every entry produces one of its halves: $x < y$ gives $x \leqslant y$ , $y < x$ gives $y \leqslant x$ , and $x = y$ gives both. The second is a conditional whose conclusion is an equality, so we suppose $x \neq y$ and derive a contradiction: the definition turns both hypotheses into strict comparisons, transitivity gives $x < x$ , and anti-reflexivity forbids it. The last two are biconditionals, proved by crossing entries off the list. Denying $x \leqslant y$ deletes the first two entries and leaves $y < x$ ; conversely $y < x$ deletes those same two, since trichotomy allows only one. Denying $x < y$ deletes just the first entry, leaving $x = y$ or $y < x$ , and that pair is what $y \leqslant x$ abbreviates.

Proof.

For the first, trichotomy gives $x < y$ , $x = y$ or $y < x$ , and in each case the definition gives $x \leqslant y$ or $y \leqslant x$ .

For the second, suppose $x \leqslant y$ and $y \leqslant x$ , and suppose $x \neq y$ . Then the definition gives $x < y$ and $y < x$ , so transitivity gives $x < x$ , contradicting anti-reflexivity. Hence $x = y$ .

For the third, if $\neg(x \leqslant y)$ then neither $x < y$ nor $x = y$ holds, so trichotomy leaves $y < x$ . Conversely $y < x$ excludes both $x < y$ and $x = y$ , so $\neg(x \leqslant y)$ .

For the fourth, trichotomy says $\neg(x < y)$ holds precisely when $x = y$ or $y < x$ , which by the definition is precisely $y \leqslant x$ .

From transitivity of $<$ we get two further transitivity laws, one for each way of mixing the two relations.

Proposition 6.12 (Mixed transitivity).

Let $<$ be a strict linear order on $S$ with associated $\leqslant$ , and let $x, y, z \in S$ . Then

if $x < y$ and $y \leqslant z$ then $x < z$ ;
if $x \leqslant y$ and $y < z$ then $x < z$ .

Discussion.

Each part has one strict hypothesis and one non-strict one, and the definition of $\leqslant$ splits the non-strict hypothesis into two cases, so each part has two cases. In the equality case no order reasoning is needed at all: the conclusion is the strict hypothesis with one letter renamed. In the strict case both hypotheses are strict, so the transitivity law applies.

Proof.

For the first, if $y = z$ then $x < z$ is the hypothesis $x < y$ with $z$ written for $y$ ; and if $y < z$ then transitivity gives $x < z$ .

For the second, if $x = y$ then $x < z$ is the hypothesis $y < z$ with $x$ written for $y$ ; and if $x < y$ then transitivity again gives $x < z$ .

Corollary 6.13 (Transitivity of the non-strict order).

Let $<$ be a strict linear order on $S$ with associated $\leqslant$ . If $x \leqslant y$ and $y \leqslant z$ then $x \leqslant z$ .

Proof.

If $x = y$ and $y = z$ then $x = z$ . If $x = y$ and $y < z$ then $x < z$ ; if $x < y$ and $y = z$ then $x < z$ ; and if $x < y$ and $y < z$ then mixed transitivity gives $x < z$ . In every case the definition of $\leqslant$ gives $x \leqslant z$ .

Definition 6.14 (Reverse orders).

Let $<$ be a strict linear order on $S$ . We write $x > y$ to mean $y < x$ , and say $x$ is strictly greater than $y$ ; and $x \geqslant y$ to mean $(x > y) \lor (x = y)$ , and say $x$ is greater than or equal to $y$ .

Since $<$ and $\leqslant$ are transitive, so are $>$ and $\geqslant$ .

Definition 6.15 (Chains).

The notation $a < b < c$ abbreviates $(a < b) \land (b < c)$ , and similarly for $>$ , $\leqslant$ and $\geqslant$ , for more than three terms, and for mixtures: $a \leqslant b < c < d$ abbreviates $(a \leqslant b) \land (b < c) \land (c < d)$ .

By transitivity, $a < b < c$ gives $a < c$ .

Problem 6.4.

What is the difference between the trichotomy law and the weak trichotomy law? Which of the two implies the other?

Problem 6.5.

Show that a strict linear order is not symmetric on any set with at least two elements, and decide whether the associated $\leqslant$ can ever be symmetric.

Partial Orders and Extrema

The order relations of the previous section were strict and linear: transitive and trichotomous, with the non-strict $\leqslant$ built afterwards. There is a second tradition, equally common, which takes the non-strict relation as primitive and weakens trichotomy so that incomparable pairs are allowed. That is the language of partial orders, and it is what one wants for inclusion of sets.

Definition 6.16 (Partial order).

A relation $R$ on a set $M$ is a partial order, and $M$ is then ordered by $R$ , if $R$ is

reflexive: $a \mathbin{R} a$ for every $a \in M$ ;
antisymmetric: $a \mathbin{R} b$ and $b \mathbin{R} a$ imply $a = b$ ;
transitive: $a \mathbin{R} b$ and $b \mathbin{R} c$ imply $a \mathbin{R} c$ .

We usually write $a \leqslant b$ for $a \mathbin{R} b$ , and then $a < b$ for $(a \leqslant b) \land (a \neq b)$ , with $b \geqslant a$ and $b > a$ as before.

Definition 6.17 (Comparable elements and total orders).

Let $M$ be partially ordered. Elements $a, b \in M$ are comparable if $a \leqslant b$ or $b \leqslant a$ . The order is a total order, or linear order, if every two elements are comparable.

Example 6.18.

The power set $\mathcal{P}(M)$ is partially ordered by $\subset$ , and also by the reverse inclusion. It is not totally ordered as soon as $M$ has two elements $a$ and $b$ , since $\{a\}$ and $\{b\}$ are incomparable: neither contains the other. The natural numbers will turn out to be totally ordered, once we put an order on them below.

Remark (Partial versus strict).

The two languages agree on total orders. If $<$ is a strict linear order and $\leqslant$ is the associated relation, then $\leqslant$ is a total order in the present sense: reflexivity is immediate, antisymmetry is one of the laws proved above, and transitivity is its corollary. Conversely, from a total order $\leqslant$ the relation $a < b \iff (a \leqslant b) \land (a \neq b)$ is a strict linear order. What partial orders add is the possibility that some pairs are incomparable.

Definition 6.19 (Minimal and maximal elements).

Let $M$ be partially ordered and $A \subset M$ non-empty. An element $a_0 \in A$ is minimal in $A$ if no element of $A$ is strictly smaller, that is, $\{a \in A \mid a < a_0\} = \emptyset$ ; and $a_1 \in A$ is maximal in $A$ if no element of $A$ is strictly larger.

A bound of a set need not lie in the set; a minimum or maximum must.

Definition 6.20 (Bounds, minima, maxima).

Let $S$ be a subset of a partially ordered set $U$ .

An element $b \in U$ is a lower bound of $S$ if $b \leqslant x$ for every $x \in S$ , and $B \in U$ is an upper bound if $x \leqslant B$ for every $x \in S$ .
An element $m \in S$ is a minimum, or least element, of $S$ if $m \leqslant x$ for every $x \in S$ ; and $M \in S$ is a maximum, or greatest element, if $x \leqslant M$ for every $x \in S$ .

The difference is membership: a minimum is a lower bound that happens to lie in $S$ . Not every subset has either, as the natural numbers will shortly illustrate: they have a least element and no greatest one.

Proposition 6.21 (A bound inside the set is an extremum).

Let $S$ be a subset of a partially ordered set. If $b$ is a lower bound of $S$ and $b \in S$ , then $b$ is a minimum of $S$ ; and if $B$ is an upper bound of $S$ with $B \in S$ , then $B$ is a maximum.

Discussion.

The two halves are dual, so it is enough to plan the first and reverse the inequalities for the second. What we are given is a lower bound of $S$ that also lies in $S$ . The bound hypothesis is $b \leqslant x$ for every $x \in S$ , which is the inequality a minimum must satisfy, and the membership is precisely the extra clause the definition of a minimum asks for. Together these are the claim; the two definitions differ only by that clause.

Proof.

If $b$ is a lower bound of $S$ then $b \leqslant x$ for every $x \in S$ , and $b \in S$ by hypothesis; those two together are the definition of a minimum. The argument for maxima is the same with the inequalities reversed.

Existence may fail, but uniqueness does not.

Proposition 6.22 (Uniqueness of extrema).

Let $S$ be a subset of a partially ordered set. The minimum of $S$ , if it exists, is unique, and likewise the maximum.

Discussion.

Suppose two elements are both minima. Each, being a minimum, is below every element of $S$ , and each, being a minimum, lies in $S$ ; so applying the defining property of the first to the second and of the second to the first gives the two inequalities antisymmetry needs. The argument for maxima is identical with the inequalities reversed. The proposition does not say that a minimum exists, only that there cannot be two. Bounds outside $S$ are under no such restriction, since a set may have many lower bounds.

Proof.

Let $m_1$ and $m_2$ both be minima of $S$ . Then $m_1 \leqslant m_2$ , since $m_1$ is a minimum and $m_2 \in S$ ; and $m_2 \leqslant m_1$ for the same reason with the roles exchanged. Antisymmetry gives $m_1 = m_2$ . The argument for maxima is identical.

Problem 6.6.

For $\mathcal{P}(M)$ ordered by inclusion, identify the minimum and the maximum. In $\mathcal{P}(M) \setminus \{\emptyset\}$ , show that every singleton is minimal, and decide whether there is a minimum.

Problem 6.7.

Let $A$ and $B$ be totally ordered sets and let $f : A \to B$ be a bijection with $f(a) < f(b)$ whenever $a < b$ . Show that $a < b$ whenever $f(a) < f(b)$ , and deduce that $f^{-1}$ has the same property.

Partitions

Consider a three-piece jigsaw. How many states of assembly are there, counting two states as different when one has two pieces interlocked that the other does not? The answer is five. Representing each cluster of interlocked pieces by the set of its numbers, each state becomes a family of such sets:

\begin{aligned} \Pi_A &= \bigl\{\{1,2,3\}\bigr\}, & \Pi_B &= \bigl\{\{1,2\}, \{3\}\bigr\}, & \Pi_C &= \bigl\{\{2,3\}, \{1\}\bigr\}, \\ \Pi_D &= \bigl\{\{1,3\}, \{2\}\bigr\}, & \Pi_E &= \bigl\{\{1\}, \{2\}, \{3\}\bigr\}. \end{aligned}

Each is a family of non-empty sets whose union is $\{1,2,3\}$ , no two of which share an element.

Definition 6.23 (Partition).

Let $S$ be a non-empty set. A partition $\Pi$ of $S$ is an indexed family $\Pi \defeq \{A_i\}_{i \in I}$ of non-empty subsets of $S$ with

\bigcup_{i \in I} A_i = S, \qquad A_i \cap A_j = \emptyset \ \text{ whenever } i \neq j.

The $A_i$ are the blocks of the partition.

Example 6.24.

The set $S = \{1,2,3,4\}$ has $\bigl\{\{1,2,3,4\}\bigr\}$ as its only one-block partition, and seven with two blocks:

\begin{aligned} &\bigl\{\{1\}, \{2,3,4\}\bigr\}, \quad \bigl\{\{2\}, \{1,3,4\}\bigr\}, \quad \bigl\{\{3\}, \{1,2,4\}\bigr\}, \quad \bigl\{\{4\}, \{1,2,3\}\bigr\}, \\ &\bigl\{\{1,2\}, \{3,4\}\bigr\}, \quad \bigl\{\{1,3\}, \{2,4\}\bigr\}, \quad \bigl\{\{1,4\}, \{2,3\}\bigr\}. \end{aligned}

It has six with three blocks, and its only four-block partition is $\bigl\{\{1\},\{2\},\{3\},\{4\}\bigr\}$ .

A partition is how we get a handle on a large set: cut it into blocks, deal with the blocks one at a time, and put the pieces back together. The blocks are usually chosen by a shared feature.

Example 6.25 (Informal).

Assuming you know the real numbers, let $S$ be a set of them and suppose a calculation requires that we treat numbers differently according to sign. Take the three blocks

A_1 \defeq \{x \in S \mid x > 0\}, \qquad A_2 \defeq \{x \in S \mid x < 0\}, \qquad A_3 \defeq \{0\}.

This is a partition only if $S$ holds a positive number, a negative number and zero, since blocks are required to be non-empty. Without that requirement every family of conditions would partition every set.

Problem 6.8.

List the six three-block partitions of $\{1,2,3,4\}$ .

Equivalence Classes

Fix an equivalence relation $\sim$ on a set $S$ . We cut $S$ into pieces by grouping together everything related to a given element.

Definition 6.26 (Equivalence class).

Let $\sim$ be an equivalence relation on a set $S$ and let $x \in S$ . The equivalence class of $x$ is

[x] \defeq \{s \in S \mid s \sim x\},

and the set of all equivalence classes is

S/{\sim} \defeq \bigl\{[x] \mid x \in S\bigr\} \subset \mathcal{P}(S),

called $S$ modulo $\sim$ , or the quotient of $S$ by $\sim$ .

Example 6.27.

On $S = \{1,2,3,4\}$ with the equivalence relation $R_1$ above, $[1] = [2] = \{1,2\}$ , $[3] = \{3\}$ and $[4] = \{4\}$ . Distinct classes here do not meet, and the next proposition shows this always happens.

Proposition 6.28 (Characterising equivalence classes).

Let $\sim$ be an equivalence relation on a set $S$ and let $a, b \in S$ . The following are equivalent.

$a \sim b$ ;
$[a] \cap [b] \neq \emptyset$ ;
$[a] = [b]$ .

Discussion.

Three statements would ordinarily need six implications, but a cycle is enough: we prove $1 \implies 3 \implies 2 \implies 1$ and let each statement reach every other by going round. For $1 \implies 3$ we assume $a \sim b$ and aim at an equality of sets, so we chase an element: $c \in [a]$ means $c \sim a$ , and the chain $c \sim a \sim b$ closes by transitivity, with symmetry supplying whichever direction the chain needs; the reverse inclusion is the same argument with $a$ and $b$ exchanged. For $3 \implies 2$ reflexivity puts $a$ in $[a]$ , so if the classes agree then $a$ itself witnesses that they meet. For $2 \implies 1$ a shared element $c$ gives $c \sim a$ and $c \sim b$ , and symmetry with transitivity gives $a \sim b$ . Only the three defining properties are used.

Proof.

$1 \implies 3$ . Suppose $a \sim b$ and let $c \in [a]$ , so $c \sim a$ . With $a \sim b$ transitivity gives $c \sim b$ , that is $c \in [b]$ ; hence $[a] \subset [b]$ . Symmetry gives $b \sim a$ , and the same argument with $a$ and $b$ exchanged gives $[b] \subset [a]$ . So $[a] = [b]$ .

$3 \implies 2$ . Reflexivity gives $a \sim a$ , so $a \in [a]$ . If $[a] = [b]$ then $a \in [a] \cap [b]$ , which is therefore non-empty.

$2 \implies 1$ . Let $c \in [a] \cap [b]$ . Then $c \sim a$ and $c \sim b$ . Symmetry gives $a \sim c$ , and transitivity with $c \sim b$ gives $a \sim b$ .

Remark (Disjoint decomposition).

It follows that $S$ is the union of the classes and that distinct classes are disjoint:

S = \bigcup_{A \in S/\sim} A, \qquad A_1 \cap A_2 = \emptyset \ \text{ whenever } A_1 \neq A_2 \text{ in } S/{\sim}.

An equivalence relation therefore breaks $S$ into pairwise disjoint non-empty pieces. Some authors write $S = \bigsqcup_{A \in S/\sim} A$ for a union whose parts are known to be disjoint.

Problem 6.9.

Let $\sim$ be an equivalence relation on $S$ . Show that $x \in [x]$ for every $x \in S$ , and that if $y \in [x]$ then $[y] = [x]$ .

Problem 6.10 (Informal).

Assuming you know the real numbers and the whole numbers, let $x \sim y$ mean that $x - y$ is a whole number. Check that $\sim$ is an equivalence relation on the real numbers, and describe $[0]$ and $\bigl[\tfrac{1}{2}\bigr]$ .

Problem 6.11.

Let $f : S \to T$ be a map and declare $x \sim y$ on $S$ when $f(x) = f(y)$ . Show that $\sim$ is an equivalence relation, and that $S/{\sim}$ is in bijection with the image $f(S)$ .

From Equivalence Relations to Partitions

An equivalence relation gives a partition, and a partition gives an equivalence relation. We prove each direction in turn.

Theorem 6.29 (Equivalence relations induce partitions).

Let $\sim$ be an equivalence relation on a non-empty set $S$ . Then the family of its equivalence classes is a partition of $S$ , called the partition induced by $\sim$ .

Discussion.

We check the three conditions in the definition of a partition: blocks non-empty, blocks pairwise disjoint, blocks covering $S$ . Non-emptiness is immediate, since reflexivity puts $x$ in $[x]$ . Disjointness is the previous proposition read contrapositively: if two classes meet they are equal, so distinct classes cannot meet. For the cover, the same reflexivity observation puts every $s$ in the block $[s]$ , while the reverse inclusion is automatic because every class is a subset of $S$ . One class may have many labels, since $[x] = [y]$ whenever $x \sim y$ , so we index the family by the distinct classes and do not count one block twice.

Proof.

Write the distinct classes as an indexed family $\{C_i\}_{i \in I}$ .

Each $C_i$ is non-empty: if $C_i = [x]$ then $x \in [x]$ by reflexivity. If $i \neq j$ then $C_i \neq C_j$ , so $C_i \cap C_j = \emptyset$ by the previous proposition. Every $C_i$ is a subset of $S$ , so the union of the $C_i$ is contained in $S$ ; and if $s \in S$ then $s \in [s]$ by reflexivity, and $[s]$ is one of the $C_i$ , so $s$ lies in the union. Hence the union is $S$ , and the family is a partition.

Example 6.30.

The relation $R_1$ on $\{1,2,3,4\}$ induces the partition $\bigl\{\{1,2\}, \{3\}, \{4\}\bigr\}$ , which is the list of classes computed above.

Theorem 6.31 (Partitions induce equivalence relations).

Let $\Pi$ be a partition of a set $S$ , and for $x, y \in S$ declare $x \sim y$ to mean that $x$ and $y$ lie in the same block of $\Pi$ . Then $\sim$ is an equivalence relation on $S$ , and the partition it induces is $\Pi$ itself.

Discussion.

There are two claims. For the first, the three properties come from the three conditions on a partition. Reflexivity uses the cover, since every $x$ lies in some block. Symmetry is immediate, since “same block” does not depend on which element is named first. Transitivity uses disjointness: if $x$ and $y$ share a block and $y$ and $z$ share a block, then $y$ lies in both, so the two blocks meet and are therefore the same block, which then holds $x$ and $z$ . For the second claim, fix $x$ and let $A$ be the block holding it. Then $y \in [x]$ exactly when $y$ shares a block with $x$ , which is exactly when $y \in A$ ; so the class $[x]$ is the block $A$ , and the family of classes is $\Pi$ .

Proof.

Write $\Pi = \{A_i\}_{i \in I}$ , so the $A_i$ are non-empty, pairwise disjoint, and their union is $S$ .

For reflexivity, an $x \in S$ lies in some $A_i$ by the cover, so $x \sim x$ . For symmetry, if $x \sim y$ then both lie in some $A_i$ , so both lie in it in the other order and $y \sim x$ . For transitivity, if $x \sim y$ and $y \sim z$ then $x, y \in A_i$ and $y, z \in A_j$ for some $i, j$ ; then $y \in A_i \cap A_j$ , so $i = j$ since distinct blocks are disjoint, and $x, z \in A_i$ gives $x \sim z$ . So $\sim$ is an equivalence relation.

For the second claim, let $x \in S$ and let $A$ be the block holding $x$ , unique because distinct blocks are disjoint. Then $y \in [x]$ if and only if $y \sim x$ , which holds if and only if $y \in A$ . So $[x] = A$ , and the family of classes is exactly $\Pi$ .

Problem 6.12.

Take the partition $\bigl\{\{1,2\}, \{3\}, \{4\}\bigr\}$ of $\{1,2,3,4\}$ and write down every pair in the equivalence relation it induces. Check that you recover $R_1$ .

Problem 6.13.

How many equivalence relations are there on a set with three elements? Count the partitions, then appeal to the two theorems.

The Order on the Natural Numbers

We now define the order on the natural numbers from addition. Throughout this section we work in $\mathbb{N}_0$ , which is a Peano system under $s$ , so all five conditions and the recursion theorem apply. Addition already suggests the order: $n$ lies beyond $m$ when it is reached from $m$ by adding something positive.

Definition 6.32 (Order on the natural numbers).

For $m, n \in \mathbb{N}_0$ we define $m < n$ to hold if and only if there is a $b \in \mathbb{N}$ with $n = m + b$ ; in symbols,

m < n \iff \exists b \in \mathbb{N}\,(n = m + b).

Since $0 \notin \mathbb{N}$ , the difference $b$ is required to be positive. Admitting $b = 0$ would make every number less than itself and break trichotomy.

Corollary 6.33 (Adding a positive element moves you up).

If $x \in \mathbb{N}_0$ and $y \in \mathbb{N}$ then $x < x + y$ . In particular $m < s(m)$ for every $m \in \mathbb{N}_0$ .

Proof.

Take $n = x + y$ in the definition. For the second claim take $y = 1$ , which lies in $\mathbb{N}$ , and recall $m + 1 = s(m)$ .

The definition is stated in terms of differences, so we first show that a difference is unique.

Proposition 6.34 (Uniqueness of differences).

Let $a, b \in \mathbb{N}_0$ . Then $a + x = b$ has at most one solution $x \in \mathbb{N}_0$ .

Discussion.

Uniqueness statements are proved by assuming two solutions and forcing them to agree, so we let $A(a)$ be the statement that any two solutions for that $a$ coincide, and induct on $a$ . The base case is immediate, since $0 + x = b$ reads $x = b$ by addition from the left. For the step we take $s(a) + x_1 = b = s(a) + x_2$ ; both sides are successors, of $a + x_1$ and of $a + x_2$ , and the third Peano condition strips a successor from an equation, leaving an equation the inductive hypothesis handles.

Proof.

We induct on $a$ . For $a = 0$ , the equation $0 + x = b$ gives $x = b$ by addition from the left, so there is at most one solution.

Suppose the claim holds for $a$ and that $s(a) + x_1 = b = s(a) + x_2$ . By addition from the left, $s(a) + x_i = s(a + x_i)$ , so $s(a + x_1) = s(a + x_2)$ ; the third Peano condition gives $a + x_1 = a + x_2$ , and the inductive hypothesis gives $x_1 = x_2$ . Induction completes the proof.

To decide whether $m \leqslant n$ we do not need the case split in the definition of the associated order: one difference is enough, provided we now allow it to be $0$ .

Proposition 6.35 (The associated order on the natural numbers).

For $m, n \in \mathbb{N}_0$ we have $m \leqslant n$ if and only if $n = m + b$ for some $b \in \mathbb{N}_0$ .

Discussion.

The statement is a biconditional, so there are two conditionals. Going forwards, the definition of $\leqslant$ unfolds $m \leqslant n$ into the two cases $m < n$ and $m = n$ , and we must produce a difference in each: the first hands us one directly, and in the second $b = 0$ works. Going backwards we are given a difference and must decide which half of $\leqslant$ holds, so we split on whether $b$ is $0$ ; if it is then $m = n$ , and if it is not then $b \in \mathbb{N}$ and the definition of $<$ applies. Either way $m \leqslant n$ .

Proof.

Suppose $m \leqslant n$ . Then $m < n$ or $m = n$ . In the first case the definition of $<$ supplies $b \in \mathbb{N}$ with $n = m + b$ ; in the second, $b = 0$ serves, since $m + 0 = m = n$ . Either way $b \in \mathbb{N}_0$ .

Conversely suppose $n = m + b$ with $b \in \mathbb{N}_0$ . If $b = 0$ then $m = n$ , so $m \leqslant n$ . If $b \neq 0$ then $b \in \mathbb{N}$ , so $m < n$ and again $m \leqslant n$ .

When $m \leqslant n$ the $b$ with $m + b = n$ is unique, by the proposition on differences, and we write it $n - m$ . So subtraction is defined exactly when the answer lies in $\mathbb{N}_0$ .

Proposition 6.36 (Positive means greater than zero).

Let $n \in \mathbb{N}_0$ . Then $n > 0$ if and only if $n \in \mathbb{N}$ . Consequently $0 \leqslant n$ for every $n \in \mathbb{N}_0$ .

Discussion.

Both directions run through the one equation $n = 0 + b$ . Going forwards we must produce a difference witnessing $0 < n$ , and $b = n$ satisfies the equation, lying in $\mathbb{N}$ because that is the hypothesis. Going backwards the definition hands us a $b \in \mathbb{N}$ with $n = 0 + b$ , so $n$ and $b$ are the same element and the hypothesis transfers. The consequence follows: an element of $\mathbb{N}_0$ is either $0$ , where equality serves, or lies in $\mathbb{N}$ , where the biconditional gives $0 < n$ .

Proof.

Suppose $n \in \mathbb{N}$ . Then $n = 0 + n$ by addition from the left, and $n \in \mathbb{N}$ , so $0 < n$ by the definition of $<$ .

Conversely suppose $0 < n$ . The definition gives $b \in \mathbb{N}$ with $n = 0 + b$ , so $n = b \in \mathbb{N}$ .

Finally, an $n \in \mathbb{N}_0$ is either $0$ or a member of $\mathbb{N}$ ; in the first case $0 = n$ and in the second $0 < n$ , so $0 \leqslant n$ in both.

The proof that $<$ is an order splits $\mathbb{N}_0$ at a chosen element, and we use the same splitting later for the induction variants and for well-ordering, so we give it a name.

Definition 6.37 (Cuts).

For $a \in \mathbb{N}_0$ we write

L_a \defeq \{x \in \mathbb{N}_0 \mid x < a\}, \qquad R_a \defeq \{x \in \mathbb{N}_0 \mid x \geqslant a\}.

Theorem 6.38 (The natural numbers are strictly ordered).

The relation of the definition above is a strict linear order on $\mathbb{N}_0$ .

Discussion.

The definition of a strict linear order wants transitivity and trichotomy, and the characterisation proved earlier lets us replace the second by weak trichotomy and anti-reflexivity, so there are three things to prove. Transitivity is a calculation: two positive differences add up to a single one, which is positive by closure. Anti-reflexivity follows from uniqueness of differences, since $a < a$ would exhibit $a$ as $a + x$ with $x$ positive while $a = a + 0$ exhibits it with difference $0$ , and only one difference is permitted. Weak trichotomy is the main part; in terms of the cuts it is the set equation $L_a \cup R_a = \mathbb{N}_0$ , which we prove by induction on $a$ . The base case is the proposition just proved, since every element is at least $0$ , so $R_0$ is everything. For the step we take an arbitrary $x$ , use the hypothesis to place it on one side of $a$ , and then place it relative to $s(a)$ : an $x$ below $a$ stays below $s(a)$ , an $x$ equal to $a$ falls below $s(a)$ , and for an $x$ above $a$ the positive difference has a predecessor, which gives $s(a) \leqslant x$ .

Proof.

For transitivity, suppose $a + x = b$ and $b + y = c$ with $x, y \in \mathbb{N}$ . Then $a + (x + y) = (a + x) + y = b + y = c$ by the laws of addition, and $x + y \in \mathbb{N}$ by closure; hence $a < c$ .

For anti-reflexivity, $a < a$ would give $a + x = a = a + 0$ with $x \in \mathbb{N}$ , and uniqueness of differences forces $x = 0$ , contradicting $x \in \mathbb{N}$ .

For weak trichotomy, fix $a$ and consider the cuts. We show $L_a \cup R_a = \mathbb{N}_0$ by induction on $a$ . For $a = 0$ , every element satisfies $0 \leqslant x$ by the proposition above, so $R_0 = \mathbb{N}_0$ .

Suppose $L_a \cup R_a = \mathbb{N}_0$ and let $x \in \mathbb{N}_0$ . If $x \in L_a$ then $x < a < s(a)$ by the corollary above, so $x \in L_{s(a)}$ by transitivity. If $x \in R_a$ there are two cases. If $x = a$ then $x < s(a)$ , so $x \in L_{s(a)}$ . If $a < x$ then $x = a + b$ with $b \in \mathbb{N}$ , and $b$ has a predecessor by the theorem on predecessors, say $b = s(c) = c + 1$ ; then $x = a + (1 + c) = s(a) + c$ , so $s(a) \leqslant x$ and $x \in R_{s(a)}$ . Induction gives $L_a \cup R_a = \mathbb{N}_0$ for every $a$ , which is weak trichotomy.

The characterisation of strict linear orders now applies.

Nothing lies strictly between $n$ and its successor. This looks obvious, but it has to be proved from the definition.

Proposition 6.39 (Nothing between n and its successor).

Let $n \in \mathbb{N}_0$ . There is no $x \in \mathbb{N}_0$ with $n < x < s(n)$ .

Discussion.

This is a non-existence claim, so we suppose such an $x$ exists and derive a contradiction. The two strict inequalities give two positive differences, $x = n + b$ and $s(n) = x + c$ , and we eliminate $n$ between them. Since $b$ is positive it has a predecessor, so $b = 1 + d$ , so the $s(n)$ of the second equation appears inside the first. Substituting turns $x$ into $x + (c + d)$ , so $x$ carries the two differences $0$ and $c + d$ ; uniqueness forces $c + d = 0$ , and a sum is zero only when both parts are, so $c = 0$ , contradicting the choice of $c$ as positive.

Proof.

Suppose $n < x$ and $x < s(n)$ . The first gives $b \in \mathbb{N}$ with $x = n + b$ , the second gives $c \in \mathbb{N}$ with $s(n) = x + c$ . Since $b \neq 0$ it has a predecessor, so $b = 1 + d$ for some $d \in \mathbb{N}_0$ . Then

x = n + b = n + (1 + d) = (n + 1) + d = s(n) + d = (x + c) + d = x + (c + d).

So $x + 0 = x + (c + d)$ , and uniqueness of differences gives $c + d = 0$ ; hence $c = 0$ , contradicting $c \in \mathbb{N}$ .

Problem 6.14.

Show that the only $x \in \mathbb{N}_0$ with $x < 2$ are $x = 0$ and $x = 1$ .

Problem 6.15.

Prove that $m < n$ if and only if $s(m) \leqslant n$ , for all $m, n \in \mathbb{N}_0$ .

Problem 6.16.

Let $x, y, z \in \mathbb{N}_0$ , with multiplication as in the problems of the last chapter.

If $x \leqslant y$ then $x + z \leqslant y + z$ , and $x < y$ if and only if $x + z < y + z$ .
If $x \leqslant y$ then $x \cdot z \leqslant y \cdot z$ ; and if moreover $z > 0$ and $x < y$ , then $x \cdot z < y \cdot z$ .
If $x \cdot z < y \cdot z$ then $x < y$ .
If $x \cdot z = y \cdot z$ and $z \neq 0$ , then $x = y$ .

Problem 6.17.

Suppose $m_1 < m_2$ and $n_1 < n_2$ in $\mathbb{N}_0$ . Show that $m_1 + n_1 < m_2 + n_2$ .

Problem 6.18.

Show that $n - n = 0$ and $n - 0 = n$ for every $n \in \mathbb{N}_0$ , and that $n - m = 0$ implies $n = m$ for all $m, n \in \mathbb{N}_0$ with $m \leqslant n$ .

Problem 6.19.

Let $x, y, z, c \in \mathbb{N}_0$ .

If $y \leqslant x$ and $z \leqslant x$ , then $x - y = z$ if and only if $x - z = y$ .
If $z \leqslant y$ then $(x + y) - z = x + (y - z)$ .
If $y \leqslant x$ then $x - y = (x + c) - (y + c)$ .

Problem 6.20.

State the hypotheses on $a, b, c \in \mathbb{N}_0$ under which each of

a + (b - a) = b, \qquad (b - a) + (c - b) = c - a, \qquad c(b - a) = cb - ca

is defined, and prove all three.

Problem 6.21.

Determine all $x, y, z \in \mathbb{N}_0$ for which $(x - y) - z$ and $x - (y - z)$ are both defined, and decide whether the two agree whenever both are.
Decide whether $x - (y + z) = (x - y) - z$ whenever the left-hand side is defined.

Well-Ordering and Induction Variants

With the order we can prove the form of induction that starts at an arbitrary natural number.

Theorem 6.40 (Induction from an arbitrary starting point).

Let $n_0 \in \mathbb{N}_0$ and let $A$ be a predicate on $\mathbb{N}_0$ . If $A(n_0)$ holds, and if $A(n) \implies A(s(n))$ for every $n \geqslant n_0$ , then $A(n)$ holds for every $n \geqslant n_0$ .

Discussion.

Ordinary induction starts at $0$ , and the hypotheses say nothing below $n_0$ , so we enlarge the set where $A$ holds until the fifth Peano condition applies. Let $M$ be that set and put $M' \defeq M \cup L_{n_0}$ , adding everything strictly below the starting point. We may assume $n_0 > 0$ , since at $n_0 = 0$ ordinary induction already applies, and then $0 \in L_{n_0} \subset M'$ settles the base case without appealing to $A$ at all. For the step we take $n \in M'$ and compare $s(n)$ with $n_0$ , which trichotomy resolves into three cases: below $n_0$ it lies in the padding, equal to $n_0$ it lies in $M$ by the base hypothesis, and above $n_0$ it forces $n$ itself to be at or above $n_0$ , so $n$ lies in $M$ and the given step applies. Induction then gives $M' = \mathbb{N}_0$ ; since the two cuts are disjoint and exhaust everything, the added part $L_{n_0}$ has nothing at or above $n_0$ , and $R_{n_0} \subset M$ is what remains.

Proof.

Let $M \defeq \{n \in \mathbb{N}_0 \mid A(n)\}$ and put $M' \defeq M \cup L_{n_0}$ . It is enough to prove $M' = \mathbb{N}_0$ , for then $L_{n_0} \cup R_{n_0} = \mathbb{N}_0$ and $L_{n_0} \cap R_{n_0} = \emptyset$ give $R_{n_0} \subset M$ , which is the assertion.

We may assume $n_0 > 0$ , since $n_0 = 0$ is ordinary induction. Then $0 \in L_{n_0} \subset M'$ . Let $n \in M'$ and compare $s(n)$ with $n_0$ .

If $s(n) < n_0$ then $s(n) \in L_{n_0} \subset M'$ . If $s(n) = n_0$ then $s(n) \in M$ by the base hypothesis. If $s(n) > n_0$ then $n \geqslant n_0$ , since $n < n_0$ would put $n_0$ strictly between $n$ and $s(n)$ ; hence $n \notin L_{n_0}$ , so $n \in M$ , and the induction hypothesis gives $s(n) \in M$ .

In every case $s(n) \in M'$ , so the fifth Peano condition gives $M' = \mathbb{N}_0$ .

Theorem 6.41 (Well-ordering of the natural numbers).

Every non-empty subset $M \subset \mathbb{N}_0$ has a smallest element: there is $m_0 \in M$ with $m_0 \leqslant m$ for every $m \in M$ . We write $m_0 = \min M$ .

Discussion.

A non-empty $M$ may be unbounded, so we first cut it down to a bounded piece. Choose any $a \in M$ and put $M' \defeq L_{s(a)} \cap M$ , which holds $a$ and so is non-empty. A smallest element of $M'$ is already smallest in $M$ , since everything in $M$ outside $L_{s(a)}$ is at least $s(a)$ and therefore above it. So we may assume $M \subset L_{s(a)}$ and induct on $a$ , taking for $A(a)$ the statement that every non-empty subset of $L_{s(a)}$ has a smallest element. The base case can be checked directly: $L_1$ is $\{0\}$ , so the only non-empty subset is $\{0\}$ . For the step, take a non-empty subset of $L_{s(s(a))}$ and intersect it with $L_{s(a)}$ ; if the intersection is empty then nothing between $s(a)$ and its successor leaves only $\{s(a)\}$ , and otherwise the hypothesis supplies a smallest element of the intersection, which by the cutting-down argument is smallest in the whole set.

Proof.

Choose $a \in M$ and put $M' \defeq L_{s(a)} \cap M$ ; since $a < s(a)$ we have $a \in M'$ , so $M'$ is non-empty. If $m_0$ is smallest in $M'$ then $m_0 \leqslant a$ , while every member of $M \setminus M'$ is at least $s(a)$ and hence above $m_0$ ; so $m_0$ is smallest in $M$ . It therefore suffices to show, by induction on $a$ , that every non-empty subset of $L_{s(a)}$ has a smallest element.

For $a = 0$ , the cut $L_1$ is $\{0\}$ : an $x < 1$ cannot be positive, since $0 < x < 1$ would put $x$ strictly between $0$ and its successor. So the only non-empty subset of $L_1$ is $\{0\}$ , whose smallest element is $0$ .

Suppose every non-empty subset of $L_{s(a)}$ has a smallest element, and let $N \neq \emptyset$ be contained in $L_{s(s(a))}$ . If $N \cap L_{s(a)} = \emptyset$ then every element of $N$ is at least $s(a)$ and below $s(s(a))$ , so $N = \{s(a)\}$ and $s(a)$ is its smallest element. Otherwise $N \cap L_{s(a)}$ has a smallest element $m_0$ , and the argument of the first paragraph shows $m_0$ is smallest in $N$ . Induction completes the proof.

Corollary 6.42 (Method of infinite descent).

No non-empty subset $M \subset \mathbb{N}_0$ has the property that every $a \in M$ admits an $a' \in M$ with $a' < a$ .

Proof.

If such an $M$ existed, well-ordering would give it a smallest element $a_0$ , and the hypothesis would supply $a_0' \in M$ with $a_0' < a_0$ , contradicting minimality.

One cannot descend for ever through a non-empty set of natural numbers. The Greeks already used this as a proof technique, and we shall use it when we come to the numbers that are not fractions.

Theorem 6.43 (Strong induction).

Let $A$ be a predicate on $\mathbb{N}_0$ with $A(0)$ true, and suppose that for every $n \in \mathbb{N}_0$ the truth of $A(m)$ for all $m \leqslant n$ implies $A(s(n))$ . Then $A(n)$ holds for every $n \in \mathbb{N}_0$ .

Discussion.

In ordinary induction the step uses only the previous case; here it may use all earlier cases at once, and we prove this from well-ordering. Instead of building up the set where $A$ holds we look at the set where it fails: assume it non-empty and let $n_0$ be its smallest element. The base hypothesis rules out $n_0 = 0$ , so $n_0$ has a predecessor $p$ . Minimality says $A(m)$ holds for every $m < n_0$ , and since nothing lies strictly between $p$ and $n_0$ , those $m$ are precisely the $m \leqslant p$ , which is what the step needs at $p$ . The step therefore yields $A(s(p))$ , that is $A(n_0)$ , and $n_0$ was chosen to make $A$ fail.

Proof.

Let $M \defeq \{n \in \mathbb{N}_0 \mid A(n) \text{ fails}\}$ and suppose $M \neq \emptyset$ . Well-ordering gives it a smallest element $n_0$ , so $A(n_0)$ fails while $A(n)$ holds for every $n < n_0$ .

Since $A(0)$ holds, $n_0 \neq 0$ , so the theorem on predecessors gives $n_0 = s(p)$ for some $p$ . Every $m \leqslant p$ satisfies $m < n_0$ , since $p < s(p) = n_0$ and nothing lies strictly between; hence $A(m)$ holds for every such $m$ . The hypothesis then gives $A(s(p))$ , that is $A(n_0)$ , a contradiction. So $M = \emptyset$ .

Problem 6.22.

Deduce the fifth Peano condition from strong induction, so that the two induction principles are equivalent.

Problem 6.23.

Show that strong induction can be proved without well-ordering, by applying ordinary induction to the set $B \defeq \{n \in \mathbb{N}_0 \mid A(m) \text{ holds for every } m \leqslant n\}$ .

Problem 6.24.

Let $f : \mathbb{N}_0 \to \mathbb{N}_0$ .

Show that $f(s(n)) < f(n)$ cannot hold for every $n \in \mathbb{N}_0$ .
Show that if $f(s(n)) \leqslant f(n)$ for every $n \in \mathbb{N}_0$ , then there is an $N$ with $f(n) = f(N)$ for every $n \geqslant N$ .

Problem 6.25.

Let $n \in \mathbb{N}_0$ . Show that there is exactly one $m \in \mathbb{N}_0$ with $m \cdot m \leqslant n < s(m) \cdot s(m)$ .

Division with Remainder

A first use of well-ordering: dividing one natural number by another leaves a quotient and a remainder, both determined by the two numbers.

Theorem 6.44 (Division with remainder).

Let $n \in \mathbb{N}_0$ and $m \in \mathbb{N}$ . Then there is exactly one pair $q, r \in \mathbb{N}_0$ with

n = q \cdot m + r, \qquad r < m.

Discussion.

An “exactly one” claim asks for existence and uniqueness, and the two need different arguments. For existence we look for the remainder rather than the quotient: the set of numbers of the form $n - q \cdot m$ that remain in $\mathbb{N}_0$ is non-empty, since $q = 0$ leaves $n$ itself, so well-ordering gives us its least member $r$ , and minimality forces $r < m$ : were $r$ at least $m$ we could subtract one more $m$ and get something smaller still. For uniqueness we assume two decompositions and show that a certain multiple of $m$ must be $0$ . Subtracting gives a multiple of $m$ equal to a difference of two remainders, and that difference is below $m$ because each remainder is; but a non-zero multiple of $m$ is at least $m$ , so the multiple is $0$ , the quotients agree, and the remainders follow.

Proof.

For existence, let $R \defeq \{r \in \mathbb{N}_0 \mid n = q \cdot m + r \text{ for some } q \in \mathbb{N}_0\}$ . Taking $q = 0$ gives $n \in R$ , so $R$ is non-empty and well-ordering supplies $r \defeq \min R$ , with $n = q \cdot m + r$ for some $q$ .

Suppose $r \geqslant m$ . Then $r = m + r'$ for some $r' \in \mathbb{N}_0$ , so $n = q \cdot m + m + r' = s(q) \cdot m + r'$ , putting $r' \in R$ ; and $r' < r$ , since $m$ is positive. That contradicts minimality, so $r < m$ .

For uniqueness, suppose $q_1 \cdot m + r_1 = q_2 \cdot m + r_2$ with $r_1, r_2 < m$ , and take $q_1 \geqslant q_2$ without loss. Then $q_1 = q_2 + d$ for some $d \in \mathbb{N}_0$ , and cancelling gives $d \cdot m + r_1 = r_2$ , so $d \cdot m \leqslant r_2 < m$ . If $d$ were positive then $d \geqslant 1$ and $d \cdot m \geqslant m$ , a contradiction; so $d = 0$ , whence $q_1 = q_2$ and then $r_1 = r_2$ .

Corollary 6.45 (Remainder classes).

Let $m \in \mathbb{N}$ . Every $n \in \mathbb{N}_0$ has exactly one of the $m$ forms $k \cdot m$ , $k \cdot m + 1$ , …, $k \cdot m + (m-1)$ with $k \in \mathbb{N}_0$ .

Proof.

The theorem gives exactly one pair $k, r$ with $n = k \cdot m + r$ and $r < m$ , and the values available to $r$ are precisely $0, 1, \ldots, m-1$ .

Taking $m = 2$ splits $\mathbb{N}_0$ into the even numbers, those of the form $2k$ , and the odd ones, those of the form $2k + 1$ ; the corollary says every natural number is one or the other and none is both.

Problem 6.26.

Let $n, m \in \mathbb{N}_0$ with $m$ positive. Show that $m$ divides $n$ , in the sense that $n = k \cdot m$ for some $k \in \mathbb{N}_0$ , exactly when the remainder of the theorem is $0$ .

Problem 6.27.

Show that divisibility orders $\mathbb{N}$ as a partial order in the sense of Definition 6.16 , and find the minimum and the maximum, if they exist, of $\{n \in \mathbb{N} \mid n \text{ divides } 12\}$ in that order.

Descent in Action

Here is an application of infinite descent.

Proposition 6.46 (Odd squares are odd).

Let $n \in \mathbb{N}_0$ . If $n \cdot n$ is even then $n$ is even.

Discussion.

We prove the contrapositive, which is the easier direction to compute with: an odd $n$ is $2k+1$ by the remainder classes, and squaring it and collecting terms exhibits the result as twice something plus one. Only the laws of addition and multiplication are needed; the remainder classes turn “not even” into the usable form $2k + 1$ .

Proof.

Suppose $n$ is odd, so $n = 2k + 1$ for some $k \in \mathbb{N}_0$ by the remainder classes. Then

n \cdot n = (2k + 1)(2k + 1) = 2\bigl(2 \cdot k \cdot k + k + k\bigr) + 1,

which is odd. So an even $n \cdot n$ forces $n$ to be even.

Theorem 6.47 (Two is not a ratio of squares).

There are no $d, a \in \mathbb{N}$ with $d \cdot d = 2 \cdot (a \cdot a)$ .

Discussion.

We argue by contradiction, using infinite descent, so we build a non-empty set in which every member has a smaller member. Suppose some pair works and let $M$ be the set of all $a \in \mathbb{N}$ admitting a partner $d$ ; the supposition puts something in $M$ . The descent comes from using parity twice. From $d \cdot d = 2(a \cdot a)$ the square $d \cdot d$ is even, so $d$ is even by the proposition above; writing $d = 2d'$ and substituting makes $a \cdot a$ even, hence $a$ even too, say $a = 2a'$ . Cancelling a factor of four leaves $d' \cdot d' = 2(a' \cdot a')$ , which puts $a'$ in $M$ as well, and $a' < a$ because halving a positive number decreases it. Infinite descent forbids this.

In terms of fractions, which we have not built yet, this says that no fraction squares to two.

Proof.

Suppose such a pair exists and let $M \defeq \{a \in \mathbb{N} \mid d \cdot d = 2 \cdot (a \cdot a) \text{ for some } d \in \mathbb{N}\}$ , which is then non-empty.

Let $a \in M$ with partner $d$ . Then $d \cdot d$ is even, so $d = 2d'$ by the proposition above. Substituting gives $4(d' \cdot d') = 2(a \cdot a)$ , so $a \cdot a = 2(d' \cdot d')$ ; hence $a \cdot a$ is even and $a = 2a'$ for some $a'$ . Substituting again gives $4(d' \cdot d') = 2 \cdot 4 (a' \cdot a')$ , so $d' \cdot d' = 2(a' \cdot a')$ , putting $a' \in M$ . Since $a = 2a' = a' + a'$ with $a'$ positive, we have $a' < a$ .

So every member of $M$ has a smaller member, which infinite descent forbids. Hence no such pair exists.

Problem 6.28.

Show that there are no $d, a \in \mathbb{N}$ with $d \cdot d = 3 \cdot (a \cdot a)$ , and decide whether the same argument rules out $d \cdot d = 4 \cdot (a \cdot a)$ .

Expansions in a Base

Writing a number in base $g$ is repeated division with remainder, and the digits are the remainders.

Theorem 6.48 (Expansion in a base).

Let $g \in \mathbb{N}$ with $g \geqslant 2$ . For every $n \in \mathbb{N}$ there are uniquely determined $r \in \mathbb{N}_0$ and $a_0, \ldots, a_r \in \mathbb{N}_0$ with $a_r \neq 0$ and $a_\rho < g$ for each $\rho$ , such that

n = a_0 + a_1 g + a_2 g^2 + \cdots + a_r g^r = \sum_{\rho = 0}^{r} a_\rho g^\rho.

Discussion.

The digits come from dividing repeatedly: set $q_0 = n$ , divide by $g$ to leave the lowest digit $a_0$ and a quotient $q_1$ , divide that to leave $a_1$ , and so on. Two things must be checked. It stops because $g \geqslant 2$ makes each non-terminal quotient strictly smaller than the number it came from, and a scheme that never halted would descend for ever through $\mathbb{N}_0$ , which descent forbids. The digits add back to $n$ because each equation expresses one quotient in terms of the next, so substituting upwards leaves the displayed sum. Uniqueness runs the same divisions in reverse: division with remainder determines $a_0$ and $q_1$ from $n$ , then $a_1$ and $q_2$ from $q_1$ , and so on, fixing the leading digit and the final index along with the rest.

Proof.

Put $q_0 \defeq n$ and apply division with remainder repeatedly:

\begin{aligned} q_0 &= q_1 g + a_0, & a_0 &< g, \\ q_1 &= q_2 g + a_1, & a_1 &< g, \\ &\ \,\vdots \\ q_{r-1} &= q_r g + a_{r-1}, & a_{r-1} &< g, \\ q_r &= a_r, & 0 < a_r &< g. \end{aligned}

Each line is division with remainder applied to $q_\rho$ and $g$ . Since $g \geqslant 2$ , a non-zero quotient satisfies $q_{\rho+1} < q_\rho$ , so the scheme cannot run for ever without contradicting infinite descent; it therefore halts at some $r$ with $q_r < g$ .

Substituting each line into the one above it gives $n = a_0 + a_1 g + \cdots + a_r g^r$ . Uniqueness of the quotient and remainder at each step gives uniqueness of the list of digits, and of $r$ .

We write $n = (a_r \ldots a_0)_g$ , and call the cases $g = 2$ and $g = 10$ the binary and decimal expansions. The value can be recovered by a recursion which never computes a power separately:

n = \Bigl(\cdots\bigl((a_r g + a_{r-1})g + a_{r-2}\bigr)g + \cdots + a_1\Bigr)g + a_0.

This is Horner’s scheme for base $g$ . It uses one multiplication by $g$ at each stage, $r$ in all, where computing every power first and then forming every term uses up to twice as many.

Problem 6.29.

Write $1000$ in base $7$ , and write $(110\,1101)_2$ in base $10$ .

Problem 6.30.

Let $n \in \mathbb{N}$ and let $r$ be as in the theorem. Show that $g^r \leqslant n < g^{s(r)}$ .

The Maximum Principle

Well-ordering says every non-empty subset of $\mathbb{N}_0$ has a minimum. The dual statement is false as it stands, since $\mathbb{N}_0$ itself has no maximum, but it becomes true once an upper bound is present.

Theorem 6.49 (Maximum principle).

Let $W \subset \mathbb{N}_0$ be non-empty with an upper bound in $\mathbb{N}_0$ . Then $W$ has a maximum.

Discussion.

We cannot induct on $W$ , which is an arbitrary set, so we induct on the bound instead: let $T$ be the set of those $B$ such that every non-empty subset bounded above by $B$ has a maximum, and aim at $T = \mathbb{N}_0$ ; once that is known, any upper bound of the given $W$ lies in $T$ and gives $W$ a maximum. The base case can be checked directly, since a set bounded above by $0$ can only be $\{0\}$ , every element being at least $0$ . For the step we take a non-empty $S$ bounded above by $s(n)$ and split on whether $s(n)$ belongs to $S$ : if it does, it is a bound lying in the set and the previous proposition finishes at once; if it does not, then $n$ is itself an upper bound, because an element strictly above $n$ and at or below $s(n)$ but different from it would sit strictly between $n$ and its successor, which nothing does.

Proof.

Let $T$ be the set of $B \in \mathbb{N}_0$ such that every non-empty $S \subset \mathbb{N}_0$ with upper bound $B$ has a maximum.

For $0 \in T$ : let $S$ be non-empty with upper bound $0$ , and take $k \in S$ . Then $k \leqslant 0$ , while $0 \leqslant k$ , so $k = 0$ by antisymmetry. Hence $S = \{0\}$ , and $0$ is a bound lying in $S$ , so it is the maximum.

Suppose $n \in T$ and let $S$ be non-empty with upper bound $s(n)$ . If $s(n) \in S$ then $s(n)$ is an upper bound lying in $S$ , hence the maximum. If $s(n) \notin S$ , we claim $n$ bounds $S$ : an element $k \in S$ with $n < k$ satisfies $k \leqslant s(n)$ and $k \neq s(n)$ , so $n < k < s(n)$ , which nothing does. So $n$ is an upper bound of $S$ , and $n \in T$ supplies a maximum. Hence $s(n) \in T$ .

By induction $T = \mathbb{N}_0$ . Given $W$ non-empty with upper bound $B$ , we have $B \in T$ , so $W$ has a maximum.

A partially ordered set in which every non-empty subset has a minimum is called well-ordered. By the theorem above $\mathbb{N}_0$ is well-ordered, and so is $\mathbb{N}$ , whose non-empty subsets are non-empty subsets of $\mathbb{N}_0$ . $\mathbb{N}_0$ has two further order properties: every element beyond the minimum has an immediate predecessor, and there is no greatest element. Together with well-ordering these determine the order completely.

Theorem 6.50 (Characterisation of the natural numbers).

Let $M$ be a non-empty well-ordered set in which every element other than $\min M$ has an immediate predecessor, meaning an element $p < a$ with nothing of $M$ strictly between, and which has no greatest element. Then there is an order-preserving bijection from $\mathbb{N}_0$ to $M$ .

Discussion.

The conclusion asks for the order of $\mathbb{N}_0$ inside $M$ , so the plan is to build a successor map on $M$ from the three hypotheses and then check that it reaches everything. In $\mathbb{N}_0$ the successor of $n$ is the least element above $n$ , and that phrase makes sense in $M$ : the absence of a greatest element keeps the set of elements above $a$ non-empty, and well-ordering gives it a minimum, which we call $\nu(a)$ . The predecessor hypothesis makes $\nu$ behave like a successor map, since nothing is reached from below except through an immediate predecessor, so $\min M$ is never a value of $\nu$ and two elements cannot share one. With $\nu$ in hand the recursion theorem builds the candidate: start at $\min M$ and keep applying $\nu$ . As $\nu(a) > a$ , the values climb, which already gives order-preservation and injectivity; the main work is surjectivity, where we take the least element never reached, which cannot exist because its immediate predecessor is reached and $\nu$ carries that to it.

Proof.

Write $m_0 \defeq \min M$ and for $a \in M$ put $\nu(a) \defeq \min\{m \in M \mid a < m\}$ , which exists because $M$ has no greatest element and is well-ordered. By construction $\nu(a) > a$ and nothing of $M$ lies strictly between $a$ and $\nu(a)$ .

An element $c \neq m_0$ has exactly one immediate predecessor: two of them, say $p < q$ , would put $q$ strictly between $p$ and $c$ . Hence $\nu$ is injective, since $\nu(a) = \nu(b) = c$ makes $a$ and $b$ both immediate predecessors of $c$ ; and $m_0$ is not a value of $\nu$ , since $\nu(a) > a \geqslant m_0$ .

The recursion theorem, applied with $a = m_0$ and $g = \nu$ , gives exactly one $f : \mathbb{N}_0 \to M$ with $f(0) = m_0$ and $f(s(n)) = \nu(f(n))$ . Since $\nu(x) > x$ for every $x$ , induction gives $f(a) < f(b)$ whenever $a < b$ ; in particular $f$ is injective.

For surjectivity, let $A$ be the range of $f$ . Then $m_0 \in A$ , and $A$ is closed under $\nu$ , since $\nu(f(n)) = f(s(n))$ . Suppose $M \setminus A$ is non-empty and let $c$ be its least element. Since $m_0 \in A$ we have $c \neq m_0$ , so $c$ has an immediate predecessor $p$ ; by minimality of $c$ we have $p \in A$ , and $c = \nu(p)$ because nothing lies strictly between. Closure of $A$ under $\nu$ gives $c \in A$ , a contradiction. Hence $f$ is a bijection.

Remark.

The proof never carries out an induction inside $M$ : surjectivity uses well-ordering through a least counterexample, and $f$ itself comes from the recursion theorem on $\mathbb{N}_0$ . One may therefore take well-ordering, the predecessor property and the absence of a greatest element as axioms in place of Peano’s; the theorem says the two systems describe the same ordered set.

Problem 6.31.

Let $S \subset \mathbb{N}_0$ be non-empty. Prove that $S$ has a minimum by applying the maximum principle to the set of lower bounds of $S$ , rather than by appealing to well-ordering.

Problem 6.32.

State and prove the maximum principle for non-empty subsets of $\mathbb{N}$ bounded above in $\mathbb{N}$ , starting the induction at $1$ .

Problem 6.33.

Let $M$ be an ordered set in which every two-element subset has a minimum. Show that the order is total, and deduce that every well-ordered set is totally ordered.

Problem 6.34.

Let $S \subset \mathbb{N}_0$ be such that for every $c \in \mathbb{N}_0$ there is an $s \in S$ with $s > c$ . Prove that there is exactly one bijection $f : \mathbb{N}_0 \to S$ with $f(m) < f(n)$ whenever $m < n$ .

Problem 6.35.

Order $\mathbb{N}_0 \times \mathbb{N}_0$ by declaring $(a, b) < (c, d)$ when $a < c$ , or when $a = c$ and $b < d$ . Show that this is a strict linear order in which every non-empty subset has a least element.

Problem 6.36.

Theorem 6.50 asks three things of $M$ . Exhibit a well-ordered set with no greatest element in which some element other than the minimum has no immediate predecessor, and a well-ordered set in which every element other than the minimum has an immediate predecessor but a greatest element exists. For the first, name the element that has no immediate predecessor.

Problem 6.37.

Exhibit a totally ordered set $L$ with no greatest element in which every element has an immediate predecessor, and which is not well-ordered.
Show that there is no bijection $f : \mathbb{N}_0 \to L$ with $f(m) < f(n)$ whenever $m < n$ , and say where the proof of Theorem 6.50 uses the hypothesis that $L$ fails.
Order $L \times \mathbb{N}_0$ as in Problem 6.35 . Show that the result is totally ordered but not well-ordered.

Sequences

A list is a function whose inputs are positions. The cuts defined for the natural numbers give us the positions, so nothing new is needed.

Definition 6.51 (Sequence).

Let $A$ be a set. A finite sequence in $A$ of length $n$ is a function from $L_n$ to $A$ , written $(a_i)_{i < n}$ or $(a_0, a_1, \ldots)$ , in the notation already used for tuples; the unique sequence of length $0$ is the empty function, written $(\,)$ . An infinite sequence in $A$ is a function from $\mathbb{N}_0$ to $A$ . We write

\operatorname{Seq}(A) \defeq \bigcup_{n \in \mathbb{N}_0} A^{L_n}

for the set of all finite sequences in $A$ , using the Cartesian power of the last chapter but one.

That $\operatorname{Seq}(A)$ is a set takes a moment. Each $A^{L_n}$ consists of functions from $L_n$ to $A$ , hence of subsets of $\mathbb{N}_0 \times A$ , so all of them lie inside $\mathcal{P}(\mathbb{N}_0 \times A)$ ; comprehension collects them and the union axiom pools them.

Remark.

Some texts say instead that a finite sequence is a function whose domain is a natural number. That is the same definition read through the construction of the last chapter, where $n$ was built as the set holding exactly its predecessors, so that $n = L_n$ . We use the cut, which says the same thing without leaning on which sets the numerals happen to be.

Restriction gives the initial pieces of a sequence: for $f$ defined on $\mathbb{N}_0$ and $n \in \mathbb{N}_0$ , the restriction $f|_{L_n}$ is the finite sequence $(f_0, \ldots)$ of length $n$ , recording everything before position $n$ and nothing else.

The General Recursion Theorem

The recursion theorem of the last chapter lets the next value depend on the previous one, and its parametrised form lets it depend on the index as well. Neither is enough when the next value depends on all the earlier ones at once. With sequences we can state such a rule, and with the order we can prove that it defines a function.

Theorem 6.52 (General recursion).

Let $S$ be a set and let $g : \operatorname{Seq}(S) \to S$ . Then there is exactly one $f : \mathbb{N}_0 \to S$ with

f(n) = g\bigl(f|_{L_n}\bigr) \quad\text{for every } n \in \mathbb{N}_0.

Discussion.

The rule wants the whole history at each step, so we recurse on histories rather than on values: instead of building $f$ directly we build the sequence of its initial pieces. Let $G$ send a pair $(t, n)$ to $t$ extended by the value $g(t)$ at position $n$ , whenever $t$ has length $n$ , and to the empty sequence otherwise; that second clause is bookkeeping only, since the case never arises. Parametrised recursion then gives a map $F$ on $\mathbb{N}_0$ with $F(0)$ the empty sequence and $F(s(n)) = G(F(n), n)$ , and an induction shows $F(n)$ has length exactly $n$ , so the bookkeeping clause is never reached. Each $F(n)$ extends its predecessors, so their union is a single function $f$ on $\mathbb{N}_0$ whose restriction to $L_n$ is $F(n)$ , and reading the construction back gives the required identity. Uniqueness uses strong induction: two solutions agreeing everywhere below $n$ have the same restriction to $L_n$ , so $g$ returns the same value at $n$ , and ordinary induction would not deliver that hypothesis.

Proof.

Define $G : \operatorname{Seq}(S) \times \mathbb{N}_0 \to \operatorname{Seq}(S)$ by

G(t, n) \defeq \begin{cases} t \cup \bigl\{(n, g(t))\bigr\} & \text{if } t \text{ has length } n, \\ (\,) & \text{otherwise.} \end{cases}

By parametrised recursion there is a unique $F : \mathbb{N}_0 \to \operatorname{Seq}(S)$ with $F(0) = (\,)$ and $F(s(n)) = G(F(n), n)$ .

We claim $F(n)$ has length $n$ . This holds at $0$ . If $F(n)$ has length $n$ , then $F(s(n)) = F(n) \cup \{(n, g(F(n)))\}$ , whose domain is $L_n \cup \{n\}$ ; and $L_n \cup \{n\} = L_{s(n)}$ , since $x < s(n)$ holds exactly when $x < n$ or $x = n$ , nothing lying strictly between $n$ and $s(n)$ . So $F(s(n))$ has length $s(n)$ , and induction gives the claim.

Each $F(n)$ is therefore a function extending all the earlier ones, so their union $f \defeq \bigcup_{n \in \mathbb{N}_0} F(n)$ is a function on $\mathbb{N}_0$ with $f|_{L_n} = F(n)$ for every $n$ . Then

f(n) = F(s(n))(n) = g\bigl(F(n)\bigr) = g\bigl(f|_{L_n}\bigr),

which is the required identity.

For uniqueness, let $h$ also satisfy the identity, and suppose $f(m) = h(m)$ for every $m < n$ . Then $f|_{L_n} = h|_{L_n}$ , so $f(n) = g(f|_{L_n}) = g(h|_{L_n}) = h(n)$ . Strong induction gives $f = h$ .

Problem 6.38.

Show that the recursion theorem of the last chapter is a special case of this one: given $a \in S$ and $\gamma : S \to S$ , exhibit a $g : \operatorname{Seq}(S) \to S$ whose solution is the map with $f(0) = a$ and $f(s(n)) = \gamma(f(n))$ .

Problem 6.39.

Let $S$ be a set and $g : \operatorname{Seq}(S) \to S$ . Show that the finite sequences $F(n)$ built in the proof satisfy $F(m) \subset F(n)$ whenever $m < n$ , and explain where that is used.

General Associativity

We can now prove something used without proof earlier. Addition was defined for two arguments and proved associative and commutative for two and three; the notation $a_1 + \cdots + a_n$ assumes more, namely that the brackets and the order may be chosen freely. With finite sequences and induction in hand we can say what that means and prove it.

Fix a left-associated reading once and for all: for a finite sequence $a$ of length $n$ in a Peano system with addition, define

p(a) \defeq \begin{cases} 0 & \text{if } n = 0, \\ p\bigl(a|_{L_{k}}\bigr) + a_{k} & \text{if } n = s(k). \end{cases}

So $p$ of the sequence $(a_0, a_1, a_2)$ is $(0 + a_0 + a_1) + a_2$ , brackets to the left. A full bracketing of $a_0 + \cdots + a_{n-1}$ is either the single term, when $n = 1$ , or a sum $L + R$ in which $L$ is a full bracketing of an initial piece and $R$ is a full bracketing of the rest.

Theorem 6.53 (General associativity and neighbour swaps).

Let $a$ be a finite sequence of length $n$ in a Peano system with addition.

Every full bracketing of $a_0 + \cdots + a_{n-1}$ equals $p(a)$ .
If $b$ is obtained from $a$ by exchanging two neighbouring entries, then $p(b) = p(a)$ .

Discussion.

The first part is an induction on $n$ , but a bracketing may split the sum anywhere, so the step needs one auxiliary fact: the left-associated sum of the whole is the left-associated sum of the first $k$ entries added to that of the rest. That splitting identity is itself a short induction on the length of the second block, using nothing but binary associativity, and with it the step applies the hypothesis to the two halves of the bracketing and adds the results.

Once every bracketing is identified with $p$ , the second part may ignore brackets altogether. Exchanging neighbours rearranges a two-term middle and nothing else, so binary commutativity sorts that middle and the splitting identity puts the rest of the list back around it.

Proof.

We first record the splitting identity: for $k < n$ ,

p(a) = p\bigl(a|_{L_k}\bigr) + p\bigl((a_k, \ldots, a_{n-1})\bigr).

Fix $k$ and induct on the length $m$ of the second block. If $m = 1$ this is the defining clause of $p$ . If it holds at $m$ , then appending one more entry gives

p(a) = p\bigl(a|_{L_{n-1}}\bigr) + a_{n-1} = \Bigl(p\bigl(a|_{L_k}\bigr) + p\bigl((a_k, \ldots, a_{n-2})\bigr)\Bigr) + a_{n-1},

and associativity of addition regroups this as $p(a|_{L_k}) + p((a_k, \ldots, a_{n-1}))$ , which is the identity at $m + 1$ .

For the first part, induct on $n$ . The case $n = 1$ is immediate. Let $P$ be a full bracketing of $a_0 + \cdots + a_{n-1}$ , so $P = L + R$ where $L$ brackets the first $k$ entries and $R$ the rest. The inductive hypothesis gives $L = p(a|_{L_k})$ and $R = p((a_k, \ldots, a_{n-1}))$ , so $P = p(a)$ by the splitting identity.

For the second part, let $b$ exchange the entries at positions $k$ and $s(k)$ . The two-term identity $a_k + a_{s(k)} = a_{s(k)} + a_k$ is commutativity. If the exchanged pair is the whole sequence we are done; otherwise two applications of the splitting identity write $p(a)$ as a sum in which that pair appears as a single middle block, and the same decomposition of $p(b)$ differs only in that block, which commutativity identifies with it.

Brackets therefore no longer matter, and neighbouring summands may be exchanged at will. Any rearrangement of a finite list can be reached by a string of neighbour exchanges, moving each entry past its neighbours until it sits where it should, so the value of a finite sum depends on neither the bracketing nor the order of its terms.

Problem 6.40.

Carry out the same argument for multiplication: define the left-associated product $\pi$ by the clauses $\pi((\,)) \defeq 1$ and $\pi(a) \defeq \pi(a|_{L_k}) \cdot a_k$ when $a$ has length $s(k)$ , and prove both parts of the theorem for it. Which laws replace associativity and commutativity of addition at each step?

Problem 6.41.

Show that any rearrangement of a finite sequence is a string of neighbour exchanges, so that the last sentence above is a theorem rather than an appeal to intuition. (Harder.)

Zorn’s Lemma

Finally, two statements equivalent to the axiom of choice, and the reason some mathematicians do not accept it.

Definition 6.54 (Chains and inductively ordered sets).

Let $M$ be a partially ordered set. A subset $A \subset M$ is a chain if the order $M$ induces on $A$ is a total order, that is, if any two elements of $A$ are comparable. The set $M$ is inductively ordered if every chain in $M$ has an upper bound in $M$ .

Remark.

An inductively ordered set is never empty. The empty subset is a chain, vacuously, so it too must have an upper bound, and an upper bound is an element of $M$ .

Theorem 6.55 (Zorn's lemma).

Every inductively ordered set has at least one maximal element.

Discussion.

We state this without proof. It is not a theorem of the axioms we have listed: on the basis of the rest, Zorn’s lemma is equivalent to the axiom of choice, so proving it would amount to deriving one from the other, and the derivation is long enough to belong to a course in set theory rather than here. Note the form of the statement. The hypothesis is about chains, which are the totally ordered pieces of $M$ , and the conclusion is about a maximal element, which need not be a maximum: nothing says the element found is comparable with everything, only that nothing strictly exceeds it. In a partial order a maximal element need not be a maximum, which is why the conclusion is stated this way.

Remark.

The result is due to Kuratowski, though it carries Zorn’s name; Zorn proved the equivalence with choice and was the first to put the statement to work on algebraic questions.

Theorem 6.56 (Well-ordering theorem).

Every non-empty set can be ordered so that every non-empty subset has a least element.

Discussion.

Again we state without proof, for the same reason: on the basis of the other axioms the well-ordering theorem, Zorn’s lemma and the axiom of choice are all equivalent to one another. Of the three it is the hardest to believe. Well-ordering of $\mathbb{N}_0$ was a theorem we proved, resting on the successor structure; here the claim is that any set whatever admits such an order, however little structure it carries and however unlike the natural numbers it looks.

Remark (Informal).

Assuming you know the real numbers, they are not well-ordered by their usual order: the set of those greater than $0$ has no least element, since any candidate is beaten by something smaller still above $0$ . The well-ordering theorem nevertheless asserts that some other order on them is a well-order. No such order has ever been exhibited, and none can be described explicitly; the theorem asserts that one exists and offers no way to find it.

That gap between existence and construction is what leads some mathematicians to reject the axiom of choice. Our position is to accept it, and to say plainly when we use it.

Problem 6.42.

Show that a maximal element need not be a maximum, by exhibiting a partially ordered set with two maximal elements. Show also that in a total order the two notions coincide.

Problem 6.43.

Let $M$ be a set and consider $\mathcal{P}(M) \setminus \{M\}$ ordered by inclusion. Show that it is inductively ordered when $M$ is non-empty, and identify its maximal elements.

Exercises

Exercises in Lean

The proofs below are checked in your browser. Nothing is sent anywhere, and nothing is stored but your own work. Type \to for →, \and for ∧, \< and \> for ⟨ ⟩.

The last sheet gave the checker a Peano system to compute in. This chapter needs two more things of it, and only one of them is new. A relation is a set of ordered pairs, and pairs, products and sets are all already there; the order on $\mathbb{N}_0$ is not, and arrives as notation below.

Relations

A relation on $S$ is a subset of $S \times S$ , so $x \mathbin{R} y$ is typed as the membership it abbreviates, (x, y) ∈ R. The three properties of Definition 6.2 are not names the checker knows; they are written out as the statements they are, so symmetry of $R$ enters a problem as the hypothesis

hs : ∀ a b : Obj, (a, b) ∈ R → (b, a) ∈ R

and is used by applying it to two objects and a membership, left to right: hs x y h.

One piece of notation is new. The separation axiom carves a subset out of a set by a criterion, and it is written {x ∈ A | p x}, with the membership criterion the axiom gives: y ∈ {x ∈ A | p x} is y ∈ A ∧ p y, in the way that x ∈ A ∩ B was a conjunction on an earlier sheet. So ⟨_, _⟩ builds a membership and .left and .right take one apart. This is what lets an equivalence class be written down: $[a]$ of Definition 6.26 is {s ∈ S | (s, a) ∈ R}.

Example.

The universal relation on $S$ is reflexive. Membership in a product is a conjunction, and here both halves are the same hypothesis.

Example.

Reflexivity puts $a$ in its own class. The two halves of the criterion are the two things the class asks of $a$ .

Exercise 6.11.

The empty relation is symmetric and transitive.

Exercise 6.12.

And the universal relation is transitive.

Exercise 6.13.

Symmetry and transitivity give reflexivity, but only at the elements the relation already reaches.

lean proof

1example (R x y : Obj)2    (hs : ∀ a b : Obj, (a, b) ∈ R → (b, a) ∈ R)3    (ht : ∀ a b c : Obj, (a, b) ∈ R → (b, c) ∈ R → (a, c) ∈ R)4    (h : (x, y) ∈ R) : (x, x) ∈ R ∧ (y, y) ∈ R := by

5  sorry

goalR x y : Obj hs : ∀ (a : Obj), ∀ (b : Obj), (a, b) ∈ R → (b, a) ∈ R ht : ∀ (a : Obj), ∀ (b : Obj), ∀ (c : Obj), (a, b) ∈ R → (b, c) ∈ R → (a, c) ∈ R h : (x, y) ∈ R ⊢ (x, x) ∈ R ∧ (y, y) ∈ R

Exercise 6.14.

The first half of Problem 6.3 , with $T$ written out.

lean proof

1example (S R x y : Obj)2    (hs : ∀ a b : Obj, (a, b) ∈ R → (b, a) ∈ R)3    (ht : ∀ a b c : Obj, (a, b) ∈ R → (b, c) ∈ R → (a, c) ∈ R)4    (hx : x ∈ S) (h : (x, y) ∈ R) : x ∈ {t ∈ S | (t, t) ∈ R} := by

5  sorry

goalS R x y : Obj hs : ∀ (a : Obj), ∀ (b : Obj), (a, b) ∈ R → (b, a) ∈ R ht : ∀ (a : Obj), ∀ (b : Obj), ∀ (c : Obj), (a, b) ∈ R → (b, c) ∈ R → (a, c) ∈ R hx : x ∈ S h : (x, y) ∈ R ⊢ x ∈ {t ∈ S | (t, t) ∈ R}

Exercise 6.15.

Symmetry survives intersection.

Exercise 6.16.

Proposition 6.22 , with the order written as a relation $L$ and antisymmetry as a hypothesis.

lean proof

1example (S L m n : Obj)2    (hanti : ∀ a b : Obj, (a, b) ∈ L → (b, a) ∈ L → a = b)3    (hm : m ∈ S ∧ ∀ x : Obj, x ∈ S → (m, x) ∈ L)4    (hn : n ∈ S ∧ ∀ x : Obj, x ∈ S → (n, x) ∈ L) : m = n := by

5  sorry

goalS L m n : Obj hanti : ∀ (a : Obj), ∀ (b : Obj), (a, b) ∈ L → (b, a) ∈ L → a = b hm : m ∈ S ∧ (∀ (x : Obj), x ∈ S → (m, x) ∈ L) hn : n ∈ S ∧ (∀ (x : Obj), x ∈ S → (n, x) ∈ L) ⊢ m = n

Exercise 6.17.

The step $2 \implies 1$ of Proposition 6.28 : classes that meet come from related elements.

lean proof

1example (S R a b c : Obj)2    (hs : ∀ x y : Obj, (x, y) ∈ R → (y, x) ∈ R)3    (ht : ∀ x y z : Obj, (x, y) ∈ R → (y, z) ∈ R → (x, z) ∈ R)4    (h : c ∈ {s ∈ S | (s, a) ∈ R} ∩ {s ∈ S | (s, b) ∈ R}) : (a, b) ∈ R := by

5  sorry

goalS R a b c : Obj hs : ∀ (x : Obj), ∀ (y : Obj), (x, y) ∈ R → (y, x) ∈ R ht : ∀ (x : Obj), ∀ (y : Obj), ∀ (z : Obj), (x, y) ∈ R → (y, z) ∈ R → (x, z) ∈ R h : c ∈ {s ∈ S | (s, a) ∈ R} ∩ {s ∈ S | (s, b) ∈ R} ⊢ (a, b) ∈ R

Exercise 6.18.

And the step $1 \implies 3$ , which is an equality of sets.

lean proof

1example (S R a b : Obj)2    (hs : ∀ x y : Obj, (x, y) ∈ R → (y, x) ∈ R)3    (ht : ∀ x y z : Obj, (x, y) ∈ R → (y, z) ∈ R → (x, z) ∈ R)4    (h : (a, b) ∈ R) : {s ∈ S | (s, a) ∈ R} = {s ∈ S | (s, b) ∈ R} := by

5  sorry

goalS R a b : Obj hs : ∀ (x : Obj), ∀ (y : Obj), (x, y) ∈ R → (y, x) ∈ R ht : ∀ (x : Obj), ∀ (y : Obj), ∀ (z : Obj), (x, y) ∈ R → (y, z) ∈ R → (x, z) ∈ R h : (a, b) ∈ R ⊢ {s ∈ S | (s, a) ∈ R} = {s ∈ S | (s, b) ∈ R}

The order on the natural numbers

ℕ is the carrier of the last sheet, and the order is written ≤ and <, typed \le and <. Neither is a name in the library: they are the definitions of Definition 6.32 and Proposition 6.35 outright, so

m ≤ n   is   ∃ b : ℕ, n = m + b
m < n   is   ∃ b : ℕ, b ≠ 0 ∧ n = m + b

and every tactic for an existential acts on them unchanged: use supplies the difference, obtain ⟨b, hb, he⟩ takes a strict inequality apart into a positive $b$ and the equation it satisfies, and obtain ⟨b, he⟩ does the same for a non-strict one. The reverse orders > and ≥ are these two read backwards, as Definition 6.14 has them. Numerals are still spelled as the last sheet spelled them, so $1$ is succ 0.

Example.

The difference that makes an element reach itself is $0$ , and the first clause of addition checks it.

Example.

Corollary 6.33 . Once the difference is named, one half is the hypothesis and the other holds of itself.

Exercise 6.19.

The second half of Proposition 6.36 .

Exercise 6.20.

Every element is below its successor.

Exercise 6.21.

A positive difference is a difference.

Exercise 6.22.

Proposition 6.8 , proved here from uniqueness of differences rather than from trichotomy.

Exercise 6.23.

Transitivity: two positive differences compose into one.

Exercise 6.24.

Antisymmetry, which is the second law of Proposition 6.11 .

Exercise 6.25.

No comparison points both ways.

Exercise 6.26.

The first part of Problem 6.16 .

Exercise 6.27.

And the second, with the factor on the left so that distributivity applies as it stands.

Exercise 6.28.

Problem 6.15 . Both directions move a successor between the two sides of a sum.

Exercise 6.29.

Problem 6.17 .

Exercise 6.30.

Proposition 6.39 . (Harder.)

What the checker understands

Tactics

intro h	assume the hypothesis of an implication, naming it h
exact e	give the proof outright
apply f	reduce the goal to the hypotheses of f
assumption	close the goal with a hypothesis already present
trivial	close the goal True
exfalso	replace the goal with False
by_contra h	assume the negation of the goal
constructor	split ∧ into both halves, or ↔ into both directions
left / right	choose which half of a ∨ to prove
rcases h with a \| b	argue by cases on a disjunction
obtain ⟨a, b⟩ := h	take a conjunction or an existential apart
cases h	as above, keeping the name
refine e	give the proof with holes left in it
have h : p := …	record an intermediate result
show p	restate the goal in an equal form
use w	give a witness for ∃
specialize h a	instantiate a ∀ hypothesis
rw [h]	rewrite with an equation, ← to go backwards
rfl	both sides compute to the same thing
decide / norm_num	settle a closed computation
tauto	close a goal that is true by pure logic
induction n with k ih	the fifth Peano condition: prove the goal at 0, then at succ k from ih

Results you may cite

Classical.em	∀ (a : Prop), a ∨ ¬a — the law of excluded middle
Classical.byContradiction	∀ {a : Prop}, (¬a → False) → a — proof by contradiction; the tactic by_contra does this for you
Classical.byCases	∀ {a b : Prop}, (a → b) → (¬a → b) → b — split on whether a holds
not_not	∀ {a : Prop}, ¬¬a ↔ a — double negation
not_and_or	∀ {a b : Prop}, ¬(a ∧ b) ↔ ¬a ∨ ¬b — De Morgan
not_or	∀ {a b : Prop}, ¬(a ∨ b) ↔ ¬a ∧ ¬b — De Morgan
not_imp	∀ {a b : Prop}, ¬(a → b) ↔ a ∧ ¬b
and_comm	∀ {a b : Prop}, a ∧ b ↔ b ∧ a
or_comm	∀ {a b : Prop}, a ∨ b ↔ b ∨ a
Set.ext	∀ {A B : Obj}, (∀ x : Obj, x ∈ A ↔ x ∈ B) → A = B — extensionality: sets with the same elements are equal
Set.ext_iff	∀ {A B : Obj}, A = B ↔ (∀ x : Obj, x ∈ A ↔ x ∈ B) — extensionality and substitution, in one biconditional
Set.subset_antisymm	∀ {A B : Obj}, A ⊆ B → B ⊆ A → A = B — mutual inclusion is equality
Set.empty_subset	∀ {A : Obj}, ∅ ⊆ A — the empty set is a subset of every set
Set.pair_eq	∀ {a b c d : Obj}, ((a, b) = (c, d)) ↔ (a = c ∧ b = d) — two ordered pairs are equal exactly when their coordinates are
Nat.succ_inj	∀ {m n : ℕ}, succ m = succ n → m = n — the third Peano condition: the successor is injective
Nat.succ_ne_zero	∀ (n : ℕ), succ n ≠ 0 — the fourth: zero is nobody's successor
Nat.pred	∀ {n : ℕ}, n ≠ 0 → ∃ m : ℕ, n = succ m — predecessors: everything but zero is a successor
Nat.add_zero	∀ (m : ℕ), m + 0 = m — the first clause of addition
Nat.add_succ	∀ (m n : ℕ), m + succ n = succ (m + n) — the second clause of addition
Nat.zero_add	∀ (n : ℕ), 0 + n = n — addition from the left
Nat.succ_add	∀ (m n : ℕ), succ m + n = succ (m + n) — addition from the left, at a successor
Nat.add_assoc	∀ (m n p : ℕ), (m + n) + p = m + (n + p) — addition is associative
Nat.add_comm	∀ (m n : ℕ), m + n = n + m — addition is commutative
Nat.add_ne_zero	∀ {a : ℕ} (b : ℕ), a ≠ 0 → a + b ≠ 0 — positivity is absorbing
Nat.mul_zero	∀ (m : ℕ), m * 0 = 0 — the first clause of multiplication
Nat.mul_succ	∀ (m n : ℕ), m * succ n = m * n + m — the second clause of multiplication
Nat.zero_mul	∀ (m : ℕ), 0 * m = 0 — multiplication from the left
Nat.succ_mul	∀ (m n : ℕ), succ m * n = m * n + n — multiplication from the left, at a successor
Nat.add_right_cancel	∀ {m n k : ℕ}, m + k = n + k → m = n — cancellation, from the last sheet
Nat.add_eq_zero	∀ {m n : ℕ}, m + n = 0 → m = 0 ∧ n = 0 — a sum is zero only when both parts are, from the last sheet
Nat.mul_comm	∀ (m n : ℕ), m * n = n * m — multiplication is commutative, from the last sheet
Nat.mul_add	∀ (m n p : ℕ), m * (n + p) = m * n + m * p — multiplication distributes over addition, from the last sheet
Nat.mul_assoc	∀ (m n p : ℕ), (m * n) * p = m * (n * p) — multiplication associates, from the problems of the last chapter
Nat.add_mul	∀ (m n p : ℕ), (m + n) * p = m * p + n * p — distributivity on the other side
Nat.add_left_cancel	∀ {a m n : ℕ}, a + m = a + n → m = n — uniqueness of differences
Nat.lt_trichotomy	∀ (m n : ℕ), m < n ∨ m = n ∨ n < m — trichotomy, from the theorem that ℕ is strictly ordered

From the logical core

And.intro	∀ {a b : Prop}, a → b → a ∧ b
And.left	∀ {a b : Prop}, a ∧ b → a
And.right	∀ {a b : Prop}, a ∧ b → b
And.symm	∀ {a b : Prop}, a ∧ b → b ∧ a
Or.inl	∀ {a b : Prop}, a → a ∨ b
Or.inr	∀ {a b : Prop}, b → a ∨ b
Or.elim	∀ {a b c : Prop}, a ∨ b → (a → c) → (b → c) → c
Or.symm	∀ {a b : Prop}, a ∨ b → b ∨ a
Iff.intro	∀ {a b : Prop}, (a → b) → (b → a) → (a ↔ b)
Iff.mp	∀ {a b : Prop}, (a ↔ b) → a → b
Iff.mpr	∀ {a b : Prop}, (a ↔ b) → b → a
Iff.symm	∀ {a b : Prop}, (a ↔ b) → (b ↔ a)
Iff.rfl	∀ {a : Prop}, a ↔ a
Iff.trans	∀ {a b c : Prop}, (a ↔ b) → (b ↔ c) → (a ↔ c)
True.intro	True
False.elim	∀ {a : Prop}, False → a
absurd	∀ {a b : Prop}, a → ¬a → b
id	∀ {a : Prop}, a → a
mt	∀ {a b : Prop}, (a → b) → ¬b → ¬a
Eq.refl	∀ {α : Type} (a : α), a = a
Eq.symm	∀ {α : Type} {a b : α}, a = b → b = a
Eq.trans	∀ {α : Type} {a b c : α}, a = b → b = c → a = c
congrArg	∀ {α : Type} {β : Type} {a b : α} (f : α → β), a = b → f a = f b
Exists.intro	∀ {α : Type} {p : α → Prop} (w : α), p w → ∃ x : α, p x
Exists.elim	∀ {α : Type} {p : α → Prop} {b : Prop}, (∃ x : α, p x) → (∀ y : α, p y → b) → b

Lesson 7

Infinite Sets

Taught 17 Aug 2026

Finite Sets

To count a herd is to pair its animals off against $0, 1, 2$ and so on, and to report the number we stopped at. Nothing in that procedure asks what the counting numbers are; it asks for a bijection between the herd and an opening stretch of them. Both are now built, so we can count, and the notion of the same size that counting rests on still makes sense for sets far too large to count.

Comparing Sets by Functions

Definition 7.1 (Equinumerous sets).

Sets $A$ and $B$ are equinumerous, written $A \approx B$ , if there is a bijection $f : A \to B$ . We also say that $A$ and $B$ have the same size.

The definition asks for one bijection and says nothing about how many there are; two sets of three elements are matched up in six different ways, and the definition is satisfied by any of them. It also does not mention counting, so it still makes sense for sets we cannot count.

Proposition 7.2 (Equinumerosity is reflexive, symmetric and transitive).

Let $A$ , $B$ and $C$ be sets.

$A \approx A$ .
If $A \approx B$ then $B \approx A$ .
If $A \approx B$ and $B \approx C$ then $A \approx C$ .

Discussion.

Each part is an existence claim about bijections, so each is proved by producing one, and in every case the bijection we need has already been built. For the first, the identity map on $A$ is its own inverse, and a map with an inverse is a bijection by the theorem on invertibility. The second and third are conditionals, so we start from the bijections the hypothesis gives: from one bijection, the theorem that the inverse is a bijection; from two, the theorem that bijections compose. No part uses anything about the elements of the sets.

Proof.

For the first, $\mathrm{id}_A \circ \mathrm{id}_A = \mathrm{id}_A$ , so $\mathrm{id}_A$ is invertible and hence a bijection from $A$ to $A$ .

For the second, let $f : A \to B$ be a bijection. Its inverse $f^{-1} : B \to A$ is a bijection, so $B \approx A$ .

For the third, let $f : A \to B$ and $g : B \to C$ be bijections. Then $g \circ f : A \to C$ is a bijection, so $A \approx C$ .

Remark.

Those are the three properties of an equivalence relation, and yet $\approx$ is not one, because it is not a relation at all in our sense: a relation on $S$ is a subset of $S \times S$ , and there is no set of all sets for $S$ to be. Fix a set $X$ , however, and $\approx$ restricted to $\mathcal{P}(X)$ is an honest equivalence relation on an honest set, so it partitions $\mathcal{P}(X)$ into classes of subsets of the same size. That is how the notion is used in practice, and the general statement is a convenience of language rather than a claim about a set of pairs.

Example 7.3.

Write $L_n$ for the cut $\{x \in \mathbb{N}_0 \mid x < n\}$ , so that $L_3 = \{0, 1, 2\}$ . If $a$ , $b$ and $c$ are distinct then $\{a, b, c\} \approx L_3$ , witnessed by $a \mapsto 0$ , $b \mapsto 1$ , $c \mapsto 2$ . Any other listing of the three letters gives another bijection, and the definition does not prefer one.

Problem 7.1.

Show that $A \times B \approx B \times A$ for all sets $A$ and $B$ , and that $(A \times B) \times C \approx A \times (B \times C)$ .

Problem 7.2.

Show that $A \approx B$ implies $\mathcal{P}(A) \approx \mathcal{P}(B)$ . Which of the theorems on images and preimages does your bijection rest on?

Counting

The cuts are the sets we count against: $L_n$ collects the $n$ numbers below $n$ . First we record how they grow, which every induction below uses.

Proposition 7.4 (The cuts grow by one point).

$L_0 = \emptyset$ , and for every $n \in \mathbb{N}_0$ ,

L_{s(n)} = L_n \cup \{n\}, \qquad n \notin L_n.

Discussion.

There are three assertions and each is a statement about which elements a cut holds, so each unfolds to a comparison with $n$ . That $L_0$ is empty is a negative claim, ruled out by the proposition that every natural number is at least zero together with trichotomy, which forbids $x < 0$ once $0 \leqslant x$ is known. The equation is an equality of sets, so it splits into two inclusions, and both come from asking where $x$ sits relative to $n$ : to the left, $x < s(n)$ leaves the alternatives $x < n$ and $x = n$ once the proposition that nothing lies strictly between $n$ and $s(n)$ has removed the third; to the right, $x < n$ or $x = n$ gives $x < s(n)$ because $n < s(n)$ and $<$ is transitive. The last assertion is anti-reflexivity read off the definition of the cut.

Proof.

An $x \in \mathbb{N}_0$ satisfies $0 \leqslant x$ , so $x < 0$ would violate trichotomy; hence $L_0 = \emptyset$ .

Let $x < s(n)$ . Trichotomy gives $x < n$ , $x = n$ or $n < x$ , and the last would place $x$ strictly between $n$ and $s(n)$ , which nothing does. So $x \in L_n \cup \{n\}$ . Conversely $n < s(n)$ by the corollary that adding a positive element moves you up, so $x = n$ gives $x < s(n)$ , and $x < n$ gives $x < s(n)$ by transitivity. Hence $L_{s(n)} = L_n \cup \{n\}$ .

Finally $\neg(n < n)$ , so $n \notin L_n$ .

Definition 7.5 (Finite and infinite sets).

A set $A$ is finite if $A \approx L_n$ for some $n \in \mathbb{N}_0$ , and infinite otherwise.

So $\emptyset$ is finite, since $L_0 = \emptyset$ , and each $L_n$ is finite by way of its identity map. The definition offers some $n$ , and to speak of the number of elements we must know that no two cuts are matched by a bijection. The next few results prove this. The main tool is the simplest rearrangement: a map exchanging two points and leaving the rest alone.

Definition 7.6 (Transposition).

Let $S$ be a set and let $a, b \in S$ . The transposition $\tau_{a,b} : S \to S$ is the function given by

\tau_{a,b}(a) \defeq b, \qquad \tau_{a,b}(b) \defeq a, \qquad \tau_{a,b}(x) \defeq x \ \text{ for every other } x \in S.

When $a = b$ the three clauses agree and $\tau_{a,b}$ is the identity map on $S$ .

Proposition 7.7 (A transposition is its own inverse).

Let $S$ be a set and let $a, b \in S$ . Then $\tau_{a,b} \circ \tau_{a,b} = \mathrm{id}_S$ , and $\tau_{a,b}$ is a bijection.

Discussion.

Two assertions, the second following from the first. The first is an equality of functions, so by equality of functions we compare values at an arbitrary point of $S$ , and the definition splits the points into three cases, each settled by reading the definition twice. The second is then the theorem on invertibility, which makes a map with a two-sided inverse a bijection; the first assertion offers $\tau_{a,b}$ itself as that inverse, so nothing further need be built.

Proof.

Write $\tau \defeq \tau_{a,b}$ and let $x \in S$ . If $x = a$ then $\tau(\tau(a)) = \tau(b) = a$ ; if $x = b$ then $\tau(\tau(b)) = \tau(a) = b$ ; and otherwise $\tau(\tau(x)) = \tau(x) = x$ . So $\tau \circ \tau = \mathrm{id}_S$ , and $\tau$ is invertible, hence a bijection from $S$ to $S$ .

Proposition 7.8 (Bijections remove points).

Let $f : A \to B$ be a bijection and let $a \in A$ . Then $A \setminus \{a\} \approx B \setminus \{f(a)\}$ .

Discussion.

We need a bijection between the two smaller sets, and the only map we have is $f$ , so we restrict it and check the two halves of bijectivity. Injectivity is inherited, since a restriction of an injective map is injective. For the values, injectivity is again what we use: a point other than $a$ cannot be sent to $f(a)$ , so the restriction does land in $B \setminus \{f(a)\}$ ; and surjectivity of $f$ supplies, for each $y$ in that set, a preimage, which cannot be $a$ because $y \neq f(a)$ .

Proof.

Let $g$ be the restriction of $f$ to $A \setminus \{a\}$ . If $x \neq a$ then $f(x) \neq f(a)$ by injectivity, so $g$ takes its values in $B \setminus \{f(a)\}$ , and $g$ is injective because $f$ is.

Let $y \in B \setminus \{f(a)\}$ . By surjectivity $y = f(x)$ for some $x \in A$ , and $x \neq a$ since $y \neq f(a)$ . So $y = g(x)$ , and $g$ is a bijection onto $B \setminus \{f(a)\}$ .

Proposition 7.9 (Removing a point from a cut).

Let $n \in \mathbb{N}_0$ and let $k \in L_{s(n)}$ . Then $L_{s(n)} \setminus \{k\} \approx L_n$ .

Discussion.

The previous proposition removes a point and its image under a bijection, so the plan is to build a bijection of $L_{s(n)}$ with itself that carries $k$ to $n$ and then remove $n$ , where the result is the cut $L_n$ by the proposition on how cuts grow. The transposition of $k$ and $n$ does exactly that, and it is a bijection by the proposition just proved, so nothing needs checking. The case $k = n$ is the growth proposition on its own, the transposition there being the identity.

Proof.

If $k = n$ then $L_{s(n)} \setminus \{n\} = L_n$ by the proposition on how cuts grow, and a set is equinumerous with itself.

Suppose $k \neq n$ and let $\tau \defeq \tau_{k,n}$ be the transposition of $k$ and $n$ on $L_{s(n)}$ , which is a bijection by the proposition above. Since $\tau(k) = n$ , the proposition on removing points gives

L_{s(n)} \setminus \{k\} \approx L_{s(n)} \setminus \{n\} = L_n.

Theorem 7.10 (No cut injects into a shorter one).

For every $n \in \mathbb{N}_0$ there is no injection $f : L_{s(n)} \to L_n$ .

Discussion.

The claim is a universally quantified non-existence statement, so we induct on $n$ and argue each case by contradiction. At $n = 0$ the target is empty while the source is not, and a function must name a value at every point of its domain, so there is no such function. For the step we assume no injection $L_{s(n)} \to L_n$ and suppose one is given from $L_{s(s(n))}$ to $L_{s(n)}$ . The extra point of the source is $s(n)$ , and by injectivity no other point is sent to its value $c$ , so removing both leaves an injection from $L_{s(n)}$ into $L_{s(n)} \setminus \{c\}$ . The previous proposition says that set is a copy of $L_n$ , and composing with a bijection onto $L_n$ gives an injection the inductive hypothesis forbids.

Proof.

We induct on $n$ . For $n = 0$ the cut $L_1$ is $\{0\}$ and $L_0$ is empty, so a function $L_1 \to L_0$ would have to give a value $f(0) \in \emptyset$ . There is none.

Suppose there is no injection $L_{s(n)} \to L_n$ , and let $f : L_{s(s(n))} \to L_{s(n)}$ be injective. Since $s(n) < s(s(n))$ , the value $c \defeq f(s(n))$ is defined. If $x < s(n)$ then $x \neq s(n)$ , so $f(x) \neq c$ by injectivity; hence the restriction of $f$ to $L_{s(n)}$ is an injection taking its values in $L_{s(n)} \setminus \{c\}$ . Composing it with a bijection from $L_{s(n)} \setminus \{c\}$ onto $L_n$ , which the previous proposition supplies, yields an injection $L_{s(n)} \to L_n$ , contrary to the hypothesis. So no such $f$ exists, and induction completes the proof.

Corollary 7.11 (Cuts of different length are not equinumerous).

If $L_m \approx L_n$ then $m = n$ .

Proof.

Suppose $m < n$ . Then $L_m \subset L_n$ by transitivity, and $m \in L_n$ , so $L_{s(m)} = L_m \cup \{m\} \subset L_n$ . A bijection $L_n \to L_m$ restricted to $L_{s(m)}$ is then an injection $L_{s(m)} \to L_m$ , which the theorem forbids. The same argument with $m$ and $n$ exchanged rules out $n < m$ , so trichotomy leaves $m = n$ .

Definition 7.12 (Cardinality of a finite set).

Let $A$ be finite. The unique $n \in \mathbb{N}_0$ with $A \approx L_n$ is the cardinality of $A$ , written $\#A$ , and we say $A$ has $n$ elements.

Uniqueness is the corollary, and without it the notation would not be well defined. Some authors write $|A|$ for the same number.

Corollary 7.13 (Cardinality classifies finite sets).

Let $A$ and $B$ be finite. Then $A \approx B$ if and only if $\#A = \#B$ .

Proof.

Write $m \defeq \#A$ and $n \defeq \#B$ , so $A \approx L_m$ and $B \approx L_n$ . If $A \approx B$ then $L_m \approx A \approx B \approx L_n$ by symmetry and transitivity, so $m = n$ by the previous corollary. Conversely if $m = n$ then $A \approx L_m = L_n \approx B$ .

Example 7.14.

$\#\emptyset = 0$ and $\#L_n = n$ . A finite set has cardinality $0$ exactly when it is empty, since a bijection onto $L_0 = \emptyset$ from a non-empty set would have to name an element of $\emptyset$ . If $a \neq b$ then $\#\{a\} = 1$ and $\#\{a, b\} = 2$ , since $L_1 = \{0\}$ and $L_2 = \{0, 1\}$ .

Proposition 7.15 (Adjoining a point).

Let $A$ be finite and let $b \notin A$ . Then $A \cup \{b\}$ is finite and $\#(A \cup \{b\}) = s(\#A)$ .

Discussion.

Both assertions come from one bijection, which we build and check. We hold a bijection $g$ from $A$ onto $L_n$ , and the set to be counted has exactly one extra point, while the cut $L_{s(n)}$ has exactly one point more than $L_n$ , namely $n$ itself. So we extend $g$ by sending $b$ to $n$ . That the extension is a function uses $b \notin A$ , which stops the two clauses from disagreeing anywhere; that it is injective uses $n \notin L_n$ , which stops the new value from repeating an old one; and that it is surjective is the growth proposition, which says $L_{s(n)}$ holds nothing beyond $L_n$ and $n$ .

Proof.

Put $n \defeq \#A$ and let $g : A \to L_n$ be a bijection. Define $g' : A \cup \{b\} \to L_{s(n)}$ by $g'(x) \defeq g(x)$ for $x \in A$ and $g'(b) \defeq n$ . Since $b \notin A$ , every point of the domain falls under exactly one clause, so $g'$ is a function, and its values lie in $L_{s(n)} = L_n \cup \{n\}$ .

For injectivity, two points of $A$ are separated by $g$ , and a point of $A$ is separated from $b$ because $g(x) \in L_n$ while $n \notin L_n$ . For surjectivity, an element of $L_{s(n)}$ is either $n$ , which is $g'(b)$ , or an element of $L_n$ , which is $g(x)$ for some $x \in A$ by surjectivity of $g$ . So $A \cup \{b\} \approx L_{s(n)}$ .

Corollary 7.16 (Removing a point).

Let $A$ be finite and non-empty and let $a \in A$ . Then $A \setminus \{a\}$ is finite and $\#A = s\bigl(\#(A \setminus \{a\})\bigr)$ .

Proof.

Let $f : A \to L_n$ be a bijection, where $n \defeq \#A$ . Since $A$ is non-empty, $n \neq 0$ , so the theorem on predecessors gives $n = s(k)$ for some $k$ . The proposition on removing points from bijections gives $A \setminus \{a\} \approx L_{s(k)} \setminus \{f(a)\}$ , and the proposition on removing a point from a cut gives $L_{s(k)} \setminus \{f(a)\} \approx L_k$ . So $A \setminus \{a\}$ is finite with cardinality $k$ , and $\#A = s(k)$ .

Subsets and Images

Theorem 7.17 (Subsets of a finite set).

Let $A$ be finite and let $B \subset A$ . Then $B$ is finite and $\#B \leqslant \#A$ . If moreover $B \subsetneq A$ then $\#B < \#A$ .

Discussion.

The first sentence quantifies over all subsets of all finite sets, so we induct on the cardinality, taking for the predicate at $n$ the statement that every subset of every set with $n$ elements is finite with cardinality at most $n$ . At $n = 0$ the ambient set is empty and so is the subset. For the step we peel off a point $a$ of the ambient set, leaving a set the hypothesis governs, and split on whether $a$ belongs to the subset: if not, the subset is already inside the smaller set; if so, we apply the hypothesis to the subset with $a$ removed and put $a$ back, which raises both cardinalities by one successor.

The second sentence uses properness. A point $a$ of $A$ outside $B$ can be adjoined to $B$ without leaving $A$ , so the first sentence applies to $B \cup \{a\}$ and yields $s(\#B) \leqslant \#A$ ; and $s(m) \leqslant n$ forces $m < n$ , since $s(m) + b = m + s(b)$ and $s(b)$ is positive.

Proof.

We first record two small facts about successors, both read off the clauses for addition. If $m \leqslant n$ then $n = m + b$ for some $b \in \mathbb{N}_0$ , so $s(n) = s(m + b) = s(m) + b$ by addition from the left, and hence $s(m) \leqslant s(n)$ . If $s(m) \leqslant n$ then $n = s(m) + b = s(m + b) = m + s(b)$ , and $s(b)$ is positive, so $m < n$ .

For the first assertion we induct on $n$ , proving that every subset of every set of cardinality $n$ is finite with cardinality at most $n$ . If $n = 0$ then the ambient set is empty, so $B = \emptyset$ and $\#B = 0$ .

Suppose the claim holds at $n$ , let $\#A = s(n)$ and let $B \subset A$ . Since $s(n) \neq 0$ , the set $A$ is non-empty; choose $a \in A$ and put $A' \defeq A \setminus \{a\}$ , which has cardinality $n$ by the corollary on removing a point. If $a \notin B$ then $B \subset A'$ , so the hypothesis makes $B$ finite with $\#B \leqslant n \leqslant s(n)$ . If $a \in B$ then $B \setminus \{a\} \subset A'$ , so $B \setminus \{a\}$ is finite with cardinality $m \leqslant n$ ; adjoining $a$ makes $B$ finite with $\#B = s(m) \leqslant s(n)$ .

For the second assertion, let $B \subsetneq A$ and choose $a \in A \setminus B$ . Then $B \cup \{a\} \subset A$ , so the first assertion gives $s(\#B) = \#(B \cup \{a\}) \leqslant \#A$ , and therefore $\#B < \#A$ .

Corollary 7.18 (No finite set is equinumerous with a proper subset).

Let $A$ be finite and $B \subsetneq A$ . Then $A \not\approx B$ .

Proof.

The theorem gives $\#B < \#A$ , so $\#B \neq \#A$ by anti-reflexivity, and finite sets of different cardinality are not equinumerous.

Proposition 7.19 (Images of a finite set).

Let $A$ be finite and let $f : A \to B$ be a function. Then the image $f(A)$ is finite with $\#f(A) \leqslant \#A$ , and $\#f(A) = \#A$ if and only if $f$ is injective.

Discussion.

The inequality is proved by induction on $\#A$ : remove a point $a$ from $A$ , apply the hypothesis to what remains, and observe that the image gains at most the single value $f(a)$ , so it grows by at most one successor.

The biconditional needs no second induction, because each direction is already available. If $f$ is injective it is a bijection onto its range, so the two cardinalities agree. If $f$ is not injective, two distinct points share a value, so deleting one of them changes nothing about the image; the inequality applied to the smaller set then gives a strict drop, since a proper subset of a finite set is strictly smaller. Those two together are the biconditional, the second read contrapositively.

Proof.

For the inequality we induct on $n = \#A$ . If $n = 0$ then $A = \emptyset$ and $f(A) = \emptyset$ .

Suppose the claim holds for sets of cardinality $n$ and let $\#A = s(n)$ . Choose $a \in A$ and put $A' \defeq A \setminus \{a\}$ , of cardinality $n$ . Since $A = A' \cup \{a\}$ we have $f(A) = f(A') \cup \{f(a)\}$ , and $f(A')$ is finite with $\#f(A') \leqslant n$ by hypothesis. If $f(a) \in f(A')$ then $f(A) = f(A')$ and $\#f(A) \leqslant n \leqslant s(n)$ . Otherwise adjoining the point gives $\#f(A) = s(\#f(A')) \leqslant s(n)$ .

If $f$ is injective then $f$ is a bijection from $A$ onto $f(A)$ , so $\#f(A) = \#A$ . If $f$ is not injective, choose $x \neq y$ in $A$ with $f(x) = f(y)$ . Every value of $f$ is then already taken on $A \setminus \{x\}$ , since $f(x) = f(y)$ and $y \neq x$ , so $f(A) = f(A \setminus \{x\})$ and

\#f(A) \leqslant \#(A \setminus \{x\}) < \#A,

the second inequality because $A \setminus \{x\}$ is a proper subset. So equality holds exactly when $f$ is injective.

The Pigeonhole Principle

Theorem 7.20 (Injections, surjections and size).

Let $A$ and $B$ be finite.

If there is an injection $f : A \to B$ then $\#A \leqslant \#B$ .
If there is a surjection $f : A \to B$ then $\#B \leqslant \#A$ .

Discussion.

Both parts turn the given function into a statement about the image, and then apply the two results just proved. An injection is a bijection onto its range, so $A$ and $f(A)$ have the same cardinality, and $f(A)$ is a subset of $B$ , which the theorem on subsets bounds by $\#B$ . A surjection has range all of $B$ , so $\#B$ is the cardinality of an image, which the proposition on images bounds by $\#A$ . Notice that no choice of preimages is made anywhere, so neither part appeals to the axiom of choice.

Proof.

For the first, $f$ is a bijection from $A$ onto $f(A)$ , so $\#A = \#f(A)$ ; and $f(A) \subset B$ , so $\#f(A) \leqslant \#B$ by the theorem on subsets.

For the second, surjectivity gives $f(A) = B$ , so $\#B = \#f(A) \leqslant \#A$ by the proposition on images.

Corollary 7.21 (Pigeonhole principle).

Let $A$ and $B$ be finite with $\#B < \#A$ . Then no function $A \to B$ is injective, and no function $B \to A$ is surjective.

Proof.

An injection $A \to B$ would give $\#A \leqslant \#B$ , and a surjection $B \to A$ would give $\#A \leqslant \#B$ as well; either contradicts $\#B < \#A$ by trichotomy.

Example 7.22.

Among thirteen people, two were born in the same month: the map sending each person to the month of their birth goes from a set of thirteen elements to one of twelve, so it cannot be injective, and two people share a value. The principle says the pair exists but gives no way to find it.

Theorem 7.23 (Injective, surjective and bijective agree on a finite set).

Let $A$ be finite and let $f : A \to A$ . Then $f$ is injective if and only if it is surjective, and either condition makes $f$ a bijection.

Discussion.

The statement is a biconditional between two conditions on the same map, so we prove each direction, and the last clause follows: a bijection is by definition an injection that is also a surjection, so once the two conditions are equivalent, each one gives both.

Going forwards, injectivity makes the image as large as the whole, and a subset of a finite set with the full cardinality cannot be proper, so the image is everything. Going backwards we argue by contradiction, because failure of injectivity is what we can compute with: a repeated value lets us delete a point without shrinking the image, so the image of a proper subset would have to be all of $A$ , and the proposition on images bounds it by something strictly smaller.

Proof.

Suppose $f$ is injective. Then $\#f(A) = \#A$ by the proposition on images, and $f(A) \subset A$ , so $f(A) = A$ by the theorem on subsets, since a proper subset would have strictly smaller cardinality. Hence $f$ is surjective.

Suppose $f$ is surjective and not injective, so $f(x) = f(y)$ for some $x \neq y$ . Then $f(A) = f(A \setminus \{x\})$ , and surjectivity makes this all of $A$ , so

\#A = \#f(A \setminus \{x\}) \leqslant \#(A \setminus \{x\}) < \#A,

which anti-reflexivity forbids. So $f$ is injective.

In either case $f$ is both injective and surjective, hence a bijection.

Remark.

Finiteness is needed. The successor map $s : \mathbb{N}_0 \to \mathbb{N}_0$ is injective by the third Peano condition and is not surjective, since $0$ is not a successor. So on $\mathbb{N}_0$ the two conditions come apart, and the theorem cannot be extended by weakening its hypothesis.

Problem 7.3.

Let $A$ and $B$ be finite with $\#A = \#B$ and let $f : A \to B$ be injective. Show that $f$ is a bijection.

Problem 7.4.

Let $A$ be finite and let $f : A \to A$ satisfy $f \circ f = f$ . Show that $f = \mathrm{id}_A$ if $f$ is injective, and describe what $f$ can be otherwise.

Sums and Products

Counting a union means laying one cut after another, so we first say how a cut breaks into an opening piece and a shifted one.

Proposition 7.24 (Splitting a cut).

Let $m, n \in \mathbb{N}_0$ and put $S \defeq \{m + i \mid i \in L_n\}$ . Then

L_{m+n} = L_m \cup S, \qquad L_m \cap S = \emptyset,

and $i \mapsto m + i$ is a bijection from $L_n$ onto $S$ .

Discussion.

Three assertions, each about the position of a number relative to $m$ . The equality is a set equality, hence two inclusions. From left to right, trichotomy puts $x$ either below $m$ , which is the first part, or at or above it, in which case $x = m + i$ for a difference $i$ , and the difference is below $n$ because subtracting $m$ from both sides of $x < m + n$ leaves $i < n$ . From right to left, both parts are checked against $m + n$ directly, using that adding a positive element moves you up. Disjointness is trichotomy once more, since the members of $S$ are at least $m$ and the members of $L_m$ are strictly below it. The last assertion is uniqueness of differences, which says $m + i = m + i'$ can happen only for $i = i'$ ; surjectivity holds because $S$ was defined as the set of such values.

Proof.

Let $x < m + n$ . If $x < m$ then $x \in L_m$ . Otherwise $m \leqslant x$ , so $x = m + i$ for some $i \in \mathbb{N}_0$ by the description of the associated order. From $m + i < m + n$ we get $m + n = (m + i) + b = m + (i + b)$ for some positive $b$ , by the laws of addition, so $n = i + b$ by uniqueness of differences and hence $i < n$ . Thus $x \in S$ .

Conversely, $m \leqslant m + n$ , so $x < m$ gives $x < m + n$ by mixed transitivity. And if $i < n$ then $n = i + b$ with $b$ positive, so $m + n = (m + i) + b$ and $m + i < m + n$ .

If $x \in L_m \cap S$ then $x < m$ and $m \leqslant x$ , which trichotomy forbids, so the intersection is empty.

Finally, $i \mapsto m + i$ is injective by uniqueness of differences and surjective onto $S$ by the definition of $S$ .

Theorem 7.25 (Cardinality of a disjoint union).

Let $A$ and $B$ be finite and disjoint. Then $A \cup B$ is finite and

\#(A \cup B) = \#A + \#B.

Discussion.

We must exhibit a bijection from $A \cup B$ onto the cut $L_{m+n}$ , and the splitting proposition cuts that target into the two pieces we need. So we define the map in two clauses, counting the points of $A$ by their own bijection and the points of $B$ by theirs, shifted up by $m$ so as to land in the second piece. Disjointness of $A$ and $B$ means each point falls under exactly one clause, so we have a function. Injectivity then has three cases, two inside a piece, which the two bijections handle, and one across the pieces, which the disjointness of the split handles. Surjectivity is the other half of the split.

Proof.

Put $m \defeq \#A$ and $n \defeq \#B$ , and let $g : A \to L_m$ and $h : B \to L_n$ be bijections. Define

k : A \cup B \to L_{m+n}, \qquad k(x) \defeq \begin{cases} g(x) & \text{if } x \in A, \\ m + h(x) & \text{if } x \in B. \end{cases}

Since $A \cap B = \emptyset$ , each point of the domain falls under exactly one clause, so $k$ is a function, and its values lie in $L_{m+n}$ by the splitting proposition.

For injectivity, two points of $A$ are separated by $g$ ; two points of $B$ are separated by $h$ together with the injectivity of $i \mapsto m + i$ ; and a point of $A$ cannot collide with a point of $B$ , since their values lie in the two disjoint pieces of the split. For surjectivity, an element of $L_{m+n}$ lies either in $L_m$ , hence is $g(x)$ for some $x \in A$ , or is $m + i$ with $i \in L_n$ , hence is $m + h(y) = k(y)$ for some $y \in B$ .

So $A \cup B \approx L_{m+n}$ , which is the assertion.

Corollary 7.26 (Cardinality of a union).

Let $A$ and $B$ be finite. Then $A \cup B$ is finite and

\#(A \cup B) + \#(A \cap B) = \#A + \#B.

Proof.

The sets $B \setminus A$ and $A \cap B$ are subsets of $B$ , hence finite. Now $A \cup B = A \cup (B \setminus A)$ with the two parts disjoint, so $\#(A \cup B) = \#A + \#(B \setminus A)$ ; and $B = (B \setminus A) \cup (A \cap B)$ with the two parts disjoint, so $\#B = \#(B \setminus A) + \#(A \cap B)$ . Adding $\#(A \cap B)$ to the first equation and substituting the second gives

\#(A \cup B) + \#(A \cap B) = \#A + \#(B \setminus A) + \#(A \cap B) = \#A + \#B,

by the laws of addition.

Remark.

The familiar form of that identity subtracts the overlap, and we have written it with everything on the right instead. Subtraction is available to us only when the answer stays in $\mathbb{N}_0$ , which it does here, but stating the law as an equation between sums spares us from checking that each time.

Problem 7.5.

Let $A$ , $B$ and $C$ be finite. Show that

\#(A \cup B \cup C) + \#(A \cap B) + \#(A \cap C) + \#(B \cap C) = \#A + \#B + \#C + \#(A \cap B \cap C).

Theorem 7.27 (Cardinality of a product).

Let $A$ and $B$ be finite. Then $A \times B$ is finite and $\#(A \times B) = \#A \cdot \#B$ , with multiplication as the problems of the natural-numbers chapter defined it.

Discussion.

Multiplication was defined by recursion on its second argument, with the clauses $x \cdot 0 = 0$ and $x \cdot s(n) = x \cdot n + x$ , so we induct on $\#B$ following them. At $0$ the set $B$ is empty and so is the product, since a pair needs a second coordinate. For the step we split $B$ into a smaller set and a single extra point $b$ , which splits $A \times B$ into two disjoint pieces, one governed by the inductive hypothesis and one a copy of $A$ . The theorem on disjoint unions adds the two cardinalities, and the resulting sum is the right-hand side of the second clause of the multiplication recursion.

Proof.

We induct on $n \defeq \#B$ . If $n = 0$ then $B = \emptyset$ , so $A \times B = \emptyset$ and $\#(A \times B) = 0 = \#A \cdot 0$ .

Suppose the claim holds for every $B$ of cardinality $n$ , and let $\#B = s(n)$ . Choose $b \in B$ and put $B' \defeq B \setminus \{b\}$ , of cardinality $n$ . A pair $(x, y)$ with $y \in B$ has either $y \in B'$ or $y = b$ , and not both, so

A \times B = (A \times B') \cup (A \times \{b\}), \qquad (A \times B') \cap (A \times \{b\}) = \emptyset,

using equality of ordered pairs to read off the second coordinate. The map $x \mapsto (x, b)$ is a bijection from $A$ onto $A \times \{b\}$ , again by equality of ordered pairs, so that piece has cardinality $\#A$ ; and $A \times B'$ has cardinality $\#A \cdot n$ by hypothesis. The theorem on disjoint unions gives

\#(A \times B) = \#A \cdot n + \#A = \#A \cdot s(n),

the last step being the second clause of the definition of multiplication.

Problem 7.6.

Show that a union of finitely many finite sets is finite. State the claim carefully first: it is an assertion about an indexed family whose index set is a cut.

Problem 7.7.

Let $A$ be finite with $\#A = n$ . Show that $\#\mathcal{P}(A) = 2^n$ , with powers as the problems of the natural-numbers chapter defined them.

Problem 7.8.

Let $A$ and $B$ be finite. Show that the Cartesian power $A^B$ , the set of functions from $B$ to $A$ , is finite with $\#(A^B) = (\#A)^{\#B}$ .

Extrema of Finite Sets

Proposition 7.28 (Finite subsets of an ordered set have extrema).

Let $M$ be a totally ordered set and let $S \subset M$ be finite and non-empty. Then $S$ has a minimum and a maximum.

Discussion.

The hypothesis is about an arbitrary finite non-empty set, so the induction runs on the cardinality and starts at $1$ rather than $0$ , which is allowed by induction from an arbitrary starting point. A set of cardinality $1$ has a single element, which is both extrema by reflexivity of the order. For the step we remove a point $a$ , apply the hypothesis to what remains, which is still non-empty, and then compare $a$ with the maximum found there: the two are comparable because the order is total, and whichever of the two is the larger is the maximum of the whole. Minima are the same argument with the inequalities reversed, so we write only one of them out.

Proof.

We induct on $n = \#S$ , beginning at $n = 1$ . Then $S \approx L_1 = \{0\}$ , so $S = \{a\}$ for a single $a$ , and $a \leqslant a$ makes it both the minimum and the maximum.

Suppose every subset of cardinality $n \geqslant 1$ has both extrema, and let $\#S = s(n)$ . Choose $a \in S$ and put $S' \defeq S \setminus \{a\}$ , which has cardinality $n \geqslant 1$ and so is non-empty. Let $M'$ be its maximum. Since $M$ is totally ordered, $a$ and $M'$ are comparable. If $a \leqslant M'$ then $M'$ is an upper bound of $S$ lying in $S$ , hence its maximum; if $M' \leqslant a$ then $a$ is such a bound, hence the maximum. The argument for the minimum reverses the inequalities.

Corollary 7.29 (The natural numbers are infinite).

$\mathbb{N}_0$ and $\mathbb{N}$ are infinite.

Proof.

The order on $\mathbb{N}_0$ is strict and linear, so the associated $\leqslant$ is a total order. Were $\mathbb{N}_0$ finite, it would be non-empty and so would have a maximum $M$ ; but $s(M) \in \mathbb{N}_0$ and $M < s(M)$ , contradicting that $M$ is an upper bound. The same argument applies to $\mathbb{N}$ , whose element $s(M)$ is positive because it is a successor.

Problem 7.9.

Show that a set with an infinite subset is infinite, and that the image of a finite set under any function is finite.

Comparing Sets

For an infinite set there is no $n \in \mathbb{N}_0$ counting it, so we cannot compare numbers. Injections and bijections still make sense, though, and for finite sets the next theorem matches each inequality with a condition on functions.

Comparing Finite Sets

Theorem 7.30 (Comparing finite sets).

Let $A$ and $B$ be finite sets. Then

$\#A \leqslant \#B$ if and only if there is an injection $A \to B$ ;
$\#A = \#B$ if and only if $A \approx B$ ;
$\#A < \#B$ if and only if there is an injection $A \to B$ but no bijection $A \to B$ .

Discussion.

Write $m \defeq \#A$ and $n \defeq \#B$ and fix bijections $f : L_m \to A$ and $g : L_n \to B$ ; every part is then a matter of conjugating a map between the cuts into a map between the sets, or the other way about.

For the first part, one direction is already the theorem on injections and size. The other builds the injection: $m \leqslant n$ gives $L_m \subset L_n$ , whose inclusion map is injective, and $g \circ \iota \circ f^{-1}$ is then an injection $A \to B$ , being a composite of injections. The second part is the corollary that cardinality classifies finite sets, read in both directions.

The third is the first two put together. If $m < n$ then the first part supplies an injection, and a bijection would force $m = n$ by the second, which trichotomy forbids. Conversely an injection gives $m \leqslant n$ and the absence of a bijection gives $m \neq n$ , and those two are what $m < n$ abbreviates.

Proof.

Write $m \defeq \#A$ and $n \defeq \#B$ , and fix bijections $f : L_m \to A$ and $g : L_n \to B$ .

For the first part, suppose $m \leqslant n$ . Then $L_m \subset L_n$ , since $x < m \leqslant n$ gives $x < n$ , and the inclusion $\iota : L_m \to L_n$ is injective. So $g \circ \iota \circ f^{-1} : A \to B$ is an injection. Conversely an injection $A \to B$ gives $\#A \leqslant \#B$ by the theorem on injections and size.

The second part is the corollary that cardinality classifies finite sets.

For the third, let $m < n$ . The first part gives an injection $A \to B$ , and a bijection would give $m = n$ by the second part, contradicting trichotomy. Conversely, an injection gives $m \leqslant n$ by the first part and the absence of a bijection gives $m \neq n$ by the second, so $m < n$ .

The three right-hand sides make sense whether or not the sets are finite, so they may be taken as definitions in general. By the theorem, the new definitions agree with the old ones for finite sets.

Definition 7.31 (Comparing sizes).

Let $A$ and $B$ be sets, finite or not. We write

$\#A \leqslant \#B$ if there is an injection $A \to B$ ;
$\#A = \#B$ if $A \approx B$ ;
$\#A < \#B$ if there is an injection $A \to B$ but no bijection $A \to B$ .

For finite sets these agree with the numerical readings, by the theorem. The symbol $\#A$ still names a natural number only when $A$ is finite; in general the three displays above are single assertions about functions, and it is the assertion, not the symbol $\#A$ standing alone, that has been defined. What $\#A$ may be taken to name in general is settled at the end of the chapter.

Read informally, $\#A < \#B$ says that $B$ is roomy enough to hold an injective copy of $A$ and still has something left over, in the strong sense that every injection misses something.

Proposition 7.32 (Reflexivity and transitivity).

For all sets $A$ , $B$ , $C$ :

$\#A \leqslant \#A$ ;
if $\#A \leqslant \#B$ and $\#B \leqslant \#C$ then $\#A \leqslant \#C$ .

Discussion.

Both parts unfold to claims about injections, and both follow from basic facts about injections. Reflexivity needs an injection $A \to A$ , and the identity map is one. Transitivity needs an injection $A \to C$ out of injections $A \to B$ and $B \to C$ , and their composite is injective: if the composite identifies two points, the outer map identifies their images and the inner map identifies them.

Proof.

The identity $\mathrm{id}_A$ is injective, so $\#A \leqslant \#A$ .

Let $f : A \to B$ and $g : B \to C$ be injections. If $g(f(x)) = g(f(y))$ then $f(x) = f(y)$ by injectivity of $g$ , and $x = y$ by injectivity of $f$ . So $g \circ f$ is an injection $A \to C$ and $\#A \leqslant \#C$ .

Problem 7.10.

Prove or refute each of the following, for arbitrary sets $A$ , $B$ and $C$ .

If $\#A \leqslant \#B$ then $\#(A \times C) \leqslant \#(B \times C)$ .
$\#(A \setminus B) \leqslant \#B$ .

Problem 7.11.

Let $B$ be non-empty. Show that $\#A \leqslant \#(A \times B)$ for every set $A$ .

The Schröder–Bernstein Theorem

The last proposition does not give antisymmetry. For finite sets it is immediate, since $m \leqslant n \leqslant m$ forces $m = n$ ; in general it is a harder theorem.

Theorem 7.33 (Schröder–Bernstein).

Let $A$ and $B$ be sets with $\#A \leqslant \#B$ and $\#B \leqslant \#A$ . Then $A \approx B$ .

Discussion.

We are given injections $f : A \to B$ and $g : B \to A$ and must build a bijection out of them. Neither alone will do: $f$ may miss part of $B$ and $g$ may miss part of $A$ . The idea is to use $f$ on some of $A$ and the inverse of $g$ on the rest, and the problem is to decide where the boundary falls. The points where we have no choice are those of $A \setminus g(B)$ , which have no $g$ -preimage at all, so $f$ must be used there; and then $f$ must be used at $g(f(x))$ for each such $x$ , since otherwise that point would be sent back to $f(x)$ , which is already taken. Iterating gives a family $A_0, A_1, A_2, \ldots$ built by the recursion theorem, and their union $C$ is the region where $f$ is used.

With the boundary fixed, three checks remain. The map is defined everywhere, because a point outside $C$ is in particular outside $A_0$ , hence has a $g$ -preimage, unique by injectivity. It is injective on each of the two regions separately, and the two cannot collide, because a collision would carry a point of $C$ one step further along the chain and so put an element of $C$ outside $C$ . It is surjective by chasing a given $b$ backwards: if $g(b)$ lies outside $C$ it is where $b$ comes from, and if it lies inside $C$ it lies in some $A_n$ with $n$ a successor, which exhibits $b$ as a value of $f$ .

Proof.

Let $f : A \to B$ and $g : B \to A$ be injections. The recursion theorem, applied in $\mathcal{P}(A)$ with the map $X \mapsto g(f(X))$ , gives exactly one family $\{A_n\}_{n \in \mathbb{N}_0}$ with

A_0 \defeq A \setminus g(B), \qquad A_{s(n)} \defeq g\bigl(f(A_n)\bigr),

and we put $C \defeq \bigcup_{n \in \mathbb{N}_0} A_n$ . The two injections run the pieces alternately into one another,

\begin{array}{ccccccc} A: & A_0 & & A_1 & & A_2 & \cdots \\ & \big\downarrow {\scriptstyle f} & \nearrow {\scriptstyle g} & \big\downarrow {\scriptstyle f} & \nearrow {\scriptstyle g} & \big\downarrow {\scriptstyle f} & \\ B: & f(A_0) & & f(A_1) & & f(A_2) & \cdots \end{array}

and $C$ is the whole of the top row. On it the bijection will follow the downward arrows; off it every point has a $g$ -preimage, and the bijection will run back up them. Define

h : A \to B, \qquad h(x) \defeq \begin{cases} f(x) & \text{if } x \in C, \\ \text{the unique } b \text{ with } g(b) = x & \text{if } x \notin C. \end{cases}

The second clause makes sense: $x \notin C$ gives $x \notin A_0$ , so $x \in g(B)$ , and the preimage is unique because $g$ is injective. So $h$ is a function.

For injectivity, suppose $h(x) = h(y)$ . If both points lie in $C$ then $f(x) = f(y)$ gives $x = y$ ; if neither does then applying $g$ to both sides gives $x = y$ . Suppose then $x \in C$ and $y \notin C$ , so $f(x) = h(y)$ and hence $y = g(f(x))$ . Since $x \in C$ we have $x \in A_n$ for some $n$ , so $y \in g(f(A_n)) = A_{s(n)} \subset C$ , contradicting $y \notin C$ .

For surjectivity, let $b \in B$ and put $x \defeq g(b)$ . If $x \notin C$ then $h(x) = b$ by the second clause. If $x \in C$ then $x \in A_n$ for some $n$ , and $n \neq 0$ because $x \in g(B)$ ; so $n = s(k)$ by the theorem on predecessors, and $x \in g(f(A_k))$ gives $x = g(f(z))$ for some $z \in A_k$ . Injectivity of $g$ gives $b = f(z)$ , and $z \in C$ , so $b = h(z)$ .

Hence $h$ is a bijection and $A \approx B$ .

The proof gives the bijection in two pieces, one for the chained points and one for the rest, rather than as a single formula. What the applications use is only that it exists.

Example 7.34 (Informal).

Assuming you know the real numbers, write $R$ for them and consider the closed interval $[0,1]$ and the open interval $(0,1)$ . The inclusion of $(0,1)$ in $[0,1]$ is an injection one way. In the other direction $x \mapsto \tfrac{x}{2} + \tfrac{1}{4}$ is injective and carries $[0,1]$ into $(0,1)$ , since $0$ goes to $\tfrac14$ and $1$ to $\tfrac34$ . Schröder–Bernstein therefore supplies a bijection between the two intervals, although neither of the injections we wrote is surjective.

Problem 7.12.

Schröder–Bernstein is sometimes read as saying that injections $f : A \to B$ and $g : B \to A$ are each bijections. Exhibit sets $A$ and $B$ with injections both ways, neither of which is surjective.

Theorem 7.35 (Transitivity for strict comparison).

Let $A$ , $B$ and $C$ be sets.

If $\#A < \#B$ and $\#B < \#C$ then $\#A < \#C$ .
If $\#A < \#B$ and $\#B \leqslant \#C$ then $\#A < \#C$ .
If $\#A \leqslant \#B$ and $\#B < \#C$ then $\#A < \#C$ .

Discussion.

Each part asserts the existence of an injection and the non-existence of a bijection, so each splits in two, and the first half is the previous proposition in every case, since a strict comparison contains a weak one. The second half is where Schröder–Bernstein is needed, and it is used contrapositively: a bijection $A \to C$ would let us pull an injection $B \to C$ back to an injection $B \to A$ , and with the injection $A \to B$ already in hand the theorem would return $A \approx B$ , which the strict hypothesis forbids. We write out the first part and leave the other two, which run the same way, to the problems.

Proof.

For the first part, the two hypotheses supply injections $A \to B$ and $B \to C$ , so $\#A \leqslant \#C$ by the previous proposition.

Suppose $A \approx C$ and let $\varphi : C \to A$ be a bijection. Composing $\varphi$ with an injection $B \to C$ gives an injection $B \to A$ , so $\#B \leqslant \#A$ . Since $\#A < \#B$ gives $\#A \leqslant \#B$ , Schröder–Bernstein yields $A \approx B$ , contradicting $\#A < \#B$ . So no bijection $A \to C$ exists and $\#A < \#C$ .

Comparability

Reflexive, transitive, antisymmetric: the comparison behaves like an order. What an order on the natural numbers also had was trichotomy, and for sizes that is a separate matter, since nothing so far rules out two sets neither of which injects into the other.

Theorem 7.36 (Comparability).

Let $A$ and $B$ be sets. Then $\#A \leqslant \#B$ or $\#B \leqslant \#A$ .

Discussion.

An injection defined on all of $A$ is what we want and cannot build directly, so we build the largest injection defined on part of $A$ and show that largest one leaves nothing out. The candidates are the injective functions whose domain is a subset of $A$ and whose values lie in $B$ ; each is a subset of $A \times B$ , so comprehension collects them into a set, which inclusion orders.

To apply Zorn’s lemma we must show every chain has an upper bound, and we take the union of the chain: it is a function because two pairs with the same first coordinate lie in a common member of the chain, since members of a chain are comparable, and it is injective for the same reason with the coordinates exchanged. The empty chain is covered by the empty function. Zorn then supplies a maximal element $p$ , and maximality is used contrapositively: if $p$ missed a point of $A$ and also missed a point of $B$ , the pair of them could be added to $p$ , giving a strictly larger candidate. So one of the two is not missed, and the two cases give the two halves of the conclusion, the second by inverting $p$ .

Proof.

Let $M$ be the set of $p \in \mathcal{P}(A \times B)$ that are injective functions whose domain is a subset of $A$ , ordered by inclusion. The empty function belongs to $M$ , and inclusion is a partial order.

Let $K \subset M$ be a chain and put $q \defeq \bigcup K$ , a subset of $A \times B$ . If $(x, b)$ and $(x, b')$ lie in $q$ , they lie in members $p$ and $p'$ of $K$ , which are comparable, so both pairs lie in the larger one; that member is a function, so $b = b'$ . Hence $q$ is a function, with domain the union of the domains. The same argument with the coordinates exchanged shows $q$ is injective. So $q \in M$ and $q$ is an upper bound of $K$ , and $M$ is inductively ordered.

Zorn’s lemma gives a maximal $p \in M$ . Suppose $\operatorname{dom} p \neq A$ and $\operatorname{im} p \neq B$ , and choose $a \in A \setminus \operatorname{dom} p$ and $b \in B \setminus \operatorname{im} p$ . Then $p \cup \{(a, b)\}$ is again an injective function with domain inside $A$ , and it properly contains $p$ , contradicting maximality.

So $\operatorname{dom} p = A$ , in which case $p$ is an injection $A \to B$ ; or $\operatorname{im} p = B$ , in which case the inverse of $p$ on its range is an injection $B \to A$ .

Corollary 7.37 (Trichotomy for sizes).

For any sets $A$ and $B$ , exactly one of $\#A < \#B$ , $\#A = \#B$ , $\#B < \#A$ holds.

Proof.

Comparability gives $\#A \leqslant \#B$ or $\#B \leqslant \#A$ . If both hold then $A \approx B$ by Schröder–Bernstein; if only the first holds then $\#A < \#B$ , and if only the second then $\#B < \#A$ . So at least one of the three holds. No two hold together: $A \approx B$ makes both strict comparisons fail by definition, and $\#A < \#B$ together with $\#B < \#A$ would give $A \approx B$ by Schröder–Bernstein, contradicting either.

Remark.

Comparability was proved from Zorn’s lemma, and it is in fact equivalent to the axiom of choice, so it is not a free consequence of the other axioms. Schröder–Bernstein, by contrast, used nothing but the recursion theorem. So comparability needs the axiom of choice, while Schröder–Bernstein does not.

Cantor’s Theorem

Theorem 7.38 (Cantor's theorem).

For every set $S$ we have $\#S < \#\mathcal{P}(S)$ .

Discussion.

By the definition the claim splits in two: an injection $S \to \mathcal{P}(S)$ , and no bijection. The injection is the map sending a point to the set holding just that point, and it is injective because a singleton determines its element.

For the second half we take an arbitrary $f : S \to \mathcal{P}(S)$ and produce a subset that is not one of its values, which denies surjectivity and so denies bijectivity. A set differs from $f(x)$ as soon as it disagrees with it at one element, and the element we use is $x$ itself, so we build the subset that disagrees with $f(x)$ at $x$ for every $x$ at once: take those $x$ that are outside their own value. Comprehension makes that a set and it is a subset of $S$ , so it is eligible to be a value. Supposing it is the value at $a$ , the question whether $a$ belongs to it has a membership criterion that turns each answer into the other, which is Russell’s argument again.

Proof.

The map $x \mapsto \{x\}$ sends $S$ into $\mathcal{P}(S)$ , and $\{x\} = \{y\}$ gives $x = y$ , so it is injective and $\#S \leqslant \#\mathcal{P}(S)$ .

Let $f : S \to \mathcal{P}(S)$ be any function and put

D \defeq \{x \in S \mid x \notin f(x)\},

a set by comprehension and a subset of $S$ , so $D \in \mathcal{P}(S)$ . Suppose $D = f(a)$ for some $a \in S$ . If $a \in D$ then the criterion gives $a \notin f(a) = D$ . If $a \notin D$ then, $a$ being an element of $S$ , the criterion gives $a \in f(a) = D$ . Both are contradictions, so $D$ is not a value of $f$ and $f$ is not surjective.

In particular no bijection $S \to \mathcal{P}(S)$ exists, so $\#S < \#\mathcal{P}(S)$ .

Remark.

The diagonal set is the construction that showed there is no set of all sets, used for a different purpose. There it produced a contradiction from an assumption we then dropped; here it produces one from the assumption that $D$ is a value, and what we drop is surjectivity. Nothing about $S$ was used, so the theorem applies to every set: iterating it gives

\#S < \#\mathcal{P}(S) < \#\mathcal{P}(\mathcal{P}(S)) < \cdots,

a strictly increasing chain with no top, by transitivity for strict comparison.

Problem 7.13.

Show that if $\#A \leqslant \#B$ and $A \approx A'$ and $B \approx B'$ , then $\#A' \leqslant \#B'$ . Transport the injection along the two bijections.

Problem 7.14.

Prove parts 2 and 3 of the theorem on transitivity for strict comparison.

Problem 7.15.

Use Schröder–Bernstein, rather than an explicit pairing, to show that $\#\mathbb{N}_0 = \#(\mathbb{N}_0 \times \mathbb{N}_0)$ once you have an injection each way.

Problem 7.16.

Show that $\#A \leqslant \#B$ implies $\#\mathcal{P}(A) \leqslant \#\mathcal{P}(B)$ .

Problem 7.17.

Deduce from Cantor’s theorem that $n < 2^n$ for every $n \in \mathbb{N}_0$ .

Infinite Sets

A set is infinite when it is not finite, which is a purely negative description: no cut matches it. Cantor found that infinite sets come in different sizes, and Cantor’s theorem above already gives a strictly increasing chain of them. We start with the smallest.

Countable Sets

Definition 7.39 (Countable and uncountable).

A set $A$ is countably infinite if $A \approx \mathbb{N}_0$ , countable if it is finite or countably infinite, and uncountable otherwise.

A bijection $f : \mathbb{N}_0 \to A$ is an infinite sequence in $A$ , so a countably infinite set is one whose elements can be written as a list

(a_0, a_1, a_2, \ldots)

in which every element appears exactly once. Such an $f$ is an enumeration of $A$ . The list is not part of the set, and a set usually admits many.

That the two cases of countability do not overlap is the corollary that $\mathbb{N}_0$ is infinite, since otherwise a finite set could be equinumerous with $\mathbb{N}_0$ .

Proposition 7.40 (Dropping zero).

$\mathbb{N} \approx \mathbb{N}_0$ , so $\mathbb{N}$ is countably infinite.

Discussion.

We must produce a bijection, and we use the successor map, checking three things against the Peano conditions. Its values are positive, since a successor is never $0$ , so it does map into $\mathbb{N}$ . It is injective, which is the third condition verbatim. It is surjective onto $\mathbb{N}$ , which is the theorem on predecessors: every element other than $0$ is a successor.

Proof.

Consider $s : \mathbb{N}_0 \to \mathbb{N}$ . Its values lie in $\mathbb{N}$ , since $s(n) \neq 0$ by the fourth Peano condition. It is injective by the third. It is surjective, since an element of $\mathbb{N}$ is not $0$ and is therefore $s(n)$ for some $n$ by the theorem on predecessors. So $s$ is a bijection and $\mathbb{N}_0 \approx \mathbb{N}$ .

Remark.

Since $\mathbb{N} \subsetneq \mathbb{N}_0$ , an infinite set can be equinumerous with a proper subset of itself, which the corollary of the last chapter shows no finite set can do. Dedekind turned the observation round and took it as the definition of infinite, and we prove below that his definition agrees with ours.

Subsets of the Natural Numbers

Theorem 7.41 (Infinite subsets of the natural numbers).

Every infinite subset $M \subset \mathbb{N}_0$ is countably infinite.

Discussion.

We must produce an enumeration of $M$ , and there is an obvious rule for one: list the smallest element first, then the smallest of those left, and so on. Two things must be checked before this is a definition. The rule at stage $n$ refers to everything listed before $n$ , not merely to the previous entry, so it is the general recursion theorem rather than the plain one that builds the function, taking for its rule the map sending a finite list to the least element of $M$ not on it. And that least element must exist: the elements listed so far form the image of a cut, hence a finite set, so they cannot exhaust the infinite $M$ , and well-ordering then supplies the minimum.

What remains is to check that the enumeration works. It is strictly increasing, because each entry is chosen from a smaller pool than the one before and differs from the entry just removed, and a strictly increasing map is injective by trichotomy. Surjectivity is a least-counterexample argument: if some $m \in M$ never appears, look at the first stage whose entry overshoots $m$ . Everything listed before that stage is below $m$ , so $m$ was still in the pool at that stage, and the minimum chosen there cannot have exceeded it.

Proof.

For a finite sequence $t$ in $\mathbb{N}_0$ , the range $\operatorname{im} t$ is finite, being the image of a cut. So $M \setminus \operatorname{im} t$ is non-empty: otherwise $M \subset \operatorname{im} t$ would make $M$ finite. Define

g : \operatorname{Seq}(\mathbb{N}_0) \to \mathbb{N}_0, \qquad g(t) \defeq \min\bigl(M \setminus \operatorname{im} t\bigr),

which exists by well-ordering. The general recursion theorem gives exactly one $f : \mathbb{N}_0 \to \mathbb{N}_0$ with $f(n) = g(f|_{L_n})$ for every $n$ , that is,

f(n) = \min\bigl(M \setminus \{f(i) \mid i < n\}\bigr).

Every value of $f$ lies in $M$ . The map is strictly increasing: the set from which $f(s(n))$ is chosen is contained in the set from which $f(n)$ is chosen, so $f(n) \leqslant f(s(n))$ by minimality, and $f(s(n)) \neq f(n)$ because $f(n)$ was removed at the later stage. Induction then gives $f(m) < f(n)$ whenever $m < n$ , and trichotomy makes $f$ injective.

Induction also gives $n \leqslant f(n)$ : this holds at $0$ , and if $n \leqslant f(n)$ then $n < f(s(n))$ , so $f(s(n)) < s(n)$ would place $f(s(n))$ strictly between $n$ and $s(n)$ , which nothing does; hence $s(n) \leqslant f(s(n))$ .

Suppose some $m \in M$ is not a value of $f$ . Since $m < s(m) \leqslant f(s(m))$ , the set of $n$ with $m < f(n)$ is non-empty, so well-ordering gives a least such $k$ . Every $i < k$ has $f(i) \leqslant m$ and $f(i) \neq m$ , so $f(i) < m$ ; hence $m$ belongs to $M \setminus \{f(i) \mid i < k\}$ , and minimality of $f(k)$ in that set gives $f(k) \leqslant m$ , contradicting $m < f(k)$ .

So $f$ is a bijection from $\mathbb{N}_0$ onto $M$ .

Characterising At Most Countable Sets

Theorem 7.42 (Characterising countable sets).

Let $M$ be a non-empty set. The following are equivalent.

$M$ is countable.
There is an injection $M \to \mathbb{N}_0$ .
There is a surjection $\mathbb{N}_0 \to M$ .

Discussion.

We prove the three conditions equivalent by a cycle of implications.

From the first to the third: a countably infinite $M$ has a bijection from $\mathbb{N}_0$ , which is a surjection; a finite non-empty $M$ has a bijection from a cut, which we extend to all of $\mathbb{N}_0$ by parking every later index on one fixed point, and parking spoils nothing because surjectivity asks only that every point be hit.

From the third to the second: a surjection has an injective right inverse, and here we can name one without appealing to choice. Each point of $M$ has a non-empty set of preimages inside $\mathbb{N}_0$ , so well-ordering picks out its least preimage, and a right inverse is injective because applying the surjection recovers the point.

From the second to the first: an injection makes $M$ equinumerous with its image, a subset of $\mathbb{N}_0$ , and a subset of $\mathbb{N}_0$ is finite or infinite; in the first case $M$ is finite, and in the second the theorem just proved makes it countably infinite.

Proof.

Suppose $M$ is countable. If $M \approx \mathbb{N}_0$ , a bijection is a surjection. If $M$ is finite and non-empty, let $f : L_n \to M$ be a bijection with $n \neq 0$ , and define $\beta : \mathbb{N}_0 \to M$ by $\beta(k) \defeq f(k)$ for $k < n$ and $\beta(k) \defeq f(0)$ for $k \geqslant n$ . Every point of $M$ is $f(k)$ for some $k < n$ , so $\beta$ is surjective.

Suppose $\beta : \mathbb{N}_0 \to M$ is surjective. For $x \in M$ the preimage $\beta^{-1}(\{x\})$ is a non-empty subset of $\mathbb{N}_0$ , so it has a least element; let $\alpha(x)$ be that element. Then $\beta(\alpha(x)) = x$ for every $x$ , so $\alpha(x) = \alpha(y)$ gives $x = y$ , and $\alpha : M \to \mathbb{N}_0$ is injective.

Suppose $\alpha : M \to \mathbb{N}_0$ is injective. Then $M \approx \alpha(M) \subset \mathbb{N}_0$ . If $\alpha(M)$ is finite then so is $M$ . If $\alpha(M)$ is infinite then $\alpha(M) \approx \mathbb{N}_0$ by the theorem on infinite subsets, so $M \approx \mathbb{N}_0$ by transitivity. Either way $M$ is countable.

Corollary 7.43 (Subsets of countable sets).

Every subset of a countable set is countable.

Proof.

Let $B \subset A$ with $A$ countable. If $B$ is empty it is finite. Otherwise $A$ is non-empty, so there is an injection $A \to \mathbb{N}_0$ by the theorem, and its restriction to $B$ is injective, so $B$ is countable by the theorem again.

Products and Unions

Theorem 7.44 (Pairs of natural numbers).

$\mathbb{N}_0 \times \mathbb{N}_0$ is countably infinite.

Discussion.

By the characterisation it is enough to find one injection into $\mathbb{N}_0$ , and then to observe that the set is not finite. For the injection we enumerate the pairs in diagonals: all pairs with $m + n = 0$ , then those with $m + n = 1$ , and so on. The $k$ th diagonal has $s(k)$ entries, so a pair $(m, n)$ on the diagonal $k = m + n$ should be given the position

j(m, n) \defeq T(k) + m, \qquad T(k) \defeq \sum_{i=1}^{k} i,

where the summation symbol is the one built by recursion in the last chapter but one. Injectivity then rests on the diagonals not overlapping, which is the inequality $T(k) + m < T(k')$ whenever $k < k'$ and $m \leqslant k$ ; it follows from the recursion clause $T(s(k)) = T(k) + s(k)$ . Once the diagonal is known, $m$ and then $n$ are determined by uniqueness of differences. Infinitude is easy: the pairs $(n, 0)$ form a copy of $\mathbb{N}_0$ inside, and a set with an infinite subset is infinite.

Proof.

Write $T(k) \defeq \sum_{i=1}^{k} i$ , so that $T(0) = 0$ and $T(s(k)) = T(k) + s(k)$ .

We first record that $T$ is non-decreasing: if $k \leqslant l$ then $l = k + b$ , and induction on $b$ gives the claim, since $T(k + s(b)) = T(s(k+b)) = T(k+b) + s(k+b)$ is at least $T(k + b)$ , which is at least $T(k)$ by the inductive hypothesis.

Define $j(m, n) \defeq T(m + n) + m$ . Laid out with $m$ down the side and $n$ across, the values run

\begin{array}{c|ccccc} & 0 & 1 & 2 & 3 & \cdots \\ \hline 0 & 0 & 1 & 3 & \boxed{6} & \cdots \\ 1 & 2 & 4 & \boxed{7} & & \\ 2 & 5 & \boxed{8} & & & \\ 3 & \boxed{9} & & & & \\ \vdots & \vdots & & & & \ddots \end{array}

The boxed entries are the pairs with $m + n = 3$ , and they fill the consecutive block from $T(3) = 6$ to $T(3) + 3 = 9$ . What follows is that observation for a general diagonal.

Suppose $j(m, n) = j(m', n')$ and put $k \defeq m + n$ , $k' \defeq m' + n'$ . If $k < k'$ then $k' = k + s(c)$ for some $c$ , so

T(k') = T(s(k + c)) = T(k + c) + k' \geqslant T(k) + k'.

Also $m \leqslant k < k'$ , so $k' = m + b$ for a positive $b$ and $T(k) + k' = (T(k) + m) + b$ , giving $T(k) + m < T(k) + k' \leqslant T(k')$ ; and $T(k') \leqslant T(k') + m'$ . Together these give $j(m, n) < j(m', n')$ , against the supposition. The same argument rules out $k' < k$ , so $k = k'$ by trichotomy. Then $T(k) + m = T(k) + m'$ gives $m = m'$ by uniqueness of differences, and $m + n = m + n'$ gives $n = n'$ likewise. So $j$ is injective and $\mathbb{N}_0 \times \mathbb{N}_0$ is countable.

Finally $n \mapsto (n, 0)$ is injective, so $\mathbb{N}_0 \times \mathbb{N}_0$ has an infinite subset and is therefore infinite.

Corollary 7.45 (Products of countable sets).

If $A$ and $B$ are countable then so is $A \times B$ .

Proof.

If either set is empty the product is empty. Otherwise let $u : A \to \mathbb{N}_0$ and $v : B \to \mathbb{N}_0$ be injections. Then $(x, y) \mapsto (u(x), v(y))$ is injective, by equality of ordered pairs, and composing it with the injection $j$ of the theorem gives an injection $A \times B \to \mathbb{N}_0$ .

Theorem 7.46 (Countable unions of countable sets).

Let $I$ be countable and let $\{A_i\}_{i \in I}$ be an indexed family of countable sets. Then $\bigcup_{i \in I} A_i$ is countable.

Discussion.

The characterisation lets us argue with surjections rather than injections, and a surjection onto the union is easy to describe: run over the indices with one surjection and over each set with another, so that the pair $(m, n)$ names the $n$ th element of the $m$ th set. Two points need care. Countability of $A_i$ says only that a surjection onto it exists, and we need one for every $i$ at once, and this needs the axiom of choice. And the domain of the resulting map is $\mathbb{N}_0 \times \mathbb{N}_0$ rather than $\mathbb{N}_0$ , which the previous theorem fixes, since a countably infinite set admits a bijection from $\mathbb{N}_0$ . Composing the two gives a surjection from $\mathbb{N}_0$ onto the union, which is then countable by the characterisation.

Proof.

Discard the indices with $A_i = \emptyset$ ; the union is unchanged and the smaller index set is still countable. If nothing is left the union is empty, hence countable, so suppose $I \neq \emptyset$ and every $A_i \neq \emptyset$ .

By the characterisation there is a surjection $\gamma : \mathbb{N}_0 \to I$ . For each $i$ the set $S_i$ of surjections $\mathbb{N}_0 \to A_i$ is a non-empty subset of $\mathcal{P}(\mathbb{N}_0 \times A_i)$ , again by the characterisation, so the axiom of choice applied to $\{S_i\}_{i \in I}$ gives a function $i \mapsto \beta_i$ with $\beta_i \in S_i$ for every $i$ .

The map $(m, n) \mapsto \beta_{\gamma(m)}(n)$ is a surjection from $\mathbb{N}_0 \times \mathbb{N}_0$ onto $\bigcup_{i \in I} A_i$ : a point $x$ of the union lies in some $A_i$ , and $i = \gamma(m)$ for some $m$ while $x = \beta_i(n)$ for some $n$ . Composing with a bijection $\mathbb{N}_0 \to \mathbb{N}_0 \times \mathbb{N}_0$ , which the previous theorem supplies, gives a surjection from $\mathbb{N}_0$ onto the union, which is therefore countable.

Remark (Hilbert's hotel).

A hotel with a room for every natural number, all of them occupied, can still take in a new guest: move the occupant of room $n$ to room $s(n)$ and give the newcomer room $0$ . It can take in countably many new guests at once: move the occupant of room $n$ to room $2 \cdot n$ and use the odd rooms, which the remainder classes say are exactly the rooms left free. It can even take in countably many coaches each carrying countably many guests, by the theorem on countable unions. Infinite sizes do not behave like finite ones, and here we have to rely on the propositions above rather than on intuition.

Remark (Informal).

Assuming you know the whole numbers and the fractions: the whole numbers are the union of the natural numbers, their negatives and zero, so the theorem on countable unions makes them countable. Every fraction is determined by a pair of whole numbers, so the corollary on products makes the fractions countable as well. Both arguments go through as soon as those systems are built.

Problem 7.18.

Let $A_1, \ldots, A_n$ be countable. Show that $A_1 \times \cdots \times A_n$ is countable, by induction on $n$ .

Problem 7.19.

Let $A$ be countable. Show that the set $\operatorname{Seq}(A)$ of finite sequences in $A$ is countable, and that the set of finite subsets of $A$ is countable.

Problem 7.20.

Show that the map $j$ of the theorem on pairs is a bijection onto $\mathbb{N}_0$ , not merely an injection. Well-ordering applied to the set of $k$ with $x < T(s(k))$ will locate the diagonal on which $x$ sits.

Problem 7.21.

Let $A$ be countably infinite and let $B$ be finite and non-empty. Show that $A \cup B$ and $A \times B$ are countably infinite.

Problem 7.22.

Write $\mathbb{N}_0 \times \mathbb{N}_0$ as the union of a countably infinite family of pairwise disjoint countably infinite sets.

Finite and Infinite

One question is still open. We defined infinite negatively, as the failure of finiteness, and observed that $\mathbb{N}_0$ has a proper subset of its own size. Whether every infinite set does is still open, and the answer needs the axiom of choice.

Theorem 7.47 (Every infinite set has a countably infinite subset).

Let $A$ be infinite. Then some subset $E \subset A$ is countably infinite.

Discussion.

The plan copies the enumeration of an infinite subset of $\mathbb{N}_0$ : pick an element, then an element not yet picked, and so on. Two things we had there are missing here. There is no order on $A$ , so nothing selects an element for us, and the axiom of choice gives a rule that selects one from every non-empty subset at once. And the rule at each stage refers to all the earlier picks, so it is the general recursion theorem that turns the rule into a function. The recursion never stalls, because the elements picked so far form a finite set and $A$ is infinite, so something is always left. Injectivity is immediate from the construction, since each value is chosen outside the earlier ones, and an injective map from $\mathbb{N}_0$ has range a countably infinite subset.

Proof.

Since $A$ is infinite it is non-empty. Apply the axiom of choice to the family of non-empty subsets of $A$ , indexed by itself, to obtain a function $c$ with $c(X) \in X$ for every non-empty $X \subset A$ .

For a finite sequence $t$ in $A$ the range $\operatorname{im} t$ is finite, so $A \setminus \operatorname{im} t$ is non-empty, since otherwise $A$ would be a subset of a finite set. Define

g : \operatorname{Seq}(A) \to A, \qquad g(t) \defeq c\bigl(A \setminus \operatorname{im} t\bigr),

and let $f : \mathbb{N}_0 \to A$ be the map the general recursion theorem produces, so that

f(n) = c\bigl(A \setminus \{f(i) \mid i < n\}\bigr)

for every $n$ . If $m < n$ then $f(m)$ is one of the elements excluded at stage $n$ , so $f(n) \neq f(m)$ ; with trichotomy this makes $f$ injective. Hence $E \defeq \operatorname{im} f$ satisfies $\mathbb{N}_0 \approx E$ and $E \subset A$ .

Theorem 7.48 (Characterising finite sets).

Let $M$ be a non-empty set. The following are equivalent.

$M$ is finite.
There is a surjection $L_n \to M$ for some $n \in \mathbb{N}$ .
There is an injection $M \to L_n$ for some $n \in \mathbb{N}$ .
There is no injection $\mathbb{N}_0 \to M$ .

When they hold, $\#M$ is the least $n$ of the second kind and equally the least $n$ of the third.

Discussion.

We run a cycle again. From the first to the second, a counting of $M$ is itself a surjection from a cut. From the second to the third, a surjection has an injective right inverse, obtained as before by sending each point to its least preimage, which well-ordering supplies inside the cut. From the third to the fourth, an injection $\mathbb{N}_0 \to M$ followed by an injection $M \to L_n$ would inject $\mathbb{N}_0$ into $L_n$ , and restricting to $L_{s(n)}$ would inject a cut into a shorter one. The last step is the theorem just proved, read contrapositively: an infinite $M$ carries a copy of $\mathbb{N}_0$ , hence an injection $\mathbb{N}_0 \to M$ .

The minimality clause is the two size theorems. A surjection $L_n \to M$ gives $\#M \leqslant n$ by the proposition on images, an injection $M \to L_n$ gives $\#M \leqslant n$ by the theorem on injections and size, and $n = \#M$ is achieved in both cases by a counting of $M$ and its inverse.

Proof.

Suppose $M$ is finite. Being non-empty, $\#M = n$ for some $n \in \mathbb{N}$ , and a bijection $L_n \to M$ is a surjection.

Suppose $\sigma : L_n \to M$ is surjective. For $x \in M$ the preimage $\sigma^{-1}(\{x\})$ is a non-empty subset of $\mathbb{N}_0$ , so it has a least element; sending $x$ to it defines $\iota : M \to L_n$ with $\sigma \circ \iota = \mathrm{id}_M$ , and $\iota$ is injective.

Suppose $\iota : M \to L_n$ is injective and let $\gamma : \mathbb{N}_0 \to M$ be injective. Then $\iota \circ \gamma : \mathbb{N}_0 \to L_n$ is injective, and its restriction to $L_{s(n)}$ is an injection $L_{s(n)} \to L_n$ , which no cut admits. So no such $\gamma$ exists.

Suppose finally that $M$ is not finite. The previous theorem gives a subset $E \subset M$ equinumerous with $\mathbb{N}_0$ , hence a bijection $\mathbb{N}_0 \to E$ , which composed with the inclusion of $E$ in $M$ is an injection $\mathbb{N}_0 \to M$ . That is the fourth condition denied, which completes the cycle.

For the last claim, a surjection $L_n \to M$ gives $\#M = \#\sigma(L_n) \leqslant \#L_n = n$ , and an injection $M \to L_n$ gives $\#M \leqslant n$ ; both bounds are attained by a counting of $M$ , which is a surjection $L_{\#M} \to M$ , and by its inverse.

Theorem 7.49 (Dedekind's characterisation of finiteness).

A set is finite if and only if it is not equinumerous with any proper subset of itself.

Discussion.

One direction is much easier than the other. One of them is the corollary of the first chapter, which said that a finite set is strictly larger than each of its proper subsets, so no bijection is available.

The other is best proved contrapositively: given an infinite $A$ , we produce a proper subset equinumerous with it. The theorem above puts a copy of $\mathbb{N}_0$ inside $A$ , and inside it we can shift along successors, as in $\mathbb{N}_0$ . So we shift inside the copy and do nothing outside it: the points of $A$ off the copy are left where they are, and the $n$ th point of the copy is sent to the $s(n)$ th. The result misses the initial point of the copy and nothing else, which is the proper subset we wanted.

Proof.

If $A$ is finite and $B \subsetneq A$ then $A \not\approx B$ , by the corollary on proper subsets.

Suppose $A$ is infinite. The theorem on countably infinite subsets gives an injection $f : \mathbb{N}_0 \to A$ ; write $E \defeq \operatorname{im} f$ and define

h : A \to A \setminus \{f(0)\}, \qquad h(x) \defeq \begin{cases} f(s(n)) & \text{if } x = f(n), \\ x & \text{if } x \notin E. \end{cases}

Each $x \in E$ is $f(n)$ for exactly one $n$ , since $f$ is injective, so $h$ is a function; and its values avoid $f(0)$ , since $f(s(n)) \neq f(0)$ by injectivity and $x \notin E$ gives $x \neq f(0)$ .

For injectivity, two points of $E$ are separated because $f$ and $s$ are injective, two points outside $E$ are unchanged, and a point of $E$ cannot collide with one outside $E$ because its value lies in $E$ . For surjectivity, a point $x \in A \setminus \{f(0)\}$ outside $E$ is $h(x)$ ; and a point $f(m)$ with $m \neq 0$ is $f(s(k))$ for the predecessor $k$ of $m$ , hence $h(f(k))$ .

So $A \approx A \setminus \{f(0)\}$ , a proper subset of $A$ .

Corollary 7.50 (Finiteness by self-maps).

A set $M$ is finite if and only if every injection $M \to M$ is surjective, and if and only if every surjection $M \to M$ is injective.

Proof.

If $M$ is finite, the theorem on finite self-maps gives both conditions. If $M$ is infinite, the map $h$ built in the last proof is an injection $M \to M$ whose range omits $f(0)$ , so it is injective and not surjective; and its inverse on that range, extended by sending $f(0)$ to itself, is a surjection $M \to M$ that is not injective.

Remark.

Only the second direction of the theorem used the axiom of choice, and used it twice, once to select elements and once through the recursion that strung the selections together. Dedekind took the property in the theorem as his definition of infinite, and without choice his definition and ours are not known to agree: there is no contradiction in a set that is infinite in our sense and yet admits no bijection with a proper subset. We accept choice, so for us the two notions are one.

Problem 7.23.

Show that a set $A$ is infinite if and only if for every $n \in \mathbb{N}_0$ there is an injection $L_n \to A$ .

Problem 7.24.

Let $A$ be infinite and $B$ countable. Show that $A \cup B \approx A$ .

Uncountable Sets

Every infinite set met so far has turned out to be countable, and the closure properties above keep it that way: subsets, products and countable unions of countable sets are all countable. Cantor’s theorem already gives sets that are not, and the same diagonal argument gives an explicit one.

Theorem 7.51 (Cantor's diagonal argument).

Let $A$ be a set with at least two elements. Then the Cartesian power $A^{\mathbb{N}_0}$ , the set of infinite sequences in $A$ , is uncountable.

Discussion.

Uncountable means neither finite nor countably infinite, and the characterisation of countable sets turns that into one statement: no surjection $\mathbb{N}_0 \to A^{\mathbb{N}_0}$ exists. So we assume a surjection, write $a^{(j)}$ for the sequence it puts at $j$ , and build a sequence it has missed.

The sequences are laid out as an infinite array whose $j$ th row is $a^{(j)}$ , and a sequence differs from the $j$ th row as soon as it differs from it in one place. The place we can always name is the $j$ th, on the diagonal of the array, so we build $\alpha$ by walking down the diagonal and disagreeing at every step. Disagreement is possible because $A$ has a second element, and we fix two elements of $A$ once and for all rather than choosing at each step, so the axiom of choice is not needed. Then $\alpha$ differs from every row, so it is not a value of the surjection.

Proof.

Fix $p, q \in A$ with $p \neq q$ , and suppose $\beta : \mathbb{N}_0 \to A^{\mathbb{N}_0}$ is a surjection. Write $a^{(j)} \defeq \beta(j)$ and $a^{(j)}_i \defeq a^{(j)}(i)$ , so the values are laid out as

\begin{array}{c|cccc} & 0 & 1 & 2 & \cdots \\ \hline a^{(0)} & \boxed{a^{(0)}_0} & a^{(0)}_1 & a^{(0)}_2 & \cdots \\ a^{(1)} & a^{(1)}_0 & \boxed{a^{(1)}_1} & a^{(1)}_2 & \cdots \\ a^{(2)} & a^{(2)}_0 & a^{(2)}_1 & \boxed{a^{(2)}_2} & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}

Define $\alpha : \mathbb{N}_0 \to A$ by $\alpha(j) \defeq p$ if $a^{(j)}_j \neq p$ , and $\alpha(j) \defeq q$ otherwise. Then $\alpha(j) \neq a^{(j)}_j$ for every $j$ , since in the second case $a^{(j)}_j = p \neq q$ .

Now $\alpha \in A^{\mathbb{N}_0}$ , so surjectivity gives $\alpha = a^{(j)}$ for some $j$ , and reading both sides at $j$ gives $\alpha(j) = a^{(j)}_j$ , which is false. So no surjection exists, and $A^{\mathbb{N}_0}$ is neither finite nor countably infinite by the characterisation of countable sets.

The same diagonal argument works for an arbitrary index set.

Theorem 7.52 (A set is smaller than its sequences).

Let $M$ be non-empty and let $A$ have at least two elements. Then $\#M < \#(A^M)$ .

Discussion.

The claim splits into an injection $M \to A^M$ and the absence of a bijection. For the injection we must attach to each point of $M$ a function on $M$ , and we use the point itself: send $m$ to the function that takes one fixed value at $m$ and the other everywhere else. Two distinct points give functions that disagree at either of them, so the assignment is injective.

For the second half the diagonal argument runs verbatim, with $M$ in place of $\mathbb{N}_0$ : given $f : M \to A^M$ , the function that disagrees with $f(m)$ at $m$ , for every $m$ , cannot be a value of $f$ . Nothing in that used the order or the countability of the index set, so only the two fixed elements of $A$ are needed.

Proof.

Fix $p, q \in A$ with $p \neq q$ . For $m \in M$ let $\chi_m : M \to A$ take the value $p$ at $m$ and $q$ elsewhere. If $m_1 \neq m_2$ then $\chi_{m_1}(m_1) = p$ while $\chi_{m_2}(m_1) = q$ , so $\chi_{m_1} \neq \chi_{m_2}$ ; hence $m \mapsto \chi_m$ is an injection $M \to A^M$ and $\#M \leqslant \#(A^M)$ .

Let $f : M \to A^M$ be any function and define $\varphi : M \to A$ by $\varphi(m) \defeq p$ if $f(m)(m) \neq p$ , and $\varphi(m) \defeq q$ otherwise, so that $\varphi(m) \neq f(m)(m)$ for every $m$ . If $\varphi = f(m)$ for some $m$ , reading both sides at $m$ gives $\varphi(m) = f(m)(m)$ , which is false. So $\varphi$ is not a value of $f$ , no $f$ is surjective, and no bijection exists.

Corollary 7.53 (Power sets as sequences).

For every set $M$ we have $\mathcal{P}(M) \approx L_2^{\,M}$ .

Proof.

Send $S \subset M$ to its characteristic function $\chi_S : M \to L_2$ , taking the value $1$ on $S$ and $0$ off it. Distinct subsets differ at some point, where their characteristic functions differ, so the map is injective; and any $\varphi : M \to L_2$ is the characteristic function of $\{x \in M \mid \varphi(x) = 1\}$ , so it is surjective.

So Cantor’s theorem is the previous theorem read at $A = L_2$ , and the diagonal argument that rules out surjections onto $L_2^{\,M}$ is the one that rules out surjections onto $\mathcal{P}(M)$ .

Corollary 7.54 (An uncountable set).

$\mathcal{P}(\mathbb{N}_0)$ is uncountable, and so is $L_2^{\,\mathbb{N}_0}$ .

Proof.

$L_2$ has two elements, so $L_2^{\,\mathbb{N}_0}$ is uncountable by the diagonal argument, and $\mathcal{P}(\mathbb{N}_0)$ is equinumerous with it by the corollary above; a set equinumerous with an uncountable set is uncountable, since countability is defined by the existence of a bijection.

Remark (Informal).

Assuming you know the real numbers, the same argument shows there are uncountably many. List candidate decimal expansions of the numbers between $0$ and $1$ and build a new expansion differing from the $j$ th in its $j$ th digit, avoiding the digit $9$ so as not to fall foul of the two expansions some numbers have; the number it names is missing from the list. Since the fractions are countable, the numbers that are not fractions must be uncountable, for otherwise the reals would be a union of two countable sets.

The same counting argument shows something less expected. The polynomials with fractional coefficients are countable, being determined by finite lists of fractions, and each has finitely many roots, so the numbers that are roots of such a polynomial form a countable set. Uncountably many real numbers are therefore roots of no such polynomial at all, although exhibiting even one takes real work.

Problem 7.25.

Let $A$ be uncountable and let $B \subset A$ be countable. Show that $A \setminus B$ is uncountable.

Problem 7.26.

Deduce from Cantor’s theorem that there is no surjection $M \to \mathcal{P}(M)$ , for any set $M$ .

Cardinal Numbers

We have been writing $\#A \leqslant \#B$ for arbitrary sets without saying what $\#A$ names when $A$ is infinite, and so far we have not needed to: the three relations were defined as statements about functions, and the symbol never occurred alone. It can be given a meaning, using the equivalence classes from the remark of the first section.

Definition 7.55 (Cardinal number).

Fix a set $X$ . The cardinal number of a subset $A \subset X$ is its equivalence class $\#A \in \mathcal{P}(X)/{\approx}$ under equinumerosity. Two subsets of $X$ have the same cardinal number exactly when they are equinumerous.

Remark.

The restriction to a fixed $X$ is needed, since the sets equinumerous with a given one do not form a set. For finite $A$ nothing is lost by reading $\#A$ as the natural number counting it, because the cardinal numbers of finite subsets of $X$ correspond to the cuts, one class for each $n$ with $L_n \subset X$ . A definition free of the ambient $X$ needs the ordinal numbers, which we have not built.

The order has the properties the notation suggests. Reflexivity and transitivity were proved directly, Schröder–Bernstein gives antisymmetry, since two subsets each injecting into the other are equinumerous and so name one class, and comparability makes any two classes comparable. So on $\mathcal{P}(X)/{\approx}$ the relation $\leqslant$ is a total order, with the strict comparison as its strict part.

Definition 7.56 (Arithmetic of cardinal numbers).

For sets $A$ and $B$ define

\#A + \#B \defeq \#\bigl((A \times \{0\}) \cup (B \times \{1\})\bigr), \qquad \#A \cdot \#B \defeq \#(A \times B),

(\#A)^{\#B} \defeq \#(A^B).

The tagged copies in the sum are there because $A$ and $B$ may share elements, and tagging makes disjoint copies of both without disturbing their sizes. That the three operations depend only on the classes and not on the sets chosen to represent them is the transport problem set above. On finite sets they agree with the arithmetic of the first chapter, by the theorems on disjoint unions and products and by the problem on $A^B$ .

Since $\mathcal{P}(M) \approx L_2^{\,M}$ and $\#L_2 = 2$ , the power set records an exponential:

2^{\#M} = \#\mathcal{P}(M),

and Cantor’s theorem reads $\#M < 2^{\#M}$ . Writing $\aleph_0 \defeq \#\mathbb{N}_0$ for the smallest infinite cardinal number, the chain

\aleph_0 < 2^{\aleph_0} < 2^{2^{\aleph_0}} < \cdots

climbs for ever, so there is no largest size.

Remark (The continuum hypothesis).

Nothing proved here says whether anything sits between two consecutive terms of that chain. The continuum hypothesis asserts that nothing sits between the first two: there is no set $M$ with $\aleph_0 < \#M < 2^{\aleph_0}$ . Gödel showed in 1938 that it cannot be refuted from the axioms we have listed, choice included, and Cohen showed in 1963 that it cannot be proved from them either, so long as those axioms are consistent at all. It is independent, in the sense the axiom of choice was said to be independent, and one may add it or its negation without introducing a contradiction that was not already there.

Remark.

Cantor’s theorem also settles again a question from the naive chapter. Were there a set $\mathfrak{M}$ holding every set, then $\mathcal{P}(\mathfrak{M})$ would be one of its subsets, so

\#\mathcal{P}(\mathfrak{M}) \leqslant \#\mathfrak{M} < \#\mathcal{P}(\mathfrak{M}),

which trichotomy forbids. That is Cantor’s own argument, and it reaches the conclusion of the proposition on no set of all sets by counting rather than by self-membership.

Problem 7.27.

Show that cardinal addition and multiplication are commutative and associative, and that $\#A \cdot (\#B + \#C) = \#A \cdot \#B + \#A \cdot \#C$ . Each identity is a bijection between the sets involved.

Problem 7.28.

Show that $\aleph_0 + \aleph_0 = \aleph_0$ and $\aleph_0 \cdot \aleph_0 = \aleph_0$ , and explain which theorems of this chapter each one is.

Problem 7.29.

Show that $A \approx B$ and $C \approx D$ imply $A^C \approx B^D$ .

Problem 7.30.

Show that $(A \times B)^C \approx A^C \times B^C$ for all sets $A$ , $B$ and $C$ .

Exercises

Exercises in Lean

The proofs below are checked in your browser. Nothing is sent anywhere, and nothing is stored but your own work. Type \to for →, \and for ∧, \< and \> for ⟨ ⟩.

The last sheet gave the checker the order on the carrier and the sets separation carves out. This chapter counts, and counting needs two things the checker has not had: a map read between two sets rather than across the whole universe, and the natural numbers as objects of that universe, so that the cuts have something to hold.

Maps between two sets

A map has so far been an arrow f : Obj → Obj on everything at once, and Injective f and Surjective f asked their conditions everywhere. A map of this chapter has a domain and a codomain, so the same conditions are written on the sets they are asserted of, and each is the statement the definitions of a surjection and an injection make it:

MapsTo f A B   is   ∀ x, x ∈ A → f x ∈ B
InjOn f A      is   ∀ x y, x ∈ A → y ∈ A → f x = f y → x = y
SurjOn f A B   is   ∀ y, y ∈ B → ∃ x, x ∈ A ∧ f x = y
BijOn f A B    is   MapsTo f A B ∧ InjOn f A ∧ SurjOn f A B

Being those statements and not merely equivalent to them, they are opened with intro and used by applying them, in the way A ⊆ B has been since the sets sheet. A BijOn splits three ways at once, so obtain ⟨hm, hi, hs⟩ takes one apart and refine ⟨?_, ?_, ?_⟩ puts one together.

Example.

A map injective everywhere is injective on any set. The converse fails, which is why the two conditions are kept apart.

Example.

The values of $f$ on $A$ lie in $B$ , which is where $g$ is known to do its work.

Exercise 7.10.

A restriction of an injection is an injection.

Exercise 7.11.

The codomain may be enlarged freely.

Exercise 7.12.

The first part of the theorem that bijections compose, read on two sets.

Exercise 7.13.

Its second part.

Exercise 7.14.

Only the first map need be injective for the composite to be.

Exercise 7.15.

The inverse of a bijection is a bijection. The checker cannot choose preimages, so the inverse arrives as a hypothesis rather than being built; what is left is the second part of Proposition 7.2 .

Comparing sizes

Two sets are compared by the functions running between them, which is what Definition 7.1 and Definition 7.31 say, so the two comparisons are typed as the existence statements they are:

A ≈ B   is   ∃ f : Obj → Obj, BijOn f A B
A ≼ B   is   ∃ f : Obj → Obj, MapsTo f A B ∧ InjOn f A

typed \approx and \preceq. The second is $\#A \leqslant \#B$ ; the strict comparison is not notation of its own, being the second together with the denial of the first. A witness is supplied by use or as the first component of refine ⟨_, ?_, ?_⟩, and it is written as a function, fun x => x for the identity.

Example.

The first part of Proposition 7.2 . Each of the three conditions is read at a point, and the identity satisfies all three without any work.

Exercise 7.16.

The empty function.

Exercise 7.17.

A subset is no larger than the set it sits in.

Exercise 7.18.

A bijection is in particular an injection.

Exercise 7.19.

The second part of Proposition 7.32 .

Exercise 7.20.

The third part of Proposition 7.2 .

Exercise 7.21.

Problem 7.13 : a comparison transported along two bijections.

Exercise 7.22.

Proposition 7.8 .

Exercise 7.23.

Problem 7.16 . The map sends a subset of $A$ to its image. (Harder.)

Cantor’s theorem

Exercise 7.24.

The first half of Cantor’s theorem.

Exercise 7.25.

And the second, which is Problem 7.26 . The diagonal set is {x ∈ S | x ∉ f x}, and it is a subset of $S$ before it is anything else.

The natural numbers as objects

The carrier ℕ of the Peano sheet is a type of its own, and the cuts are sets, so the two have to be brought together before a cut can be written down. ↑n, typed \up, is the object the number n names; ω is the set of all of them, which is $\mathbb{N}_0$ ; and L n is the cut. Their criteria are the definitions:

x ∈ ω     is   ∃ k : ℕ, x = ↑k
x ∈ L n   is   ∃ k : ℕ, k < n ∧ x = ↑k

so a membership is built with use or ⟨_, _, _⟩ and taken apart with obtain ⟨k, hk, he⟩. That distinct numbers name distinct objects is Num.inj, and Finite A is ∃ n : ℕ, A ≈ L n, which is Definition 7.5 verbatim.

Example.

The witness is the number itself, and the equation it has to satisfy is an identity.

Example.

A strict inequality is a positive difference, and no positive difference reaches $0$ . It is listed below as Nat.not_lt_zero, so the exercises may lean on it.

Exercise 7.26.

The first part of Proposition 7.4 .

Exercise 7.27.

Its last part.

Exercise 7.28.

Every cut sits inside $\mathbb{N}_0$ .

Exercise 7.29.

And the cuts grow with their index.

Exercise 7.30.

The middle part of Proposition 7.4 , which every induction of this chapter turns on.

Exercise 7.31.

A cut counts itself.

Exercise 7.32.

And the empty set is counted by the cut that holds nothing.

Exercise 7.33.

A set with one point in it. (Harder.)

Exercise 7.34.

A finite set injects into $\mathbb{N}_0$ , which is the easy half of the characterisation of countability.

What the checker understands

Tactics

intro h	assume the hypothesis of an implication, naming it h
exact e	give the proof outright
apply f	reduce the goal to the hypotheses of f
assumption	close the goal with a hypothesis already present
trivial	close the goal True
exfalso	replace the goal with False
by_contra h	assume the negation of the goal
constructor	split ∧ into both halves, or ↔ into both directions
left / right	choose which half of a ∨ to prove
rcases h with a \| b	argue by cases on a disjunction
obtain ⟨a, b⟩ := h	take a conjunction or an existential apart
cases h	as above, keeping the name
refine e	give the proof with holes left in it
have h : p := …	record an intermediate result
show p	restate the goal in an equal form
use w	give a witness for ∃
specialize h a	instantiate a ∀ hypothesis
rw [h]	rewrite with an equation, ← to go backwards
rfl	both sides compute to the same thing
decide / norm_num	settle a closed computation
tauto	close a goal that is true by pure logic
induction n with k ih	the fifth Peano condition: prove the goal at 0, then at succ k from ih

Results you may cite

Classical.em	∀ (a : Prop), a ∨ ¬a — the law of excluded middle
Classical.byContradiction	∀ {a : Prop}, (¬a → False) → a — proof by contradiction; the tactic by_contra does this for you
Classical.byCases	∀ {a b : Prop}, (a → b) → (¬a → b) → b — split on whether a holds
not_not	∀ {a : Prop}, ¬¬a ↔ a — double negation
not_and_or	∀ {a b : Prop}, ¬(a ∧ b) ↔ ¬a ∨ ¬b — De Morgan
not_or	∀ {a b : Prop}, ¬(a ∨ b) ↔ ¬a ∧ ¬b — De Morgan
not_imp	∀ {a b : Prop}, ¬(a → b) ↔ a ∧ ¬b
and_comm	∀ {a b : Prop}, a ∧ b ↔ b ∧ a
or_comm	∀ {a b : Prop}, a ∨ b ↔ b ∨ a
Set.ext	∀ {A B : Obj}, (∀ x : Obj, x ∈ A ↔ x ∈ B) → A = B — extensionality: sets with the same elements are equal
Set.ext_iff	∀ {A B : Obj}, A = B ↔ (∀ x : Obj, x ∈ A ↔ x ∈ B) — extensionality and substitution, in one biconditional
Set.subset_antisymm	∀ {A B : Obj}, A ⊆ B → B ⊆ A → A = B — mutual inclusion is equality
Set.empty_subset	∀ {A : Obj}, ∅ ⊆ A — the empty set is a subset of every set
Set.pair_eq	∀ {a b c d : Obj}, ((a, b) = (c, d)) ↔ (a = c ∧ b = d) — two ordered pairs are equal exactly when their coordinates are
Nat.succ_inj	∀ {m n : ℕ}, succ m = succ n → m = n — the third Peano condition: the successor is injective
Nat.succ_ne_zero	∀ (n : ℕ), succ n ≠ 0 — the fourth: zero is nobody's successor
Nat.pred	∀ {n : ℕ}, n ≠ 0 → ∃ m : ℕ, n = succ m — predecessors: everything but zero is a successor
Nat.add_zero	∀ (m : ℕ), m + 0 = m — the first clause of addition
Nat.add_succ	∀ (m n : ℕ), m + succ n = succ (m + n) — the second clause of addition
Nat.zero_add	∀ (n : ℕ), 0 + n = n — addition from the left
Nat.succ_add	∀ (m n : ℕ), succ m + n = succ (m + n) — addition from the left, at a successor
Nat.add_assoc	∀ (m n p : ℕ), (m + n) + p = m + (n + p) — addition is associative
Nat.add_comm	∀ (m n : ℕ), m + n = n + m — addition is commutative
Nat.add_ne_zero	∀ {a : ℕ} (b : ℕ), a ≠ 0 → a + b ≠ 0 — positivity is absorbing
Nat.mul_zero	∀ (m : ℕ), m * 0 = 0 — the first clause of multiplication
Nat.mul_succ	∀ (m n : ℕ), m * succ n = m * n + m — the second clause of multiplication
Nat.zero_mul	∀ (m : ℕ), 0 * m = 0 — multiplication from the left
Nat.succ_mul	∀ (m n : ℕ), succ m * n = m * n + n — multiplication from the left, at a successor
Nat.add_right_cancel	∀ {m n k : ℕ}, m + k = n + k → m = n — cancellation, from the last sheet
Nat.add_eq_zero	∀ {m n : ℕ}, m + n = 0 → m = 0 ∧ n = 0 — a sum is zero only when both parts are, from the last sheet
Nat.mul_comm	∀ (m n : ℕ), m * n = n * m — multiplication is commutative, from the last sheet
Nat.mul_add	∀ (m n p : ℕ), m * (n + p) = m * n + m * p — multiplication distributes over addition, from the last sheet
Nat.mul_assoc	∀ (m n p : ℕ), (m * n) * p = m * (n * p) — multiplication associates, from the problems of the last chapter
Nat.add_mul	∀ (m n p : ℕ), (m + n) * p = m * p + n * p — distributivity on the other side
Nat.add_left_cancel	∀ {a m n : ℕ}, a + m = a + n → m = n — uniqueness of differences
Nat.lt_trichotomy	∀ (m n : ℕ), m < n ∨ m = n ∨ n < m — trichotomy, from the theorem that ℕ is strictly ordered
Nat.lt_irrefl	∀ (n : ℕ), ¬(n < n) — anti-reflexivity, from the last sheet
Nat.lt_trans	∀ {m n p : ℕ}, m < n → n < p → m < p — transitivity of the strict order, from the last sheet
Nat.lt_succ_self	∀ (n : ℕ), n < succ n — every number is below its successor, from the last sheet
Nat.not_lt_zero	∀ {n : ℕ}, ¬(n < 0) — nothing lies below zero
Nat.lt_succ_iff	∀ {m n : ℕ}, m < succ n ↔ m < n ∨ m = n — nothing lies strictly between n and succ n
Num.inj	∀ {m n : ℕ}, ↑m = ↑n → m = n — distinct numbers name distinct objects of ω

From the logical core

And.intro	∀ {a b : Prop}, a → b → a ∧ b
And.left	∀ {a b : Prop}, a ∧ b → a
And.right	∀ {a b : Prop}, a ∧ b → b
And.symm	∀ {a b : Prop}, a ∧ b → b ∧ a
Or.inl	∀ {a b : Prop}, a → a ∨ b
Or.inr	∀ {a b : Prop}, b → a ∨ b
Or.elim	∀ {a b c : Prop}, a ∨ b → (a → c) → (b → c) → c
Or.symm	∀ {a b : Prop}, a ∨ b → b ∨ a
Iff.intro	∀ {a b : Prop}, (a → b) → (b → a) → (a ↔ b)
Iff.mp	∀ {a b : Prop}, (a ↔ b) → a → b
Iff.mpr	∀ {a b : Prop}, (a ↔ b) → b → a
Iff.symm	∀ {a b : Prop}, (a ↔ b) → (b ↔ a)
Iff.rfl	∀ {a : Prop}, a ↔ a
Iff.trans	∀ {a b c : Prop}, (a ↔ b) → (b ↔ c) → (a ↔ c)
True.intro	True
False.elim	∀ {a : Prop}, False → a
absurd	∀ {a b : Prop}, a → ¬a → b
id	∀ {a : Prop}, a → a
mt	∀ {a b : Prop}, (a → b) → ¬b → ¬a
Eq.refl	∀ {α : Type} (a : α), a = a
Eq.symm	∀ {α : Type} {a b : α}, a = b → b = a
Eq.trans	∀ {α : Type} {a b c : α}, a = b → b = c → a = c
congrArg	∀ {α : Type} {β : Type} {a b : α} (f : α → β), a = b → f a = f b
Exists.intro	∀ {α : Type} {p : α → Prop} (w : α), p w → ∃ x : α, p x
Exists.elim	∀ {α : Type} {p : α → Prop} {b : Prop}, (∃ x : α, p x) → (∀ y : α, p y → b) → b

Lesson 8

Permutations

Taught 18 Aug 2026

Permutations

A bijection from a set to itself maps the set onto itself and may rearrange its elements. Counting was the first thing we did with finite sets, and the second is this: fix the set and ask in how many ways its points can be rearranged among themselves. The rearranging functions can be composed, and this chapter studies that composition.

Two-Row Notation

Definition 8.1 (Permutation).

Let $A$ be a set. A permutation of $A$ is a bijection $\sigma : A \to A$ .

When $A$ is finite and non-empty a permutation is settled by naming the value it takes at each point, and there are only finitely many points to name. So the whole function fits in a table of two rows: the points along the top, their values underneath.

Definition 8.2 (Blocks and $n$ -sets).

For $n \in \mathbb{N}_0$ write

\langle n \rangle \defeq \{x \in \mathbb{N} \mid x \leqslant n\} = \{1, 2, \ldots, n\},

so that $\langle 0 \rangle = \emptyset$ . A finite set $A$ with cardinality $\#A = n$ is an $n$ -set, and a subset of a set that is itself a $k$ -set is a $k$ -subset of it.

The block $\langle n \rangle$ counts as the cut $L_n$ does: $i \mapsto i + 1$ is a bijection from $L_n$ onto $\langle n \rangle$ , injective by uniqueness of differences and surjective because every positive number is a successor. So $\#\langle n \rangle = n$ , and an $n$ -set is exactly a set equinumerous with $\langle n \rangle$ . We take the block rather than the cut as the standing domain here only because it makes the tables below read as they do everywhere else, starting at one.

Definition 8.3 (Two-row notation).

Let $A$ be an $n$ -set with $n \in \mathbb{N}$ , let $a_1, \ldots, a_n$ be a listing of its elements, and let $\sigma$ be a permutation of $A$ . Write

\sigma = \begin{pmatrix} a_1 & a_2 & \cdots & a_n \\ \sigma(a_1) & \sigma(a_2) & \cdots & \sigma(a_n) \end{pmatrix}

for the permutation sending each $a_i$ to $\sigma(a_i)$ . The columns may be reordered at will: any listing of the domain along the top, with the matching values underneath, denotes the same function.

The last sentence is equality of functions and nothing more. A table names a function by naming its value at each point, and the order in which the points are named is no part of the function.

Example 8.4 (A permutation of four points).

On $A = \{1, 2, 3, 4\}$ the rule $\sigma(1) = 4$ , $\sigma(2) = 1$ , $\sigma(3) = 3$ , $\sigma(4) = 2$ is the permutation

\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 3 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 1 & 4 & 2 \\ 3 & 4 & 2 & 1 \end{pmatrix},

the second table being the first with its columns shuffled. A two-row table fails to describe a permutation exactly when its bottom row is not a rearrangement of its top: a repeated value breaks injectivity, and a missing one breaks surjectivity.

Definition 8.5 (The symmetric group).

The set of all permutations of a set $A$ is written $S_A$ and called the symmetric group on $A$ . When $A = \langle n \rangle$ one writes $S_n$ rather than $S_{\langle n \rangle}$ .

The word group is traditional and we use it as a name only. What stands behind it is that $S_A$ carries a multiplication and is closed under composition and inverses: composition sends bijections to bijections, the identity map is a bijection since it is its own inverse, and the inverse of a bijection is a bijection. No abstract definition of a group is needed below; we use only the set $S_A$ and its composition.

Working in $S_n$ rather than in $S_A$ for an arbitrary $n$ -set loses nothing. A listing of $A$ is a bijection $h : \langle n \rangle \to A$ , and $\sigma \mapsto h^{-1} \circ \sigma \circ h$ carries permutations of $A$ to permutations of $\langle n \rangle$ , reversibly. That is what two-row notation already does silently when it writes the points in a row.

Definition 8.6 (Product of permutations).

Let $\sigma, \lambda \in S_A$ . The product $\sigma\lambda$ is the composite $\sigma \circ \lambda$ , so that

(\sigma\lambda)(x) = \sigma\bigl(\lambda(x)\bigr) \qquad \text{for every } x \in A.

One applies $\lambda$ first and $\sigma$ second. The permutations $\sigma$ and $\lambda$ are the factors of the product.

Example 8.7 (Multiplying in $S_4$ ).

Take

\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 1 & 4 & 2 \end{pmatrix}, \qquad \lambda = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 1 & 3 & 2 \end{pmatrix}

in $S_4$ . To compute $\sigma\lambda$ , follow each point through $\lambda$ and then through $\sigma$ :

\begin{aligned} 1 &\overset{\lambda}{\longmapsto} 4 \overset{\sigma}{\longmapsto} 2, \\ 2 &\overset{\lambda}{\longmapsto} 1 \overset{\sigma}{\longmapsto} 3, \\ 3 &\overset{\lambda}{\longmapsto} 3 \overset{\sigma}{\longmapsto} 4, \\ 4 &\overset{\lambda}{\longmapsto} 2 \overset{\sigma}{\longmapsto} 1, \end{aligned}

so that

\sigma\lambda = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \end{pmatrix}, \qquad \lambda\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix},

the second computed the same way in the other order. So $\sigma\lambda \neq \lambda\sigma$ : the multiplication is not commutative, though particular pairs may commute.

Remark (Authors who read left to right).

Some texts write the value of a function to the right of its argument, $(x)\sigma$ in place of $\sigma(x)$ , and then read a product left to right. We keep the order fixed for composition, so $\sigma\lambda$ always means ” $\lambda$ first, then $\sigma$ ”. When reading another source, work one product out by hand before trusting its convention.

The bottom row of a two-row table for an element of $S_n$ is a rearrangement of $1, \ldots, n$ , and conversely each rearrangement of $1, \ldots, n$ is the bottom row of exactly one element of $S_n$ , the one whose top row runs in ascending order. So $S_n$ is matched point for point with the orderings of $\langle n \rangle$ , which is why permutations are so often described as arrangements rather than as functions.

Permutations describe any situation in which $n$ objects trade $n$ places, one object to a place. Number the places $1$ through $n$ ; a process that carries whatever sits in place $i$ to place $j$ is recorded by the permutation $\sigma$ with $\sigma(i) = j$ , and performing two such processes in turn is recorded by the product.

Every set has two permutations we already know. One is the identity, a bijection because it is its own inverse. The other is the transposition $\tau_{a,b}$ of the counting chapter, which exchanges $a$ and $b$ and fixes every other point; it is its own inverse, hence a permutation of whatever set it is defined on. So $S_A$ is never empty, and holds more than one element as soon as $A$ holds two distinct points.

Problem 8.1.

Write out every element of $S_2$ and of $S_3$ in two-row notation, with the top row ascending, and check that you have two and six of them respectively.

Problem 8.2.

With $\sigma$ and $\lambda$ as in the example of multiplication in $S_4$ , compute $\sigma\sigma$ and $\lambda\lambda$ , and decide whether either is the identity. Compute $\tau_{1,2}\,\sigma$ and $\sigma\,\tau_{1,2}$ and observe that they differ.

Problem 8.3.

Let $A$ be a set and let $h : \langle n \rangle \to A$ be a bijection. Show that $\sigma \mapsto h^{-1} \circ \sigma \circ h$ is a bijection from $S_A$ onto $S_n$ , and that it carries products to products.

Counting Permutations

The theorem on cardinality of a product already suggests how many bijections run between two $n$ -sets: $n$ values are available for the first point, then $n - 1$ for the second once the first is used, and so on down. The shorthand for that descending product is the factorial.

Definition 8.8 (Factorial).

Define $n!$ for $n \in \mathbb{N}_0$ by the recursion

0! \defeq 1, \qquad s(n)! \defeq s(n) \cdot n!,

so that $n! = n(n-1)\cdots 2 \cdot 1$ for $n \in \mathbb{N}$ . The number $n!$ is the factorial of $n$ .

That the recursion defines exactly one function is the recursion theorem, applied as it was for addition. The value $0! = 1$ is the empty product, chosen for the same reason the empty sum is $0$ : it starts the recursion, and it spares us a separate case in every count below.

Theorem 8.9 (Number of bijections between $n$ -sets).

Let $A$ and $B$ be $n$ -sets for some $n \in \mathbb{N}_0$ . Then there are exactly $n!$ bijections from $A$ to $B$ . In particular $\#S_n = n!$ , and the empty set has exactly one permutation, in agreement with $0! = 1$ .

Discussion.

The claim is a counting statement about a set we have not yet named: writing $J$ for the set of bijections $A \to B$ , we must show $\#J = n!$ . The factorial was defined by a recursion on $n$ , so the proof is an induction on $n$ whose step must produce the factor $s(n)$ . At $n = 0$ both sets are empty and the empty function is the only function between them, so there is one, which is $0!$ . For the step we split $J$ according to where a chosen point $a$ of $A$ goes: there are $s(n)$ possible values, the pieces of the split are pairwise disjoint and cover $J$ , and each piece is matched with the bijections between two $n$ -sets by restricting to what is left of $A$ . By the inductive hypothesis each piece has $n!$ elements, and the cardinality of a product gives the total $s(n) \cdot n!$ , which is the recursion clause for the factorial.

Proof.

We induct on $n$ . If $n = 0$ then $A = B = \emptyset$ , and the empty function is the only function $\emptyset \to \emptyset$ ; it is a bijection, so there is $1 = 0!$ of them.

Suppose the claim holds for all pairs of $n$ -sets, and let $\#A = \#B = s(n)$ . Fix $a \in A$ and, for each $b \in B$ , let $J_b$ be the set of bijections $\sigma : A \to B$ with $\sigma(a) = b$ . Every bijection $A \to B$ lies in exactly one $J_b$ , namely the one indexed by its value at $a$ , so the $J_b$ are pairwise disjoint and their union is $J$ .

Fix $b$ . Restriction to $A \setminus \{a\}$ carries a member of $J_b$ to a bijection $A \setminus \{a\} \to B \setminus \{b\}$ , by the proposition that bijections remove points; and every such bijection extends to exactly one member of $J_b$ , by sending $a$ to $b$ . So $J_b$ is matched with the set of bijections between two $n$ -sets, which the inductive hypothesis counts as $n!$ , and $\#J_b = n!$ .

The index set $B$ is an $s(n)$ -set, so $J$ is a union of $s(n)$ pairwise disjoint sets each of cardinality $n!$ ; matching it with $B \times \langle n! \rangle$ and applying the cardinality of a product gives

\#J = s(n) \cdot n! = s(n)!,

which is the claim at $s(n)$ . Taking $A = B = \langle n \rangle$ gives $\#S_n = n!$ .

Example 8.10 (The sets $S_1$ , $S_2$ and $S_3$ ).

$S_1$ holds only the identity on $\{1\}$ . $S_2$ holds the two elements

\begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix}, \qquad \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix},

and $S_3$ holds $3! = 6$ :

\begin{aligned} &\begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix}, &&\begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix}, &&\begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix}, \\[1ex] &\begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix}, &&\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix}, &&\begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix}. \end{aligned}

Without the theorem one would need care to be sure such a list is complete; with it, six is enough.

Example 8.11 (Selections and orderings).

A competition offers five water events, six running events, four cycling events and seven self-defence events. One event may be chosen from each category in $5 \cdot 6 \cdot 4 \cdot 7 = 840$ ways, by the cardinality of a product applied three times. If a competitor also fixes the order in which to attempt the four chosen events, each selection admits $4! = 24$ orders, so there are $840 \cdot 24 = 20{,}160$ full programmes.

Example 8.12 (Non-attacking rooks).

Eight rooks stand on a chessboard with no two attacking, which means no two share a row or a column, since a rook commands both. Each row holds at most one rook and there are eight rooks for eight rows, so every row holds exactly one; the same count puts exactly one in each column. Let $\sigma(i)$ be the column of the rook in row $i$ . The column condition says precisely that $\sigma$ is injective, hence a permutation of $\langle 8 \rangle$ by the theorem that injective and bijective agree on a finite set; and conversely each $\sigma \in S_8$ places one rook in each row, in the columns $\sigma(1), \ldots, \sigma(8)$ , no two equal. So the acceptable positions are matched with $S_8$ , and there are $8! = 40{,}320$ of them.

If the rooks are told apart, say by eight colours, choose the squares first in $8!$ ways and then distribute the colours over them in $8!$ ways, for $(8!)^2 = 1{,}625{,}702{,}400$ arrangements.

Example 8.13 (Necklaces).

Seven beads of distinct colours are strung on a cord whose ends are then tied, and two necklaces count as the same when one can be turned or flipped into the other. Before the knot the beads lie in a row, and there are $7! = 5040$ rows. Fix one necklace and ask how many rows display it: choose which bead is to sit at the left end, then which of that bead’s two neighbours follows it, and the rest of the row is forced, so $7 \cdot 2 = 14$ rows, all different because the colours are. Every row displays exactly one necklace, so fourteen times the number of necklaces is $5040$ , and there are $360$ necklaces.

Said the other way, call two rows equivalent when one is a rotation of the other or of its reversal; that is an equivalence relation, its classes are the necklaces, and each class holds fourteen rows.

Problem 8.4.

Show that $s(n)! = n! + n \cdot n!$ for every $n \in \mathbb{N}_0$ , and that $n! \geqslant n$ for every $n$ , with equality only at $n = 1$ and $n = 2$ .

Problem 8.5.

Give a second proof of the theorem on the number of bijections that does not partition $J$ , but instead builds a bijection one value at a time and appeals to the product of cardinalities directly. Which of the two proofs makes the appearance of $n!$ more transparent?

Problem 8.6.

Let $A$ be an $n$ -set. Show that the permutations of $A$ fixing a chosen point of $A$ number $(n-1)!$ , and that those moving every point of $A$ number fewer than $n!$ as soon as $n \geqslant 1$ .

Symmetries of Regular Polygons

Permutations record how $n$ labelled places trade their occupants. The symmetries of a regular polygon are an example: each rigid motion carrying the polygon onto itself shuffles the corners, and the motion is settled by the shuffle.

Remark (What is borrowed from geometry).

The plane, the length of a segment and the measure of an angle are borrowed here in the same spirit as the rules of school algebra were borrowed earlier: openly, and without being built. Two geometric facts are used and not proved: that a symmetry of a polygon carries corners to corners, and that a symmetry is settled by what it does to the corners. Everything else in this section is a statement about permutations, and is proved.

Definition 8.14 (Symmetry of a figure).

A figure is a subset $F$ of the plane. A symmetry of $F$ is a bijection $s$ of the plane onto itself such that $s(F) = F$ and such that the distance from $s(x)$ to $s(y)$ equals the distance from $x$ to $y$ , for all points $x$ and $y$ . Briefly: a rigid motion of the plane carrying $F$ onto itself.

Definition 8.15 (Polygon).

A polygon with $n$ corners, where $n \geqslant 3$ , is a figure consisting of points $v_1, \ldots, v_n$ , the corners, together with the segments joining $v_i$ to $v_{i+1}$ for $i \in \langle n \rangle$ , where $v_{n+1}$ means $v_1$ ; those segments are the edges. A polygon with $n$ corners is an $n$ -gon, and it is regular if all its edges have the same length and all its interior angles the same measure.

Label the corner positions of an $n$ -gon by $1, \ldots, n$ . A symmetry $s$ then determines a permutation $\sigma \in S_n$ : let $\sigma(i)$ be the label of the position to which $s$ carries the corner that started at position $i$ . Distinct symmetries determine distinct permutations, since a symmetry is settled by what it does to the corners, so the map $s \mapsto \sigma$ is injective and an $n$ -gon has at most $\#S_n = n!$ symmetries. Composing symmetries multiplies the permutations: if $s_1$ and $s_2$ answer to $\sigma_1$ and $\sigma_2$ , then $s_2 \circ s_1$ answers to $\sigma_2\sigma_1$ , because both sides send a corner to the same place. Questions of the form “what happens if these motions are performed in turn?” therefore become products in $S_n$ .

Example 8.16 (Symmetries of an equilateral triangle).

Let $P$ be an equilateral triangle with its corner positions labelled $1, 2, 3$ , and let $\ell_i$ be the line through position $i$ and the midpoint of the opposite edge. Writing $s_1, \ldots, s_6$ for the six symmetries of $P$ and $\sigma_1, \ldots, \sigma_6$ for the permutations they determine, the dictionary is

\begin{aligned} \sigma_1 &= \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix} && \text{identity}, \\ \sigma_2 &= \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix} && \text{reflection in } \ell_1, \\ \sigma_3 &= \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix} && \text{reflection in } \ell_2, \\ \sigma_4 &= \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix} && \text{reflection in } \ell_3, \\ \sigma_5 &= \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} && \text{turn by a third}, \\ \sigma_6 &= \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} && \text{turn by two thirds}. \end{aligned}

The six permutations listed are distinct and $\#S_3 = 6$ , so the list is complete in two senses at once: the triangle has no further symmetries, and every element of $S_3$ comes from one.

For the triangle every permutation of the corners comes from a symmetry. For larger polygons this fails. For a square with corners labelled $1, 2, 3, 4$ in order round the boundary, the permutation

\begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 3 & 2 & 4 \end{pmatrix}

holds position $1$ still while exchanging the two positions next to it, and no motion preserving distance can do that, since the two exchanged corners are at different distances from the first. For a regular $n$ -gon the count is exactly $2n$ .

Proposition 8.17 (Number of symmetries of a regular $n$ -gon).

A regular $n$ -gon has exactly $2n$ symmetries. Equivalently, the permutations in $S_n$ determined by those symmetries form a $2n$ -subset of $S_n$ .

Discussion.

The claim is a count, and the object counted is the set of symmetries, which we have already matched injectively with a subset of $S_n$ ; so it is enough to count the permutations that arise. A symmetry is settled by what it does to the corners, and in fact by what it does to two neighbouring corners, so the count is a product of two choices in the manner of the cardinality of a product. The first corner may go to any of the $n$ positions, which is the first factor. Its neighbour is one edge away from it, and a motion preserving distance must leave it one edge away from wherever the first corner went, so only the two positions adjacent to that one are available, which is the second factor. That those two choices determine the rest is the geometric fact we have assumed: the remaining corners are reached by stepping round the boundary, and each step is forced once the direction and the starting position are fixed.

Proof.

Label the corner positions $1, \ldots, n$ in order round the boundary and let $\sigma \in S_n$ come from a symmetry $s$ .

There are $n$ possible values for $\sigma(1)$ . The corners at positions $1$ and $2$ are joined by an edge, so $s$ carries them to corners joined by an edge, and hence $\sigma(2)$ is one of the two positions adjacent to $\sigma(1)$ : two possible values. Once $\sigma(1)$ and $\sigma(2)$ are fixed, so is the direction in which $s$ runs round the boundary, and every further corner is reached from the previous one by one edge in that direction; so $\sigma(3), \ldots, \sigma(n)$ are determined.

Each of the $n \cdot 2$ pairs of choices is realised, by a turn when the direction is preserved and by a reflection when it is reversed, and distinct pairs give distinct permutations, hence distinct symmetries. So there are exactly $2n$ symmetries.

Remark.

The same $2n$ can be counted a second way, by kind rather than by choice: there are $n$ turns, through none, one, …, $n-1$ steps round the boundary, and $n$ reflections. When $n$ is odd every reflection has its line through one corner and the midpoint of the opposite edge; when $n$ is even, half of the lines pass through two opposite corners and half through the midpoints of two opposite edges. Either way the total is $2n$ .

Definition 8.18 (Dihedral group).

The set of permutations in $S_n$ determined by the symmetries of a regular $n$ -gon is written $D_n$ and called the dihedral group of degree $n$ . By the proposition, $\#D_n = 2n$ and $D_n \subset S_n$ .

Working with the motions of a regular $n$ -gon is therefore the same as computing inside $D_n$ , since composing motions is multiplying permutations. The triangle gives $D_3 = S_3$ , as the example showed; for $n \geqslant 4$ the inclusion is proper, since $2n < n!$ .

Example 8.19 (A regular nine-gon).

A regular nine-gon has $2 \cdot 9 = 18$ symmetries. With the positions labelled $1$ through $9$ round the boundary, the two symmetries carrying position $1$ to position $5$ are

\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 5 & 6 & 7 & 8 & 9 & 1 & 2 & 3 & 4 \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 5 & 4 & 3 & 2 & 1 & 9 & 8 & 7 & 6 \end{pmatrix},

a turn through four steps and a reflection in the line through position $3$ and the centre. The reflection is recognisable from its table without any picture: it fixes $3$ and pairs off the remaining positions, which is what an involution with one fixed point looks like.

Example 8.20 (Composing motions by multiplying).

Take the six symmetries $s_1, \ldots, s_6$ of the equilateral triangle and their permutations $\sigma_1, \ldots, \sigma_6$ as listed above. The composite $s_4 \circ s_6 \circ s_2 \circ s_5 \circ s_3$ answers to the product

\sigma_4 \sigma_6 \sigma_2 \sigma_5 \sigma_3 = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix}.

Following $1$ , $2$ and $3$ through the five factors from the right gives $1 \mapsto 3$ , $2 \mapsto 2$ and $3 \mapsto 1$ , so the product is $\sigma_3$ and the composite motion is the reflection $s_3$ . Once the dictionary between motions and permutations is fixed, the computation needs no picture.

Problem 8.7.

For the square with corners labelled $1, 2, 3, 4$ round the boundary, list the eight elements of $D_4$ in two-row notation, and name an element of $S_4$ that is not one of them.

Problem 8.8.

With the triangle dictionary above, compute $\sigma_5\sigma_5$ and $\sigma_5\sigma_5\sigma_5$ and read the answers as repeated turns.

Problem 8.9.

Show that $D_n$ is closed under products and under inverses. Argue geometrically, from the fact that a composite of rigid motions carrying the polygon onto itself is one, and that the inverse of such a motion is one.

Cycles and Parity

Two-row notation is complete but bulky: it names $n$ values to describe a function that may move only two points. Cycle notation is shorter. It records a permutation by its orbits, and the order and the parity of a permutation can both be read off the same decomposition.

Throughout, $n \in \mathbb{N}$ is fixed unless said otherwise, and $\mathrm{id}$ is the identity permutation of $\langle n \rangle$ .

Inverses, Powers and Order

A permutation is a bijection, so it has an inverse, and that inverse is again a bijection of the same set: $\sigma \in S_n$ gives $\sigma^{-1} \in S_n$ , with $\sigma^{-1}$ sending each $\sigma(i)$ back to $i$ . In two-row notation this is easy: exchange the rows, then reorder the columns so that the top row ascends again.

Example 8.21 (Computing an inverse).

\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 1 & 4 & 2 \end{pmatrix},

then exchanging the rows gives

\begin{pmatrix} 3 & 1 & 4 & 2 \\ 1 & 2 & 3 & 4 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 1 & 3 \end{pmatrix} = \sigma^{-1}.

In practice one writes a blank table and, under each $j$ along the top, records the unique $i$ with $\sigma(i) = j$ ; it is unique because $\sigma$ is a bijection.

Definition 8.22 (Powers of a permutation).

Let $\sigma \in S_n$ . Define $\sigma^t$ for $t \in \mathbb{N}_0$ by the recursion

\sigma^0 \defeq \mathrm{id}, \qquad \sigma^{s(t)} \defeq \sigma^t \sigma,

so that $\sigma^1 = \sigma$ and $\sigma^t$ is the $t$ -fold product $\sigma\sigma\cdots\sigma$ . Write also

\sigma^{-t} \defeq (\sigma^{-1})^t

for $t \in \mathbb{N}$ , so that $\sigma^{-1}$ has its usual meaning.

Example 8.23 (Powers both ways).

\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 5 & 1 & 2 & 4 \end{pmatrix},

then

\sigma^2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 1 & 4 & 3 & 5 & 2 \end{pmatrix}, \qquad \sigma^{-1} = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 4 & 1 & 5 & 2 \end{pmatrix}, \qquad \sigma^{-2} = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 1 & 5 & 3 & 2 & 4 \end{pmatrix}.

Comparing the first and last tables shows $\sigma^{-2}$ to be the inverse of $\sigma^2$ , which the theorem below predicts.

Theorem 8.24 (Laws of exponents).

Let $\sigma \in S_n$ and let $s, t \in \mathbb{N}_0$ . Then

$\sigma^s \sigma^t = \sigma^{s+t}$ ;
$(\sigma^s)^t = \sigma^{st}$ ;
$(\sigma^s)^{-1} = (\sigma^{-1})^s$ , that is, $(\sigma^s)^{-1} = \sigma^{-s}$ when $s$ is positive.

Discussion.

Three identities between permutations, each of them an equality of functions on $\langle n \rangle$ . The powers were defined by a recursion on the exponent, so each part is an induction on one exponent with the other held fixed, and each inductive step is the recursion clause together with associativity of composition. For the first we fix $s$ and induct on $t$ ; the base is the identity law $\sigma^s \mathrm{id} = \sigma^s$ , and the step moves one factor across the bracket. The second is the same induction on $t$ , now using the first part at each step to add exponents. For the third, to say that a permutation is the inverse of $\sigma^s$ is to say that it composes with $\sigma^s$ to the identity on both sides, and $(\sigma^{-1})^s \sigma^s$ cancels from the inside out. We induct on $s$ for that too.

Proof.

For the first, fix $s$ and induct on $t$ . At $t = 0$ we have $\sigma^s \sigma^0 = \sigma^s \mathrm{id} = \sigma^s = \sigma^{s+0}$ , by the identity laws. Suppose $\sigma^s \sigma^t = \sigma^{s+t}$ . Then

\sigma^s \sigma^{s(t)} = \sigma^s (\sigma^t \sigma) = (\sigma^s \sigma^t) \sigma = \sigma^{s+t} \sigma = \sigma^{s(s+t)} = \sigma^{s + s(t)},

the second step by associativity of composition, the fourth and fifth by the recursion clauses for powers and for addition.

For the second, fix $s$ and induct on $t$ . At $t = 0$ both sides are $\mathrm{id}$ , since $s \cdot 0 = 0$ . Suppose $(\sigma^s)^t = \sigma^{st}$ . Then

(\sigma^s)^{s(t)} = (\sigma^s)^t \sigma^s = \sigma^{st} \sigma^s = \sigma^{st + s} = \sigma^{s \cdot s(t)},

using the first part and the recursion clause for multiplication.

For the third, induct on $s$ . At $s = 0$ both sides are $\mathrm{id}$ . Suppose $(\sigma^{-1})^s \sigma^s = \mathrm{id} = \sigma^s (\sigma^{-1})^s$ . The first part gives $\sigma^{s(s)} = \sigma^{1 + s} = \sigma \sigma^s$ , so

(\sigma^{-1})^{s(s)} \sigma^{s(s)} = \bigl((\sigma^{-1})^s \sigma^{-1}\bigr)\bigl(\sigma \sigma^s\bigr) = (\sigma^{-1})^s \bigl(\sigma^{-1} \sigma\bigr) \sigma^s = (\sigma^{-1})^s \sigma^s = \mathrm{id},

the second step by associativity, the third because $\sigma^{-1}\sigma = \mathrm{id}$ , and the last by the inductive hypothesis. In the other order, the first part applied to $\sigma^{-1}$ gives $(\sigma^{-1})^{s(s)} = \sigma^{-1} (\sigma^{-1})^s$ , and

\sigma^{s(s)} (\sigma^{-1})^{s(s)} = \bigl(\sigma^s \sigma\bigr)\bigl(\sigma^{-1} (\sigma^{-1})^s\bigr) = \sigma^s \bigl(\sigma \sigma^{-1}\bigr) (\sigma^{-1})^s = \sigma^s (\sigma^{-1})^s = \mathrm{id}.

So $(\sigma^{-1})^s$ is a two-sided inverse of $\sigma^s$ , and inverses are unique.

Remark.

With the third part in hand the notation $\sigma^{-t}$ is consistent: it does not matter whether one inverts and then takes the power or takes the power and then inverts. All three laws then hold for exponents of either sign, once the statements are read with the convention $\sigma^{-t} = (\sigma^{-1})^t$ ; we have written them for $\mathbb{N}_0$ because that is where we have arithmetic, and a negative exponent is here an abbreviation rather than a number.

Definition 8.25 (Order of a permutation).

Let $\sigma \in S_n$ . The order of $\sigma$ , written $\mathrm{ord}\,\sigma$ , is the least $k \in \mathbb{N}$ with $\sigma^k = \mathrm{id}$ , if there is one.

Proposition 8.26 (Every permutation has an order).

Let $\sigma \in S_n$ . Then $\sigma^k = \mathrm{id}$ for some $k \in \mathbb{N}$ , so $\mathrm{ord}\,\sigma$ is defined and lies in $\mathbb{N}$ .

Discussion.

The claim is an existence statement, and since all we know is that $S_n$ is finite, we use the pigeonhole principle. We have infinitely many powers $\sigma^0, \sigma^1, \sigma^2, \ldots$ and only $n!$ permutations for them to be, so two of the powers coincide. Cancelling the smaller from the larger (multiplying by an inverse, using the laws of exponents) leaves a positive power equal to the identity. Once the set of positive $k$ with $\sigma^k = \mathrm{id}$ is known to be non-empty, well-ordering supplies its least member, which is what the definition asks for.

Proof.

The map $t \mapsto \sigma^t$ from $\langle s(n!) \rangle$ to $S_n$ has a domain of $s(n!)$ elements and a range inside a set of $n!$ elements, so it is not injective by the pigeonhole principle: there are $i < j$ in that block with $\sigma^i = \sigma^j$ . Write $j = i + d$ with $d \in \mathbb{N}$ . Composing with $(\sigma^{-1})^i$ on the left and using the laws of exponents,

\mathrm{id} = (\sigma^{-1})^i \sigma^i = (\sigma^{-1})^i \sigma^{i + d} = \bigl((\sigma^{-1})^i \sigma^i\bigr)\sigma^d = \sigma^d .

So $K \defeq \{k \in \mathbb{N} \mid \sigma^k = \mathrm{id}\}$ is non-empty, and well-ordering gives it a least member, which is $\mathrm{ord}\,\sigma$ .

Theorem 8.27 (Powers repeat with the order).

Let $\sigma \in S_n$ with $\mathrm{ord}\,\sigma = k$ . Then $\sigma^0, \sigma^1, \ldots, \sigma^{k-1}$ are pairwise distinct, and every power $\sigma^t$ with $t \in \mathbb{N}_0$ is one of them.

Discussion.

Two assertions. The second says that the list exhausts the powers, and the tool is division with remainder: writing $t = qk + r$ with $r < k$ , the laws of exponents split $\sigma^t$ into $(\sigma^k)^q \sigma^r$ , and the first factor is the identity by the definition of the order, leaving $\sigma^r$ with $r$ in range. The first assertion is a uniqueness claim and goes by contradiction: two equal powers among the first $k$ would cancel to give a positive power below $k$ equal to the identity, and the order was chosen least, so no such power exists.

Proof.

Let $t \in \mathbb{N}_0$ . Division with remainder gives $q, r \in \mathbb{N}_0$ with $t = q \cdot k + r$ and $r < k$ , so by the laws of exponents

\sigma^t = \sigma^{qk + r} = (\sigma^k)^q \sigma^r = \mathrm{id}^q \sigma^r = \sigma^r,

and $\sigma^t$ is one of $\sigma^0, \ldots, \sigma^{k-1}$ .

Suppose $\sigma^i = \sigma^j$ with $i < j$ and $j \leqslant k - 1$ . Writing $j = i + d$ with $d \in \mathbb{N}$ and composing with $(\sigma^{-1})^i$ as before gives $\sigma^d = \mathrm{id}$ with $d \leqslant j < k$ , contradicting the leastness of $k$ . So the $k$ powers listed are pairwise distinct.

Problem 8.10.

Let

\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 5 & 3 & 4 & 2 & 1 \end{pmatrix}.

Compute $\sigma^2, \sigma^3, \ldots$ until the identity appears, and state $\mathrm{ord}\,\sigma$ .

Problem 8.11.

Show that if $\sigma^m = \mathrm{id}$ for some $m \in \mathbb{N}$ , then $\mathrm{ord}\,\sigma$ divides $m$ , in the sense that $m = q \cdot \mathrm{ord}\,\sigma$ for some $q$ . Division with remainder is the whole of the argument.

Problem 8.12.

Show that $\mathrm{ord}\,\sigma = \mathrm{ord}\,\sigma^{-1}$ for every $\sigma \in S_n$ , and that $\mathrm{ord}\,\sigma = 1$ exactly when $\sigma = \mathrm{id}$ .

Cycles and Disjoint Decomposition

Draw a point for each $i \in \langle n \rangle$ and an arrow from $i$ to $\sigma(i)$ . Because $\sigma$ is a function, exactly one arrow leaves each point; because it is injective, exactly one arrow arrives at each. A picture of that kind can only be a collection of separate closed loops, some of them loops from a point to itself. The notation that matches the picture is the following.

Definition 8.28 (Cycle).

Let $r \in \mathbb{N}$ with $r \leqslant n$ . A permutation $\sigma \in S_n$ is an $r$ -cycle, or a cycle of length $r$ , if there are distinct $a_1, \ldots, a_r \in \langle n \rangle$ with

\sigma(a_i) = \begin{cases} a_{i+1} & \text{if } i < r, \\ a_1 & \text{if } i = r, \end{cases}

and $\sigma(x) = x$ for every $x \in \langle n \rangle \setminus \{a_1, \ldots, a_r\}$ . The set $\{a_1, \ldots, a_r\}$ is the orbit of the cycle, written $\mathrm{orb}\,\sigma$ , and one writes

\sigma = (a_1 \ a_2 \ \cdots \ a_r)

for this permutation. A point $x$ with $\sigma(x) \neq x$ is said to be moved by $\sigma$ .

A $1$ -cycle $(a_1)$ fixes $a_1$ and fixes everything else, so it is the identity; in a product one leaves such factors out, since they do nothing. For $r \geqslant 2$ the listing inside the brackets may start wherever one likes,

(a_1 \ a_2 \ \cdots \ a_r) = (a_2 \ \cdots \ a_r \ a_1) = \cdots = (a_r \ a_1 \ \cdots \ a_{r-1}),

since all $r$ of those tables send each $a_i$ to the same place; and the inverse of a cycle is the same cycle read backwards,

(a_1 \ a_2 \ \cdots \ a_r)^{-1} = (a_r \ a_{r-1} \ \cdots \ a_1).

Example 8.29 (A five-cycle).

The permutation

\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 7 & 3 & 5 & 4 & 1 & 6 & 2 \end{pmatrix}

sends $1 \mapsto 7 \mapsto 2 \mapsto 3 \mapsto 5 \mapsto 1$ and fixes $4$ and $6$ , so $\sigma = (1 \ 7 \ 2 \ 3 \ 5)$ , which is the same cycle as $(7 \ 2 \ 3 \ 5 \ 1)$ .

Definition 8.30 (Disjoint cycles).

A family of cycles in $S_n$ is disjoint if their orbits are pairwise disjoint: no cycle in the family moves a point moved by another.

Example 8.31 (Two disjoint cycles).

In $S_7$ the cycles $\alpha = (2 \ 6 \ 4 \ 5)$ and $\beta = (3 \ 7)$ are disjoint, since $\{2, 6, 4, 5\} \cap \{3, 7\} = \emptyset$ . In two-row notation,

\alpha = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 1 & 6 & 3 & 5 & 2 & 4 & 7 \end{pmatrix}, \qquad \beta = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 1 & 2 & 7 & 4 & 5 & 6 & 3 \end{pmatrix}.

Products of permutations do not commute in general, but disjoint cycles do, because each acts inside its own orbit and fixes everything outside it. The argument has nothing to do with cycles, so we state it for arbitrary maps.

Theorem 8.32 (Maps with disjoint supports commute).

Let $A$ and $B$ be sets and let $f, g : A \cup B \to A \cup B$ satisfy

$f(A) \subset A$ , and $f(x) = x$ for every $x \in B$ ;
$g(B) \subset B$ , and $g(x) = x$ for every $x \in A$ .

Then $g \circ f = f \circ g$ .

Discussion.

The claim is an equality of two functions with domain $A \cup B$ , so by equality of functions we compare values at an arbitrary point, and the domain is a union, so the comparison splits into the two cases $x \in A$ and $x \in B$ . Each case is the same short argument run with the hypotheses exchanged: one of the two maps fixes $x$ , so one composite is immediately the other map’s value at $x$ ; and that value stays in the set where the first map fixes everything, so the other composite is the same. Only the two hypotheses are used.

Proof.

Let $x \in A \cup B$ .

If $x \in A$ then $g(x) = x$ by the second hypothesis, so $(f \circ g)(x) = f(x)$ ; and $f(x) \in A$ by the first hypothesis, so $g$ fixes it and $(g \circ f)(x) = g(f(x)) = f(x)$ .

If $x \in B$ then $f(x) = x$ by the first hypothesis, so $(g \circ f)(x) = g(x)$ ; and $g(x) \in B$ by the second, so $f$ fixes it and $(f \circ g)(x) = f(g(x)) = g(x)$ .

In both cases the two composites agree at $x$ , so they are equal.

Corollary 8.33 (Disjoint cycles commute).

If $\alpha$ and $\beta$ are disjoint cycles in $S_n$ , then $\alpha\beta = \beta\alpha$ .

Proof.

Put $A \defeq \mathrm{orb}\,\alpha$ and $B \defeq \langle n \rangle \setminus A$ . Then $\alpha(A) = A$ and $\alpha$ fixes every point of $B$ , by the definition of a cycle. Disjointness puts $\mathrm{orb}\,\beta$ inside $B$ , so $\beta(B) \subset B$ , and $\beta$ fixes every point of $A$ . The theorem applies with $f = \alpha$ and $g = \beta$ .

Example 8.34 (A product of disjoint cycles).

In $S_{10}$ ,

(3 \ 1 \ 5 \ 7 \ 9)(2 \ 8 \ 6) = (2 \ 8 \ 6)(3 \ 1 \ 5 \ 7 \ 9) = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 5 & 8 & 1 & 4 & 7 & 2 & 9 & 6 & 3 & 10 \end{pmatrix}.

Neither $4$ nor $10$ is moved by either factor, and the table records that by fixing them.

Every permutation is such a product.

Theorem 8.35 (Cycle decomposition).

Every permutation in $S_n$ is a product of pairwise disjoint cycles. The identity is the empty product, or equally a product of $1$ -cycles, which one omits from the written expression.

Discussion.

The proof is a construction together with two checks. The construction follows the arrows of the picture: start at a point, apply $\sigma$ repeatedly, and see that the trail must return to its start. That it returns at all is the pigeonhole principle, since $\langle n \rangle$ is finite; that it returns to the start rather than to some later point of the trail uses injectivity, and we get it by taking the first repetition and cancelling. The trail is then the orbit of a cycle on which $\sigma$ agrees with that cycle. The first check is that a second trail, begun at a point not yet used, is disjoint from the first, which is again cancellation. The second is that the product of all the cycles obtained equals $\sigma$ : on each orbit the product acts as the one factor that moves that orbit, by disjointness, and off all of them both sides fix every point. The process stops because each round uses at least one new point of the finite set $\langle n \rangle$ .

Proof.

Let $\sigma \in S_n$ and let $b \in \langle n \rangle$ . Among the $s(n)$ values $\sigma^0(b), \sigma^1(b), \ldots, \sigma^n(b)$ , lying in the $n$ -set $\langle n \rangle$ , two coincide by the pigeonhole principle. Let $t$ be least such that $\sigma^t(b) = \sigma^i(b)$ for some $i < t$ . If $i$ were positive, composing both sides with $(\sigma^{-1})^i$ would give $\sigma^{t - i}(b) = b = \sigma^0(b)$ with $t - i < t$ , contradicting leastness. So $i = 0$ and $\sigma^t(b) = b$ , and the points $b, \sigma(b), \ldots, \sigma^{t-1}(b)$ are pairwise distinct, again by leastness. Write

\gamma_b \defeq \bigl(b \ \ \sigma(b) \ \ \cdots \ \ \sigma^{t-1}(b)\bigr),

a cycle whose orbit is that set and on which $\sigma$ agrees with $\gamma_b$ .

Now build the decomposition. Put $\gamma_1 \defeq \gamma_{b_1}$ for $b_1 \defeq 1$ . If every point outside $\mathrm{orb}\,\gamma_1$ is fixed by $\sigma$ , then $\sigma = \gamma_1$ and we are done. Otherwise choose a point $b_2$ moved by $\sigma$ and lying outside $\mathrm{orb}\,\gamma_1$ , and form $\gamma_2 \defeq \gamma_{b_2}$ . The two orbits are disjoint: if $\sigma^i(b_1) = \sigma^j(b_2)$ then composing with $(\sigma^{-1})^j$ or $(\sigma^{-1})^i$ , whichever exponent is the smaller, expresses $b_2$ as a power of $\sigma$ applied to $b_1$ , and the theorem on powers repeating puts that power inside $\mathrm{orb}\,\gamma_1$ , contrary to the choice of $b_2$ .

Repeat. Each round adds at least one point to the union of the orbits, and $\langle n \rangle$ is finite, so after finitely many rounds every point moved by $\sigma$ lies in some orbit. Let $\gamma_1, \ldots, \gamma_m$ be the cycles obtained; they are pairwise disjoint by the argument just given, applied to each pair. Their product agrees with $\sigma$ at every point: a point in $\mathrm{orb}\,\gamma_i$ is fixed by every factor but $\gamma_i$ , which sends it where $\sigma$ does, and a point in no orbit is fixed by every factor and by $\sigma$ . So $\sigma = \gamma_1 \cdots \gamma_m$ , and the order of the factors is immaterial by the corollary on disjoint cycles.

Remark (Reading off the decomposition).

The proof is an algorithm, and it is the one used in practice. Given $\sigma$ in two-row form, begin at any point not yet written down, apply $\sigma$ until the starting point returns, close the bracket, and start again elsewhere. Only cycles of length at least two need be written. In the picture of arrows, the cycles are exactly the separate loops.

Remark (Uniqueness).

The decomposition is unique up to the order of the factors and up to where each bracket starts. For suppose $\sigma = \gamma_1 \cdots \gamma_m = \lambda_1 \cdots \lambda_t$ are two products of pairwise disjoint cycles of length at least two, and let $a \in \mathrm{orb}\,\gamma_1$ . Then $\sigma(a) \neq a$ , so $a$ lies in the orbit of exactly one $\lambda_j$ , say $\lambda_1$ after renumbering. Both $\gamma_1$ and $\lambda_1$ agree with $\sigma$ on the orbit they contain $a$ in, so the two orbits are the same set — each is the trail of $a$ under $\sigma$ — and the two cycles agree there and fix everything else, hence $\gamma_1 = \lambda_1$ . Matching the remaining factors the same way shows the two collections coincide.

Example 8.36 (Three decompositions).

$\displaystyle \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 4 & 9 & 2 & 5 & 1 & 7 & 6 & 8 & 3 \end{pmatrix} = (1 \ 4 \ 5)(2 \ 9 \ 3)(6 \ 7).$
$\displaystyle \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ 9 & 6 & 8 & 7 & 2 & 5 & 4 & 1 & 3 & 10 & 12 & 11 \end{pmatrix} = (1 \ 9 \ 3 \ 8)(2 \ 6 \ 5)(4 \ 7)(11 \ 12).$
If $\alpha = (1 \ 2 \ 3)$ and $\beta = (2 \ 3 \ 4)$ , the product $\alpha\beta$ is not written as a product of disjoint cycles, since the factors share the points $2$ and $3$ . Following each point through $\beta$ and then $\alpha$ gives $\alpha\beta = (1 \ 2)(3 \ 4)$ , which is.

Problem 8.13.

Show that an $r$ -cycle has order $r$ .

Problem 8.14.

Write each of the following as a product of disjoint cycles and give its order.

$\displaystyle \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 1 & 4 & 5 & 6 & 3 \end{pmatrix}$
$(1 \ 3 \ 5)(2 \ 4 \ 6 \ 1)$ , as an element of $S_6$ .

Problem 8.15.

Let $\sigma$ be a product of pairwise disjoint cycles of lengths $r_1, \ldots, r_m$ . Show that $\sigma^k = \mathrm{id}$ exactly when every $r_i$ divides $k$ , in the sense of the problem on divisibility, and deduce that $\mathrm{ord}\,\sigma$ is the least such $k$ . Say where disjointness is used.

Problem 8.16.

Show that $\sigma \in S_n$ is a transposition if and only if it is a $2$ -cycle, and that $\alpha\gamma\alpha^{-1} = (\alpha(a_1) \ \cdots \ \alpha(a_r))$ for every cycle $\gamma = (a_1 \ \cdots \ a_r)$ and every $\alpha \in S_n$ .

Problem 8.17.

Let $n \geqslant 3$ and let $\sigma \in S_n$ satisfy $\sigma\lambda = \lambda\sigma$ for every $\lambda \in S_n$ . Show that $\sigma = \mathrm{id}$ . Which elements of $S_2$ have the same property?

Transpositions and the Sign

A $2$ -cycle $(a \ b)$ exchanges $a$ and $b$ and fixes every other point of $\langle n \rangle$ , which is exactly the transposition $\tau_{a,b}$ written in cycle notation. In particular $(a \ b)^2 = \mathrm{id}$ and $(a \ b)^{-1} = (a \ b)$ , which is the proposition that a transposition is its own inverse read in the new notation. Transpositions move as few points as a permutation other than the identity can, and every permutation is a product of them.

Theorem 8.37 (Factorisation into transpositions).

Let $n \geqslant 2$ . Then every permutation in $S_n$ is a product of transpositions.

Discussion.

The statement is universally quantified over $S_n$ , and the cycle decomposition has already reduced any such statement to a statement about a single cycle: if each factor of a disjoint decomposition is a product of transpositions then so is the whole, by substitution. So there are two things to do. The identity is not covered by the decomposition, since its decomposition is empty, and it is handled separately by writing it as a transposition composed with itself. A cycle of length at least two is handled by exhibiting the factorisation outright, and the exhibited product is then checked point by point against the cycle.

Proof.

If $\sigma = \mathrm{id}$ then $\sigma = (1 \ 2)(1 \ 2)$ , which is available since $n \geqslant 2$ .

Otherwise the cycle decomposition writes $\sigma$ as a product of cycles of length at least two, so it is enough to factor one such cycle. We claim

(a_1 \ a_2 \ \cdots \ a_r) = (a_1 \ a_r)(a_1 \ a_{r-1}) \cdots (a_1 \ a_3)(a_1 \ a_2).

Evaluate the right-hand side from the right. The point $a_1$ is sent to $a_2$ by the first factor, and $a_2$ is fixed by all the others, so $a_1 \mapsto a_2$ . For $1 < i < r$ , the point $a_i$ is fixed by every factor until $(a_1 \ a_i)$ sends it to $a_1$ , and the next factor $(a_1 \ a_{i+1})$ sends $a_1$ to $a_{i+1}$ , which the remaining factors fix; so $a_i \mapsto a_{i+1}$ . The point $a_r$ is fixed until the last factor $(a_1 \ a_r)$ sends it to $a_1$ , and nothing follows, so $a_r \mapsto a_1$ . Any point outside the orbit is fixed by every factor. So the two sides agree everywhere.

Substituting these factorisations into the decomposition writes $\sigma$ as a product of transpositions.

Remark (Non-uniqueness).

The transposition factors are not disjoint in general, and cannot be: a $3$ -cycle moves three points, while a product of disjoint transpositions moves an even number. The factorisation is not unique either, since $(1 \ 2)(1 \ 2)$ may be inserted anywhere, and the order of non-disjoint factors matters, $(1 \ 3)(1 \ 2)$ and $(1 \ 2)(1 \ 3)$ being different permutations. What is unique is the evenness or oddness of the number of factors, and we prove this next.

To prove it we count, for a given permutation, the pairs of points whose order it reverses.

Definition 8.38 (Reversals, parity and sign).

Let $\sigma \in S_n$ and let $\{i, j\}$ be a $2$ -subset of $\langle n \rangle$ , written so that $i < j$ . Then $\sigma$ reverses $\{i, j\}$ if $\sigma(j) < \sigma(i)$ . Write $r(\sigma)$ for the number of $2$ -subsets of $\langle n \rangle$ reversed by $\sigma$ .

The permutation $\sigma$ is even if $r(\sigma)$ is even and odd if $r(\sigma)$ is odd; that alternative is the parity of $\sigma$ . The sign of $\sigma$ is

\mathrm{sgn}\,\sigma \defeq \begin{cases} 1 & \text{if } \sigma \text{ is even}, \\ -1 & \text{if } \sigma \text{ is odd}. \end{cases}

The count $r(\sigma)$ exists because the $2$ -subsets of $\langle n \rangle$ form a finite set, and each is reversed or not by the trichotomy of the order. The two values $1$ and $-1$ are just labels for the parity, and the only property of the labels we shall use is that multiplying them behaves as adding parities does, with $(-1)(-1) = 1$ recording that two odd numbers add to an even one.

Example 8.39 (Counting reversals).

The identity reverses nothing, so $r(\mathrm{id}) = 0$ and the identity is even. For

\sigma = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix}

the three $2$ -subsets are $\{1, 2\}$ , with $2 < 3$ , not reversed; $\{1, 3\}$ , with $1 < 2$ , reversed; and $\{2, 3\}$ , with $1 < 3$ , reversed. So $r(\sigma) = 2$ and $\sigma$ is even.

The proof uses two facts about $r$ : multiplying by a transposition of neighbours changes it by exactly one, and every permutation is a product of such transpositions. The second was set as a problem in the chapter on sequences, where it justified the claim that rearranging a sum does not change it; here it is proved, and in the language of $S_n$ rather than of rearrangements.

Proposition 8.40 (A neighbour swap changes one reversal).

Let $\sigma \in S_n$ , let $k \in \langle n \rangle$ with $k < n$ , and put $\tau \defeq (k \ \ k+1)$ . Then exactly one of

r(\sigma\tau) = r(\sigma) + 1, \qquad r(\sigma) = r(\sigma\tau) + 1

holds. In particular $r(\sigma\tau)$ and $r(\sigma)$ have different parities.

Discussion.

Both $r(\sigma)$ and $r(\sigma\tau)$ are counts over the same index set, the $2$ -subsets of $\langle n \rangle$ , so the claim is that the two counts differ by one, and we sort the $2$ -subsets into three kinds and compare the counts kind by kind. The permutation $\sigma\tau$ agrees with $\sigma$ except that its values at $k$ and $k+1$ are exchanged: a $2$ -subset avoiding both $k$ and $k+1$ sees the same two values in the same positions, so its status is unchanged; a $2$ -subset meeting exactly one of them is paired with the $2$ -subset meeting the other, and the two statuses are exchanged between the members of the pair, leaving their total unchanged; and the single $2$ -subset $\{k, k+1\}$ has its status reversed, because the two values are exchanged while the two positions are not. Summing the three kinds, the totals agree except for one, and trichotomy makes the two displayed alternatives exclusive. The last sentence is then the remainder classes at $m = 2$ : a number and its successor never have the same parity.

Proof.

Write $\rho \defeq \sigma\tau$ , so that $\rho(k) = \sigma(k+1)$ , $\rho(k+1) = \sigma(k)$ , and $\rho(i) = \sigma(i)$ for every other $i$ . Sort the $2$ -subsets of $\langle n \rangle$ into three kinds.

Neither point in $\{k, k+1\}$ . Both values are the same for $\rho$ as for $\sigma$ , so the subset is reversed by one exactly when it is reversed by the other.

Exactly one point in $\{k, k+1\}$ . Such subsets come in pairs $\{i, k\}$ and $\{i, k+1\}$ with $i \notin \{k, k+1\}$ . Suppose $i < k$ , so that $i < k < k+1$ and $i$ is the smaller point in both subsets. Then $\rho$ reverses $\{i, k\}$ exactly when $\sigma(k+1) < \sigma(i)$ , which is exactly when $\sigma$ reverses $\{i, k+1\}$ ; and $\rho$ reverses $\{i, k+1\}$ exactly when $\sigma$ reverses $\{i, k\}$ . So the two statuses are exchanged within the pair and the number of reversed subsets among the two is the same for $\rho$ as for $\sigma$ . The case $k + 1 < i$ is the same argument with $i$ the larger point.

The subset $\{k, k+1\}$ . Here $\rho$ reverses it exactly when $\rho(k+1) < \rho(k)$ , that is when $\sigma(k) < \sigma(k+1)$ , which is exactly when $\sigma$ does not reverse it.

Adding the three kinds, the counts agree on the first two and differ by exactly one on the third. So $r(\rho)$ is $r(\sigma)$ increased by one or $r(\sigma)$ is $r(\rho)$ increased by one, and not both, by trichotomy. A number and its successor fall in different remainder classes for the divisor $2$ , so the parities differ.

Proposition 8.41 (Neighbour swaps suffice).

Let $n \geqslant 2$ . Then every permutation in $S_n$ is a product of transpositions of the form $(k \ \ k+1)$ with $k < n$ .

Discussion.

By the theorem on factorisation into transpositions the claim reduces to a single transposition $(a \ b)$ with $a < b$ , since substituting a factorisation of each factor factorises the product. We induct on the gap $d$ between them, where $b = a + d$ . At $d = 1$ the transposition is already a neighbour swap. For the step, conjugating a transposition by a neighbour swap moves one of its two points one place along: $(a \ b)$ is $(b-1 \ \ b)(a \ \ b-1)(b-1 \ \ b)$ , whose middle factor has a smaller gap and whose outer factors are neighbour swaps. Checking that identity is a comparison of values at the three points involved, everything else being fixed by all three factors.

Proof.

By the theorem on factorisation into transpositions it is enough to write a single transposition $(a \ b)$ , with $a < b$ , as a product of neighbour swaps. Write $b = a + d$ with $d \in \mathbb{N}$ and induct on $d$ .

If $d = 1$ then $(a \ b)$ is itself a neighbour swap.

Suppose the claim holds for $d$ , and let $b = a + s(d)$ , so that $b - 1 = a + d$ and $a < b - 1 < b$ . Put $\pi \defeq (b-1 \ \ b)$ . We claim $(a \ b) = \pi (a \ \ b-1) \pi$ . Evaluating the right-hand side from the right: $a$ is fixed by $\pi$ , sent to $b-1$ by the middle factor, and sent to $b$ by the last, so $a \mapsto b$ . The point $b$ is sent to $b-1$ by the first factor, then to $a$ by the middle, and $a$ is fixed by the last, so $b \mapsto a$ . The point $b-1$ is sent to $b$ by the first factor, fixed by the middle, and returned to $b-1$ by the last, so $b - 1 \mapsto b-1$ . Every other point is fixed by all three factors. So the two sides agree everywhere.

The middle factor $(a \ \ b-1)$ has gap $d$ and is a product of neighbour swaps by the inductive hypothesis, and $\pi$ is a neighbour swap, so $(a \ b)$ is a product of neighbour swaps.

Theorem 8.42 (Properties of the sign).

Let $\sigma, \lambda \in S_n$ with $n \geqslant 2$ .

$\mathrm{sgn}(\sigma\lambda) = (\mathrm{sgn}\,\sigma)(\mathrm{sgn}\,\lambda)$ .
Every transposition is odd.
If $\sigma$ is a product of $m$ transpositions, then $\mathrm{sgn}\,\sigma = 1$ when $m$ is even and $\mathrm{sgn}\,\sigma = -1$ when $m$ is odd.

Discussion.

Three claims about the parity of $r$ . The first says that the parity of $r(\sigma\lambda)$ is settled by the parities of $r(\sigma)$ and $r(\lambda)$ , and we use the proposition on neighbour swaps together with the proposition that neighbour swaps suffice: writing $\lambda$ as a product of $m$ neighbour swaps and multiplying them onto $\sigma$ one at a time flips the parity $m$ times, so the parity of $r(\sigma\lambda)$ is that of $r(\sigma)$ shifted by $m$ ; taking $\sigma$ to be the identity, whose count is $0$ , identifies the parity of $m$ with the parity of $\lambda$ , and the two statements together are the claim. The second is a direct count of reversals for $(a \ b)$ , sorting the $2$ -subsets by whether they meet $\{a, b\}$ and where their other point lies; the answer is one more than an even number. The third is then an induction on $m$ using the first two.

Proof.

For the first, write $\lambda = \pi_1 \pi_2 \cdots \pi_m$ as a product of neighbour swaps. Then

\sigma\lambda = \bigl(\cdots((\sigma\pi_1)\pi_2)\cdots\bigr)\pi_m,

by associativity, and each of the $m$ steps changes the parity of the reversal count, by the proposition on neighbour swaps. So $r(\sigma\lambda)$ has the parity of $r(\sigma)$ when $m$ is even and the opposite parity when $m$ is odd. Taking $\sigma = \mathrm{id}$ , where $r(\mathrm{id}) = 0$ is even, shows that $\lambda$ is even exactly when $m$ is even. Combining the two: $\sigma\lambda$ has the parity of $\sigma$ when $\lambda$ is even and the opposite when $\lambda$ is odd, which in the sign notation is the stated product rule.

For the second, let $\tau = (a \ b)$ with $a < b$ , and sort the $2$ -subsets of $\langle n \rangle$ . One meeting neither $a$ nor $b$ is not reversed, since $\tau$ fixes both its points. The subset $\{a, b\}$ is reversed, since $\tau(b) = a < b = \tau(a)$ . A subset $\{c, a\}$ or $\{c, b\}$ with $c \notin \{a, b\}$ is reversed exactly when $c$ lies strictly between $a$ and $b$ : if $c < a$ or $b < c$ then $c$ keeps its position relative to both, and if $a < c < b$ then $\{a, c\}$ is reversed because $\tau(c) = c$ while $\tau(a) = b > c$ , and $\{c, b\}$ is reversed because $\tau(b) = a < c$ . So each such $c$ contributes two reversals and nothing else contributes, giving

r(\tau) = 1 + 2 \cdot \#\{c \in \langle n \rangle \mid a < c < b\},

which is odd.

For the third, induct on $m$ . At $m = 1$ the claim is the second part. Suppose it holds for $m$ and let $\sigma = \tau_1 \cdots \tau_{s(m)}$ . Then $\sigma = (\tau_1 \cdots \tau_m)\tau_{s(m)}$ , so by the first two parts

\mathrm{sgn}\,\sigma = \mathrm{sgn}(\tau_1 \cdots \tau_m) \cdot (-1),

which changes the value from $1$ to $-1$ or from $-1$ to $1$ exactly as the parity of $m$ changes to the parity of $s(m)$ .

Corollary 8.43 (The number of transposition factors has a fixed parity).

If $\sigma = \tau_1 \cdots \tau_m = \tau'_1 \cdots \tau'_{m'}$ are two factorisations of the same $\sigma \in S_n$ into transpositions, then $m$ and $m'$ are both even or both odd.

Proof.

The third part of the theorem computes $\mathrm{sgn}\,\sigma$ from each factorisation, giving $1$ when the number of factors is even and $-1$ when it is odd. Since $\mathrm{sgn}\,\sigma$ is one value, $m$ and $m'$ cannot have different parities.

An $r$ -cycle is a product of $r - 1$ transpositions, by the factorisation exhibited above, so

\mathrm{sgn}\,(a_1 \ \cdots \ a_r) = 1 \text{ if } r \text{ is odd}, \qquad \mathrm{sgn}\,(a_1 \ \cdots \ a_r) = -1 \text{ if } r \text{ is even}.

A cycle of odd length is even and a cycle of even length is odd, which reads perversely until one remembers that the length counts points and the sign counts swaps. For a product of cycles of lengths $r_1, \ldots, r_t$ , disjoint or not, the first part of the theorem multiplies the signs together, so the product is even exactly when an even number of the $r_i$ are even.

Example 8.44 (Computing the sign).

For

\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 1 & 5 & 2 & 4 \end{pmatrix} = (1 \ 3 \ 5 \ 4 \ 2),

a single cycle of length five, the sign is $1$ and $\sigma$ is even. For

\lambda = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ 3 & 1 & 9 & 7 & 8 & 4 & 6 & 5 & 2 & 12 & 11 & 10 \end{pmatrix} = (1 \ 3 \ 9 \ 2)(4 \ 7 \ 6)(5 \ 8)(10 \ 12),

three of the four lengths are even, namely $4$ , $2$ and $2$ , so three of the four factors are odd and $\mathrm{sgn}\,\lambda = -1$ .

Remark (Parity as an obstruction).

Any sequence of exchanges of two of $n$ labelled objects is a product of transpositions in $S_n$ . If the rearrangement one is aiming at is odd, then no even number of exchanges reaches it, and if it is even, no odd number does. So parity rules out many proposed sequences of exchanges without examining the intermediate states: one only computes the sign of the target rearrangement.

Problem 8.18.

Write

\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 6 & 7 & 10 & 5 & 8 & 2 & 1 & 4 & 3 & 9 \end{pmatrix}

as a product of disjoint cycles and then as a product of transpositions, and compute $\mathrm{sgn}\,\sigma$ in both ways.

Problem 8.19.

Show that the even permutations in $S_n$ are closed under products and under inverses, and that the odd ones are closed under neither.

Problem 8.20.

Let $n \geqslant 2$ . Show that the even permutations in $S_n$ and the odd ones are equinumerous, by fixing a transposition $\tau$ and considering the map $\sigma \mapsto \tau\sigma$ . Conclude that $\#S_n$ is twice the number of even permutations.

Problem 8.21.

Let $\sigma \in S_n$ . Show that $\sigma$ is a product of $r(\sigma)$ transpositions of the form $(k \ \ k+1)$ with $k < n$ , and that no product of fewer such transpositions equals $\sigma$ .

Problem 8.22.

Let $n \geqslant 2$ . Show that every permutation in $S_n$ is a product of transpositions of the form $(1 \ k)$ with $1 < k \leqslant n$ , and that every permutation in $S_n$ is a product of factors each equal to $(1 \ 2)$ or to $(1 \ 2 \ \cdots \ n)$ .

Problem 8.23.

Fifteen tiles numbered $1$ to $15$ lie in a $4 \times 4$ frame with one cell empty, and a move slides into the empty cell a tile from a cell sharing an edge with it. Decide whether a sequence of moves carries the first arrangement below to the second, and prove your answer.

\begin{array}{|c|c|c|c|} \hline 1 & 2 & 3 & 4 \\ \hline 5 & 6 & 7 & 8 \\ \hline 9 & 10 & 11 & 12 \\ \hline 13 & 14 & 15 & \phantom{00} \\ \hline \end{array} \qquad\qquad \begin{array}{|c|c|c|c|} \hline 15 & 14 & 13 & 12 \\ \hline 11 & 10 & 9 & 8 \\ \hline 7 & 6 & 5 & 4 \\ \hline 3 & 2 & 1 & \phantom{00} \\ \hline \end{array}

Binomial Coefficients

Counting the permutations of an $n$ -set asked in how many ways its points can be arranged. The remaining question of the chapter asks in how many ways they can be chosen: not how a set may be reordered, but how many subsets of a given size it has. The two questions are linked, because choosing $k$ points and then arranging them is the same as arranging $k$ of the $n$ points, and this gives the count of subsets as a quotient of factorials.

Counting Subsets

Definition 8.45 (Binomial coefficient).

Let $n, k \in \mathbb{N}_0$ and let $S$ be an $n$ -set. The binomial coefficient $\dbinom{n}{k}$ , read ” $n$ choose $k$ ”, is the number of $k$ -subsets of $S$ :

\binom{n}{k} \defeq \#\{A \subset S \mid \#A = k\}.

The definition names a set $S$ and the notation does not, so the first thing to check is that the count does not depend on which $n$ -set was taken. It does not: two $n$ -sets are equinumerous, and a bijection $f : S \to S'$ carries $k$ -subsets to $k$ -subsets in both directions, since the image of a $k$ -subset under an injection is a $k$ -subset and $f^{-1}$ undoes it. The count is also finite, since the $k$ -subsets form a subset of the finite set $\mathcal{P}(S)$ .

Theorem 8.46 (Basic properties of binomial coefficients).

Let $n, k \in \mathbb{N}_0$ . Then

$\dbinom{n}{0} = 1$ and $\dbinom{n}{n} = 1$ ;
$\dbinom{n}{k} = 0$ whenever $n < k$ ;
$\dbinom{n}{k} = \dbinom{n}{n - k}$ whenever $k \leqslant n$ ;
$\dbinom{n}{1} = n$ , and $\dbinom{n}{n-1} = n$ when $n \geqslant 1$ ;
$\displaystyle\sum_{k=0}^{n} \binom{n}{k} = 2^n$ .

Discussion.

The first four parts follow from the definition, by naming the subsets counted. For the first, the empty set is the one $0$ -subset of anything and $S$ is the one $n$ -subset of itself, both by cardinality classifying finite sets. The second is the theorem on subsets of a finite set, which forbids a subset larger than the whole. The third is an equality of two counts, so it asks for a bijection between the two collections, and complementation is one, being its own inverse. The fourth counts singletons and then applies the third. The fifth is different: it is a statement about a sum, so we partition $\mathcal{P}(S)$ by cardinality — the classes are pairwise disjoint and cover it, since every subset of a finite set is finite with exactly one cardinality — and add the pieces up with the cardinality of a disjoint union, the total being $2^n$ by the problem on power sets.

Proof.

Fix an $n$ -set $S$ .

For the first, a $0$ -subset is equinumerous with $\emptyset$ and hence empty, so $\emptyset$ is the only one; an $n$ -subset $A \subset S$ has $\#A = \#S$ , so $A = S$ by the theorem on subsets of a finite set, which makes a proper subset strictly smaller.

For the second, a $k$ -subset $A \subset S$ would give $k = \#A \leqslant \#S = n$ by that theorem, contradicting $n < k$ ; so there are none.

For the third, $A \mapsto S \setminus A$ sends $k$ -subsets to $(n-k)$ -subsets, by the cardinality of a disjoint union applied to $S = A \cup (S \setminus A)$ , and it is its own inverse, hence a bijection between the two collections. Equinumerous finite sets have equal cardinality.

For the fourth, $x \mapsto \{x\}$ is a bijection from $S$ onto the collection of $1$ -subsets, so $\dbinom{n}{1} = n$ ; the third part then gives $\dbinom{n}{n-1} = n$ .

For the fifth, every subset of $S$ is finite with exactly one cardinality, and that cardinality is at most $n$ by the theorem on subsets of a finite set, so the collections of $k$ -subsets for $k = 0, 1, \ldots, n$ are pairwise disjoint and their union is $\mathcal{P}(S)$ . Adding their cardinalities gives

\sum_{k=0}^{n} \binom{n}{k} = \#\mathcal{P}(S) = 2^n,

the last equality being the problem on the cardinality of a power set.

The link between choosing and arranging is made by counting the injections between two finite sets, so we count them first.

Proposition 8.47 (Counting injections).

Let $A$ be a $k$ -set and $B$ an $n$ -set with $k \leqslant n$ . Then the number of injections $A \to B$ is

n(n-1)\cdots(n-k+1),

the product of the $k$ numbers running down from $n$ .

Discussion.

A count again. An injection is built by choosing values one point at a time, and each choice removes one candidate from the target, so we induct on $k$ , with $n$ fixed and $B$ fixed. At $k = 0$ the domain is empty, the empty function is the only function and it is injective, and the empty product is $1$ . For the step we split the injections from a $s(k)$ -set according to the value taken at a chosen point $a$ : the pieces are pairwise disjoint and cover, there are $n$ of them, and each is matched by restriction with the injections from a $k$ -set into a set with one point removed, which the inductive hypothesis counts. Adding $n$ equal pieces is the cardinality of a product, and the arithmetic gives the next factor down because the removed point shrinks the target from $n$ to $n - 1$ and shifts every factor.

Proof.

Induct on $k$ , the claim being taken for all $n \geqslant k$ and all $n$ -sets $B$ at once.

If $k = 0$ then $A = \emptyset$ , the empty function is the unique function $A \to B$ and is injective, and the empty product is $1$ .

Suppose the claim holds for $k$ and let $\#A = s(k) \leqslant n$ . Fix $a \in A$ and, for $b \in B$ , let $I_b$ be the set of injections $f : A \to B$ with $f(a) = b$ . Every injection lies in exactly one $I_b$ . Restriction to $A \setminus \{a\}$ matches $I_b$ with the injections from that $k$ -set into $B \setminus \{b\}$ , an $(n-1)$ -set: an injection cannot take the value $b$ anywhere else, so the restriction lands there, and every injection $A \setminus \{a\} \to B \setminus \{b\}$ extends to exactly one member of $I_b$ . By the inductive hypothesis,

\#I_b = (n-1)(n-2)\cdots(n-1-k+1) = (n-1)\cdots(n-k).

Summing over the $n$ values of $b$ , the number of injections $A \to B$ is

n \cdot (n-1)\cdots(n-k) = n(n-1)\cdots\bigl(n - s(k) + 1\bigr),

which is the claim at $s(k)$ .

Theorem 8.48 (Factorial formula for binomial coefficients).

Let $n, k \in \mathbb{N}_0$ with $k \leqslant n$ . Then

\binom{n}{k} \cdot k! \cdot (n-k)! = n!.

Discussion.

The claim is an identity between products of natural numbers, and the method is to count one set in two ways and equate the answers. The set is the collection of injections from a $k$ -set $D$ into an $n$ -set $S$ . Counted directly, the previous proposition gives the descending product. Counted by what an injection is made of, an injection is an image together with a bijection onto it: the image is a $k$ -subset of $S$ , of which there are $\dbinom{n}{k}$ , and the bijections from $D$ onto a fixed $k$ -subset number $k!$ by the theorem on the number of bijections. Equating the two answers gives $\dbinom{n}{k}\, k!$ equal to the descending product, and multiplying by $(n-k)!$ completes the descending product to $n!$ , which is the displayed identity.

Proof.

Fix an $n$ -set $S$ and a $k$ -set $D$ , and let $I$ be the set of injections $D \to S$ . By the proposition on counting injections,

\#I = n(n-1)\cdots(n-k+1).

Count $I$ a second way. Each $f \in I$ determines its image $f(D)$ , a $k$ -subset of $S$ , and is a bijection from $D$ onto that image, by an injection onto its range. Conversely a $k$ -subset $B \subset S$ together with a bijection $D \to B$ determines an $f \in I$ , and different pairs give different injections. There are $\dbinom{n}{k}$ choices of $B$ and, for each, exactly $k!$ bijections $D \to B$ by the theorem on the number of bijections. So

\#I = \binom{n}{k} \cdot k!.

Equating and multiplying both sides by $(n-k)!$ ,

\binom{n}{k} \cdot k! \cdot (n-k)! = n(n-1)\cdots(n-k+1) \cdot (n-k)! = n!,

the last step because the descending product and $(n-k)!$ between them use each of the factors $n, n-1, \ldots, 1$ exactly once. When $k = 0$ both sides are $n!$ , so the identity holds there too.

Remark.

The identity is normally written as a quotient,

\binom{n}{k} = \frac{n!}{k!\,(n-k)!},

and we shall write it that way below. The division is exact by the theorem, and we allow it on the same terms as subtraction: whenever the answer lies in $\mathbb{N}_0$ . For computing a single coefficient by hand the descending product is faster,

\binom{n}{k} = \frac{n(n-1)\cdots(n-k+1)}{k!},

while the two-factorial form is the better one to manipulate.

Theorem 8.49 (Pascal's identity).

Let $n, k \in \mathbb{N}$ with $k \leqslant n$ . Then

\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}.

Discussion.

The claim is that one count equals a sum of two counts, and a sum of counts comes from a partition of the set being counted. So we fix an $n$ -set $S$ , single out a point $x$ of it, and sort the $k$ -subsets of $S$ by whether they contain $x$ : two collections, disjoint and exhaustive. Those avoiding $x$ are exactly the $k$ -subsets of the $(n-1)$ -set $S \setminus \{x\}$ , which is the first term. Those containing $x$ are matched with the $(k-1)$ -subsets of $S \setminus \{x\}$ by removing $x$ , a map undone by putting it back, which is the second. The cardinality of a disjoint union adds them.

Proof.

Let $S$ be an $n$ -set, fix $x \in S$ and put $S_0 \defeq S \setminus \{x\}$ , an $(n-1)$ -set.

A $k$ -subset of $S$ either contains $x$ or does not, and not both. Those that do not are precisely the $k$ -subsets of $S_0$ , and there are $\dbinom{n-1}{k}$ of them.

Those that do are precisely the sets $A \cup \{x\}$ with $A$ a $(k-1)$ -subset of $S_0$ : removing $x$ from such a subset leaves a subset of $S_0$ of cardinality $k - 1$ , by the corollary on removing a point, and adjoining $x$ to a $(k-1)$ -subset of $S_0$ returns it. So there are $\dbinom{n-1}{k-1}$ of them.

The two collections are disjoint and their union is the collection of all $k$ -subsets of $S$ , so their cardinalities add to $\dbinom{n}{k}$ .

Pascal’s identity together with the boundary values $\dbinom{n}{0} = \dbinom{n}{n} = 1$ determines every binomial coefficient without any factorials at all, each from two earlier ones. Setting the values out in rows indexed by $n$ , with $k$ running left to right, gives Pascal’s triangle:

\begin{array}{cccccccc} 1 & & & & & & & \\ 1 & 1 & & & & & & \\ 1 & 2 & 1 & & & & & \\ 1 & 3 & 3 & 1 & & & & \\ 1 & 4 & 6 & 4 & 1 & & & \\ 1 & 5 & 10 & 10 & 5 & 1 & & \\ 1 & 6 & 15 & 20 & 15 & 6 & 1 & \\ 1 & 7 & 21 & 35 & 35 & 21 & 7 & 1 \end{array}

Each interior entry is the sum of the two entries above it, one directly above and one to the left of that, which is Pascal’s identity. The rows add to $1, 2, 4, 8, \ldots$ , which is the fifth part of the theorem on basic properties, and each row is a palindrome, which is the third.

Example 8.50 (Two small counts).

The number of $2$ -subsets of a $10$ -set is

\binom{10}{2} = \frac{10 \cdot 9}{2} = 45,

and the number of $3$ -subsets is

\binom{10}{3} = \frac{10 \cdot 9 \cdot 8}{6} = 120.

Example 8.51 (Shortest routes).

On a grid of streets $m$ blocks tall and $n$ blocks wide, a shortest walk from the bottom left corner to the top right uses $m + n$ blocks, of which $m$ are walked upwards and $n$ across. Such a walk is settled by saying which $n$ of its $m + n$ steps are the ones across, so there are $\dbinom{m+n}{n}$ shortest routes. On a square grid $n$ blocks each way there are $\dbinom{2n}{n}$ .

Problem 8.24.

Compute $\dbinom{12}{3}$ , $\dbinom{12}{9}$ and $\dbinom{12}{0}$ from the factorial formula, and check the first two against the symmetry $\dbinom{n}{k} = \dbinom{n}{n-k}$ .

Problem 8.25.

Prove the factorial formula a second time, by induction on $n$ with Pascal’s identity as the inductive step, treating $k = 0$ and $k = n$ separately.

Problem 8.26.

How many $5$ -subsets does a $52$ -set have? If the $52$ -set is partitioned into four $13$ -sets, how many of those $5$ -subsets lie inside a single part?

Identities and the Binomial Theorem

Theorem 8.52 (Absorption).

Let $n, k \in \mathbb{N}$ with $k \leqslant n$ . Then

k \binom{n}{k} = n \binom{n-1}{k-1}.

Discussion.

An identity between two products, so again we count one set in two ways. The set is the collection of pairs $(x, B)$ in which $B$ is a $k$ -subset of an $n$ -set $S$ and $x$ is a point of $B$ — a committee together with its chair, if one likes. Choosing the committee first and then the chair from within it gives $\dbinom{n}{k}$ followed by $k$ , which is the left-hand side; choosing the chair first from all of $S$ and then the rest of the committee from what remains gives $n$ followed by $\dbinom{n-1}{k-1}$ , which is the right. Each count is an application of the cardinality of a product to a partition of the same collection, sorted the two different ways. The factorial formula gives a second proof by cancellation, and we record that as well.

Proof.

Let $S$ be an $n$ -set and let $P$ be the set of pairs $(x, B)$ with $B$ a $k$ -subset of $S$ and $x \in B$ .

Sorting $P$ by its second coordinate, each of the $\dbinom{n}{k}$ possible $B$ occurs in exactly $k$ pairs, one for each of its points, so $\#P = \dbinom{n}{k} \cdot k$ .

Sorting $P$ by its first coordinate, each of the $n$ possible $x$ occurs in one pair for each $k$ -subset containing $x$ ; those are the sets $A \cup \{x\}$ with $A$ a $(k-1)$ -subset of $S \setminus \{x\}$ , as in the proof of Pascal’s identity, so there are $\dbinom{n-1}{k-1}$ of them. Hence $\#P = n \cdot \dbinom{n-1}{k-1}$ .

Equating the two gives the identity. Alternatively, from the factorial formula,

k \binom{n}{k} = \frac{k \cdot n!}{k!\,(n-k)!} = \frac{n \cdot (n-1)!}{(k-1)!\,(n-k)!} = n\binom{n-1}{k-1}.

Theorem 8.53 (The hockey-stick identity).

Let $m, n \in \mathbb{N}_0$ . Then

\sum_{k=0}^{n} \binom{m+k}{k} = \binom{m+n+1}{n}.

Discussion.

The left-hand side is a sum whose number of terms depends on $n$ , so the proof is an induction on $n$ with $m$ held fixed. At $n = 0$ both sides are $1$ , by the first part of the theorem on basic properties. The step adds one term to the sum: the inductive hypothesis replaces everything before it by a single coefficient, and what is left is a sum of two coefficients to which Pascal’s identity applies. So the whole argument is one application of the inductive hypothesis followed by one application of Pascal, and the only care needed is in matching the indices.

Proof.

Fix $m$ and induct on $n$ . At $n = 0$ the sum has the single term $\dbinom{m}{0} = 1$ , and the right-hand side is $\dbinom{m+1}{0} = 1$ .

Suppose the identity holds at $n$ . Then

\begin{aligned} \sum_{k=0}^{n+1} \binom{m+k}{k} &= \left(\sum_{k=0}^{n} \binom{m+k}{k}\right) + \binom{m+n+1}{n+1} \\ &= \binom{m+n+1}{n} + \binom{m+n+1}{n+1} \\ &= \binom{m+n+2}{n+1}, \end{aligned}

the second line by the inductive hypothesis and the third by Pascal’s identity, applied with upper index $m+n+2$ and lower index $n+1$ .

Remark.

The name records the shape the summed entries make in Pascal’s triangle: they run down one diagonal and the answer sits one place off the end, like the blade at the foot of a stick.

A binomial is a sum of two terms, and expanding a power of one produces the binomial coefficients as the coefficients of the resulting terms. That is where the name comes from.

Theorem 8.54 (The binomial theorem).

Let $a, b \in \mathbb{N}_0$ and let $n \in \mathbb{N}_0$ . Then

(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k}.

Discussion.

Behind the statement is a count: multiplying out $n$ copies of $a + b$ produces one term for each way of taking $a$ from some of the copies and $b$ from the rest, so $a^k b^{n-k}$ appears once for each $k$ -subset of the copies, which is $\dbinom{n}{k}$ times. That is the reason the theorem is true, but it is not yet a proof, because “multiplying out” is not among our operations. What we have is the recursion defining powers, so the proof is an induction on $n$ : multiply the inductive hypothesis by $a + b$ , distribute, and reassemble. Pascal’s identity appears in the reassembly: after shifting the index of one of the two sums so that both run over the same power of $a$ , the two coefficients standing in front of each term are $\dbinom{n}{k-1}$ and $\dbinom{n}{k}$ , which Pascal’s identity adds to the coefficient wanted at $n+1$ .

Proof.

Induct on $n$ . At $n = 0$ both sides are $1$ , the left because $x^0 = 1$ and the right because the only term is $\dbinom{0}{0}a^0b^0$ .

Suppose the identity holds at $n$ . Then, distributing,

(a+b)^{n+1} = (a + b)\sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k} = \sum_{k=0}^{n} \binom{n}{k} a^{k+1} b^{n-k} + \sum_{k=0}^{n} \binom{n}{k} a^{k} b^{n-k+1}.

In the first sum put $j \defeq k + 1$ , so that $j$ runs from $1$ to $n+1$ :

\sum_{k=0}^{n} \binom{n}{k} a^{k+1} b^{n-k} = \sum_{j=1}^{n+1} \binom{n}{j-1} a^{j} b^{n+1-j}.

Renaming $j$ as $k$ and separating the terms $a^{n+1}$ and $b^{n+1}$ , which occur in only one of the two sums each,

\begin{aligned} (a+b)^{n+1} &= a^{n+1} + \sum_{k=1}^{n}\left(\binom{n}{k-1} + \binom{n}{k}\right) a^k b^{n+1-k} + b^{n+1} \\ &= a^{n+1} + \sum_{k=1}^{n} \binom{n+1}{k} a^k b^{n+1-k} + b^{n+1} \\ &= \sum_{k=0}^{n+1} \binom{n+1}{k} a^k b^{n+1-k}, \end{aligned}

by Pascal’s identity, and the two separated terms are the cases $k = n+1$ and $k = 0$ since $\dbinom{n+1}{n+1} = \dbinom{n+1}{0} = 1$ .

Remark (Where the theorem applies).

The proof uses nothing about $a$ and $b$ beyond the commutative, associative and distributive laws and the recursion defining powers. So it holds verbatim wherever those laws hold — for the real numbers of school algebra, and for expressions in an unknown — and that is how it is used in practice. We have stated it in $\mathbb{N}_0$ because that is the arithmetic built so far.

Corollary 8.55 (Sum of the binomial coefficients, again).

For every $n \in \mathbb{N}_0$ ,

\sum_{k=0}^{n} \binom{n}{k} = 2^n.

Proof.

Put $a = b = 1$ in the binomial theorem: every term $a^k b^{n-k}$ is $1$ , and the left-hand side is $2^n$ .

That is the fifth part of the theorem on basic properties, proved a second time and by quite a different route: the first proof partitioned the power set, and this one multiplies out a product. The first proof and the corollary together give a proof that $\#\mathcal{P}(S) = 2^n$ , running in the opposite direction to the one the problem asked for.

Example 8.56 (A single coefficient).

Working in school algebra, where the theorem applies by the remark above, the coefficient of $x^6$ in $(x - 3)^{11}$ is the coefficient of $x^6$ in

\sum_{k=0}^{11} \binom{11}{k} x^k (-3)^{11-k},

namely $\dbinom{11}{6}(-3)^5 = 462 \cdot (-243) = -112{,}266$ . Only one term of twelve is needed.

Problem 8.27.

Show that for every $n \in \mathbb{N}$ the binomial coefficients with even lower index add to the same total as those with odd lower index, and that each total is $2^{n-1}$ .

Problem 8.28.

Expand $(1+x)^n(1+x)^m$ in two ways and compare the coefficients of $x^r$ to obtain Vandermonde’s identity

\binom{n+m}{r} = \sum_{k=0}^{r} \binom{n}{k}\binom{m}{r-k},

for $n, m \in \mathbb{N}_0$ and $r \leqslant n + m$ , reading $\dbinom{a}{b}$ as $0$ when $a < b$ . Then prove the same identity by counting the $r$ -subsets of a set split into an $n$ -set and an $m$ -set.

Problem 8.29.

Show that $\dbinom{n}{k} < \dbinom{n}{k+1}$ exactly when $2k + 1 < n$ , and read off from that where a row of Pascal’s triangle attains its largest entry, treating $n = 2m$ and $n = 2m+1$ separately. The absorption identity compares consecutive entries of a row.

Problem 8.30.

Show that

n\binom{2n-1}{n-1} = \sum_{k=1}^{n} k\binom{n}{k}^2

for every $n \in \mathbb{N}$ .

Multinomial Coefficients

A binomial coefficient counts the ways of cutting an $n$ -set into two labelled pieces of prescribed sizes: a $k$ -subset and everything else. The same works with more than two pieces.

Definition 8.57 (Multinomial coefficient).

Let $n \in \mathbb{N}_0$ , let $t \in \mathbb{N}$ , and let $r_1, \ldots, r_t \in \mathbb{N}_0$ with $r_1 + \cdots + r_t = n$ . Let $S$ be an $n$ -set. The multinomial coefficient

\binom{n}{r_1, r_2, \ldots, r_t}

is the number of $t$ -tuples $(S_1, \ldots, S_t)$ of pairwise disjoint subsets of $S$ with $S_1 \cup \cdots \cup S_t = S$ and $\#S_i = r_i$ for each $i$ . Empty pieces are allowed, when some $r_i$ is $0$ .

The pieces are labelled by their position in the tuple, so two tuples listing the same pieces in a different order are different objects; the same convention was already in force for binomial coefficients, where the $k$ -subset was distinguished from its complement. Equivalently, such a tuple assigns each point of $S$ to one of $t$ labelled bins, with bin $i$ receiving $r_i$ points.

At $t = 2$ the definition returns the binomial coefficient, since a pair $(S_1, S_2)$ of the required kind is settled by $S_1$ alone:

\binom{n}{k, \, n-k} = \binom{n}{k}.

Theorem 8.58 (Factorial formula for multinomial coefficients).

Let $r_1 + \cdots + r_t = n$ with each $r_i \in \mathbb{N}_0$ . Then

\binom{n}{r_1, r_2, \ldots, r_t} \cdot r_1!\, r_2! \cdots r_t! = n!.

Discussion.

Once more we count one set two ways, and the set is the collection of listings of an $n$ -set $S$ — that is, of bijections $\langle n \rangle \to S$ . Counted directly there are $n!$ of them, by the theorem on the number of bijections. Counted by construction, a listing is assembled from a tuple of the kind the coefficient counts, together with an ordering inside each piece: read the first $r_1$ entries of the listing as the first piece in order, the next $r_2$ as the second, and so on. That correspondence is reversible, so the number of listings is the number of tuples multiplied by $r_1!$ through $r_t!$ , one factor for each piece by the theorem on the number of bijections again. Equating the two counts is the identity.

Proof.

Let $S$ be an $n$ -set. The listings of $S$ , meaning the bijections $\langle n \rangle \to S$ , number $n!$ .

Given a listing $h$ , cut $\langle n \rangle$ into the consecutive blocks of lengths $r_1, \ldots, r_t$ and let $S_i$ be the image of the $i$ -th block. The blocks are pairwise disjoint and cover $\langle n \rangle$ , so the $S_i$ are pairwise disjoint and cover $S$ , and $\#S_i = r_i$ since $h$ is injective. The listing also determines an ordering of each $S_i$ , namely the order in which its points appear.

Conversely a tuple $(S_1, \ldots, S_t)$ of the required kind, together with an ordering of each $S_i$ , reassembles into exactly one listing, by writing the pieces out one after another. So the listings correspond to such data, and there are $\dbinom{n}{r_1, \ldots, r_t}$ tuples with $r_i!$ orderings of the $i$ -th piece for each. Hence

n! = \binom{n}{r_1, \ldots, r_t} \cdot r_1! \cdots r_t!,

where a piece with $r_i = 0$ contributes the factor $0! = 1$ .

Remark.

As before we write the identity as a quotient,

\binom{n}{r_1, \ldots, r_t} = \frac{n!}{r_1!\,r_2! \cdots r_t!},

the division being exact by the theorem. The multinomial coefficient can also be built from binomial ones by choosing the pieces in turn, which is the content of a problem below.

Example 8.59 (Assigning students to projects).

Nine students are to be assigned to three named projects needing four, two and three students. The assignments number

\binom{9}{4, 2, 3} = \frac{9!}{4!\,2!\,3!} = 1260.

Had the projects been unnamed and only their sizes fixed, the count would be different, since the pieces would no longer be distinguished by position.

Example 8.60 (A bag of shopping).

A bag is packed with four bananas, five tins of tuna, two boxes of cereal, four lemons, three bottles of cola and six light bulbs, twenty-four items in all, alike within each kind and distinguishable between kinds. The orders in which the bag can be packed number

\binom{24}{4, 5, 2, 4, 3, 6} = \frac{24!}{4!\,5!\,2!\,4!\,3!\,6!},

since an order is settled by saying which of the twenty-four positions each kind occupies.

Theorem 8.61 (The multinomial theorem).

Let $x_1, \ldots, x_t \in \mathbb{N}_0$ and let $n \in \mathbb{N}_0$ . Then

(x_1 + \cdots + x_t)^n = \sum \binom{n}{r_1, \ldots, r_t}\, x_1^{r_1} \cdots x_t^{r_t},

the sum running over all $t$ -tuples $(r_1, \ldots, r_t)$ in $\mathbb{N}_0$ with $r_1 + \cdots + r_t = n$ .

Discussion.

The statement generalises the binomial theorem from two summands to $t$ , and it is proved the same way, by induction on $n$ with the recursion for powers supplying the step. The bookkeeping is heavier because the terms are indexed by tuples rather than by a single number, so we describe the step first: multiplying by $x_1 + \cdots + x_t$ and distributing turns each term indexed by a tuple summing to $n$ into $t$ terms, one for each coordinate raised by one, and every tuple summing to $n+1$ arises this way from exactly the tuples obtained by lowering one of its positive coordinates. Collecting like terms, the coefficient wanted at $n+1$ is the sum of the coefficients at $n$ over those predecessors, and that sum identity is Pascal’s identity in its multinomial form, which the factorial formula supplies directly.

Proof.

Write $\mathcal{R}_n$ for the set of $t$ -tuples in $\mathbb{N}_0$ summing to $n$ . We first record the identity, for $(r_1, \ldots, r_t) \in \mathcal{R}_{n+1}$ ,

\binom{n+1}{r_1, \ldots, r_t} = \sum_{i \,:\, r_i \geqslant 1} \binom{n}{r_1, \ldots, r_i - 1, \ldots, r_t},

which follows from the factorial formula: the $i$ -th summand is $\dbinom{n+1}{r_1, \ldots, r_t}$ multiplied by $r_i$ and divided by $n+1$ , so the whole sum is $\dbinom{n+1}{r_1, \ldots, r_t}$ multiplied by $r_1 + \cdots + r_t = n+1$ and divided by $n+1$ .

Now induct on $n$ . At $n = 0$ the only tuple is $(0, \ldots, 0)$ and both sides are $1$ .

Suppose the identity holds at $n$ . Multiplying by $x_1 + \cdots + x_t$ and distributing,

(x_1 + \cdots + x_t)^{n+1} = \sum_{(r_1, \ldots, r_t) \in \mathcal{R}_n} \binom{n}{r_1, \ldots, r_t} \sum_{i=1}^{t} x_i \, x_1^{r_1} \cdots x_t^{r_t}.

Each inner term is $x_1^{q_1} \cdots x_t^{q_t}$ for the tuple $(q_1, \ldots, q_t) \in \mathcal{R}_{n+1}$ got from $(r_1, \ldots, r_t)$ by raising the $i$ -th coordinate by one; and conversely each tuple in $\mathcal{R}_{n+1}$ arises exactly once from each tuple got by lowering one of its positive coordinates. Collecting the terms belonging to a fixed $(q_1, \ldots, q_t)$ , its coefficient is

\sum_{i \,:\, q_i \geqslant 1} \binom{n}{q_1, \ldots, q_i - 1, \ldots, q_t} = \binom{n+1}{q_1, \ldots, q_t}

by the recorded identity, which is the claim at $n+1$ .

Remark.

At $t = 2$ the theorem is the binomial theorem and the recorded identity is Pascal’s, since a tuple $(k, n-k)$ has two coordinates to lower. The remark about where the binomial theorem applies carries over unchanged: the proof uses only the commutative, associative and distributive laws.

Example 8.62 (A term with no unknown in it).

Working in school algebra, a term of the expansion of

\left(x + y + z + \frac{1}{xyz}\right)^{n}

has the form

x^{r_1} y^{r_2} z^{r_3} \left(\frac{1}{xyz}\right)^{r_4} = x^{r_1 - r_4} y^{r_2 - r_4} z^{r_3 - r_4}

with $r_1 + r_2 + r_3 + r_4 = n$ . All three exponents vanish exactly when $r_1 = r_2 = r_3 = r_4$ , which forces $n = 4k$ and each $r_i = k$ . So the constant term is $\dbinom{4k}{k, k, k, k}$ when $n$ is four times $k$ , and $0$ otherwise.

Problem 8.31.

Show that

\binom{n}{r_1, \ldots, r_t} = \binom{n}{r_1}\binom{n - r_1}{r_2} \cdots \binom{n - r_1 - \cdots - r_{t-1}}{r_t}

whenever $r_1 + \cdots + r_t = n$ , by choosing the pieces one after another. Then check the identity a second time from the factorial formulas.

Problem 8.32.

How many distinct arrangements are there of the eleven letters of $\mathrm{MISSISSIPPI}$ ? State the count as a multinomial coefficient before evaluating it.

Problem 8.33.

Expand $(x + y + z)^3$ by the multinomial theorem, and check the answer by multiplying the three factors out directly. How many terms does the expansion of $(x_1 + \cdots + x_t)^n$ have, before like terms are collected, and how many after?

Problem 8.34.

Let $n \in \mathbb{N}$ . By counting the partitions of a $2n$ -set into $2$ -subsets, show that $(n+1)(n+2)\cdots(2n)$ is divisible by $2^n$ and not by $2^{n+1}$ .

Exercises

Exercises in Lean

The proofs below are checked in your browser. Nothing is sent anywhere, and nothing is stored but your own work. Type \to for →, \and for ∧, \< and \> for ⟨ ⟩.

This chapter asks four new things of the checker: the factorial and the binomial coefficient on $\mathbb{N}_0$ , the powers of a map, a name for a permutation, and the transposition as a map in its own right. The arithmetic comes first, written the way the Peano sheet wrote addition.

Factorials and binomial coefficients

The factorial of Definition 8.8 is written n !, with a space, since n! is a single name to Lean. Like addition, it is not computed, and its two clauses are equations to rewrite with:

Nat.factorial_zero   0 ! = succ 0
Nat.factorial_succ   (succ n) ! = succ n * n !

The binomial coefficient choose n k of Definition 8.45 is a count of subsets, and the checker has no counts, so it takes instead what Theorem 8.46 and Theorem 8.49 proved about that count:

Nat.choose_zero_right     choose n 0 = succ 0
Nat.choose_eq_zero_of_lt  n < k → choose n k = 0
Nat.choose_succ_succ      choose (succ n) (succ k) = choose n k + choose n (succ k)

The last is Pascal’s identity with both indices moved up by one, so that no subtraction appears.

Example.

$1! = 1$ , from the two clauses and the arithmetic of the Peano sheet.

Example.

A $1$ -subset of the empty set would be larger than the set it sits in.

Exercise 8.6.

The first part of Problem 8.4 .

Exercise 8.7.

The second half of the first part of Theorem 8.46 , this time from Pascal’s identity rather than by naming the one $n$ -subset.

Exercise 8.8.

The first half of its fourth part.

Powers

The power $\sigma^t$ of Definition 8.22 is written σ^[t], which is Lean’s own notation for a map composed with itself $t$ times. The recursion is not computed either, and its clauses are

Function.iterate_zero_apply   σ^[0] x = x
Function.iterate_succ_apply   σ^[succ t] x = σ^[t] (σ x)

the second being $\sigma^{s(t)} = \sigma^t \sigma$ read at a point. The checker cannot form an inverse, so the laws set below are the ones that need none.

Example.

The first part of Theorem 8.24 , by induction on $t$ . The point $x$ is left under the quantifier so that the inductive hypothesis can be used at $\sigma x$ . The result is listed below as Function.iterate_add_apply.

lean worked

1example (σ : Obj → Obj) (s t : ℕ) : ∀ x : Obj, σ^[s] (σ^[t] x) = σ^[s + t] x := by

2  induction t with k ih3  · intro x4    rw [Function.iterate_zero_apply, Nat.add_zero]5  · intro x6    rw [Function.iterate_succ_apply, ih (σ x), Nat.add_succ, Function.iterate_succ_apply]

verified

goalGoals accomplished.

Exercise 8.9.

The second part.

Exercise 8.10.

The first half of Theorem 8.27 , with the division already made: $k$ is any exponent at which $\sigma$ is the identity.

Permutations and transpositions

Definition 8.1 is the last sheet’s BijOn with the two sets equal, and it is written

Perm σ A   is   BijOn σ A A

so obtain ⟨hm, hi, hs⟩ takes a permutation apart and refine ⟨?_, ?_, ?_⟩ builds one. An equation between two maps on a set is read at each point of the set, as equality of functions reads it.

The transposition $(a \ b)$ is written swap a b. Its value at $x$ depends on whether $x$ is $a$ , $b$ or neither, and the checker cannot decide an equation between two objects, so its definition comes as three equations:

swap_apply_left          swap a b a = b
swap_apply_right         swap a b b = a
swap_apply_of_ne_of_ne   x ≠ a → x ≠ b → swap a b x = x

An argument about swap a b x for an unknown x goes by cases, and excluded middle supplies them: rcases Classical.em (x = a) with h | h leaves one goal with h : x = a and one with h : ¬x = a.

Example.

A transposition is its own inverse. It is listed below as swap_swap.

lean worked

1example (a b x : Obj) : swap a b (swap a b x) = x := by

2  rcases Classical.em (x = a) with h | h3  · rw [h, swap_apply_left, swap_apply_right]4  · rcases Classical.em (x = b) with h' | h'5    · rw [h', swap_apply_right, swap_apply_left]6    · rw [swap_apply_of_ne_of_ne h h', swap_apply_of_ne_of_ne h h']

verified

goalGoals accomplished.

Exercise 8.11.

Theorem 8.32 , read at each point of $A \cup B$ .

lean proof

1example (f g : Obj → Obj) (A B : Obj) (hf : MapsTo f A A) (hfB : ∀ x : Obj, x ∈ B → f x = x)2    (hg : MapsTo g B B) (hgA : ∀ x : Obj, x ∈ A → g x = x) :3    ∀ x : Obj, x ∈ A ∪ B → (g ∘ f) x = (f ∘ g) x := by

4  sorry

goalf g : Obj → Obj A B : Obj hf : MapsTo f A A hfB : ∀ (x : Obj), x ∈ B → f x = x hg : MapsTo g B B hgA : ∀ (x : Obj), x ∈ A → g x = x ⊢ ∀ (x : Obj), x ∈ A ∪ B → (g ∘ f) x = (f ∘ g) x

Exercise 8.12.

A transposition is a permutation of any set that holds its two points.

Exercise 8.13.

The second part of Problem 8.16 for a cycle of length two, with $\beta$ in the place of $\alpha^{-1}$ .

Exercise 8.14.

The identity $(a \ b) = \pi \, (a \ \ b-1) \, \pi$ from the proof of Proposition 8.41 , with $c$ in the place of $b - 1$ .

Exercise 8.15.

Part of Problem 8.3 : $h^{-1} \circ \sigma \circ h$ is a permutation of the domain of $h$ . The checker cannot invert a bijection, so $h^{-1}$ arrives as $k$ , together with the two equations that make it the inverse.

What the checker understands

Tactics

intro h	assume the hypothesis of an implication, naming it h
exact e	give the proof outright
apply f	reduce the goal to the hypotheses of f
assumption	close the goal with a hypothesis already present
trivial	close the goal True
exfalso	replace the goal with False
by_contra h	assume the negation of the goal
constructor	split ∧ into both halves, or ↔ into both directions
left / right	choose which half of a ∨ to prove
rcases h with a \| b	argue by cases on a disjunction
obtain ⟨a, b⟩ := h	take a conjunction or an existential apart
cases h	as above, keeping the name
refine e	give the proof with holes left in it
have h : p := …	record an intermediate result
show p	restate the goal in an equal form
use w	give a witness for ∃
specialize h a	instantiate a ∀ hypothesis
rw [h]	rewrite with an equation, ← to go backwards
rfl	both sides compute to the same thing
decide / norm_num	settle a closed computation
tauto	close a goal that is true by pure logic
induction n with k ih	the fifth Peano condition: prove the goal at 0, then at succ k from ih

Results you may cite

Classical.em	∀ (a : Prop), a ∨ ¬a — the law of excluded middle
Classical.byContradiction	∀ {a : Prop}, (¬a → False) → a — proof by contradiction; the tactic by_contra does this for you
Classical.byCases	∀ {a b : Prop}, (a → b) → (¬a → b) → b — split on whether a holds
not_not	∀ {a : Prop}, ¬¬a ↔ a — double negation
not_and_or	∀ {a b : Prop}, ¬(a ∧ b) ↔ ¬a ∨ ¬b — De Morgan
not_or	∀ {a b : Prop}, ¬(a ∨ b) ↔ ¬a ∧ ¬b — De Morgan
not_imp	∀ {a b : Prop}, ¬(a → b) ↔ a ∧ ¬b
and_comm	∀ {a b : Prop}, a ∧ b ↔ b ∧ a
or_comm	∀ {a b : Prop}, a ∨ b ↔ b ∨ a
Set.ext	∀ {A B : Obj}, (∀ x : Obj, x ∈ A ↔ x ∈ B) → A = B — extensionality: sets with the same elements are equal
Set.ext_iff	∀ {A B : Obj}, A = B ↔ (∀ x : Obj, x ∈ A ↔ x ∈ B) — extensionality and substitution, in one biconditional
Set.subset_antisymm	∀ {A B : Obj}, A ⊆ B → B ⊆ A → A = B — mutual inclusion is equality
Set.empty_subset	∀ {A : Obj}, ∅ ⊆ A — the empty set is a subset of every set
Set.pair_eq	∀ {a b c d : Obj}, ((a, b) = (c, d)) ↔ (a = c ∧ b = d) — two ordered pairs are equal exactly when their coordinates are
Nat.succ_inj	∀ {m n : ℕ}, succ m = succ n → m = n — the third Peano condition: the successor is injective
Nat.succ_ne_zero	∀ (n : ℕ), succ n ≠ 0 — the fourth: zero is nobody's successor
Nat.pred	∀ {n : ℕ}, n ≠ 0 → ∃ m : ℕ, n = succ m — predecessors: everything but zero is a successor
Nat.add_zero	∀ (m : ℕ), m + 0 = m — the first clause of addition
Nat.add_succ	∀ (m n : ℕ), m + succ n = succ (m + n) — the second clause of addition
Nat.zero_add	∀ (n : ℕ), 0 + n = n — addition from the left
Nat.succ_add	∀ (m n : ℕ), succ m + n = succ (m + n) — addition from the left, at a successor
Nat.add_assoc	∀ (m n p : ℕ), (m + n) + p = m + (n + p) — addition is associative
Nat.add_comm	∀ (m n : ℕ), m + n = n + m — addition is commutative
Nat.add_ne_zero	∀ {a : ℕ} (b : ℕ), a ≠ 0 → a + b ≠ 0 — positivity is absorbing
Nat.mul_zero	∀ (m : ℕ), m * 0 = 0 — the first clause of multiplication
Nat.mul_succ	∀ (m n : ℕ), m * succ n = m * n + m — the second clause of multiplication
Nat.zero_mul	∀ (m : ℕ), 0 * m = 0 — multiplication from the left
Nat.succ_mul	∀ (m n : ℕ), succ m * n = m * n + n — multiplication from the left, at a successor
Nat.add_right_cancel	∀ {m n k : ℕ}, m + k = n + k → m = n — cancellation, from the last sheet
Nat.add_eq_zero	∀ {m n : ℕ}, m + n = 0 → m = 0 ∧ n = 0 — a sum is zero only when both parts are, from the last sheet
Nat.mul_comm	∀ (m n : ℕ), m * n = n * m — multiplication is commutative, from the last sheet
Nat.mul_add	∀ (m n p : ℕ), m * (n + p) = m * n + m * p — multiplication distributes over addition, from the last sheet
Nat.mul_assoc	∀ (m n p : ℕ), (m * n) * p = m * (n * p) — multiplication associates, from the problems of the last chapter
Nat.add_mul	∀ (m n p : ℕ), (m + n) * p = m * p + n * p — distributivity on the other side
Nat.add_left_cancel	∀ {a m n : ℕ}, a + m = a + n → m = n — uniqueness of differences
Nat.lt_trichotomy	∀ (m n : ℕ), m < n ∨ m = n ∨ n < m — trichotomy, from the theorem that ℕ is strictly ordered
Nat.lt_irrefl	∀ (n : ℕ), ¬(n < n) — anti-reflexivity, from the last sheet
Nat.lt_trans	∀ {m n p : ℕ}, m < n → n < p → m < p — transitivity of the strict order, from the last sheet
Nat.lt_succ_self	∀ (n : ℕ), n < succ n — every number is below its successor, from the last sheet
Nat.not_lt_zero	∀ {n : ℕ}, ¬(n < 0) — nothing lies below zero
Nat.lt_succ_iff	∀ {m n : ℕ}, m < succ n ↔ m < n ∨ m = n — nothing lies strictly between n and succ n
Num.inj	∀ {m n : ℕ}, ↑m = ↑n → m = n — distinct numbers name distinct objects of ω
swap_apply_left	∀ (a b : Obj), swap a b a = b — the transposition sends a to b
swap_apply_right	∀ (a b : Obj), swap a b b = a — and b to a
swap_apply_of_ne_of_ne	∀ {a b x : Obj}, x ≠ a → x ≠ b → swap a b x = x — and fixes every other point
swap_swap	∀ (a b x : Obj), swap a b (swap a b x) = x — a transposition is its own inverse
Function.iterate_zero_apply	∀ (f : Obj → Obj) (x : Obj), f^[0] x = x — the first clause of the powers of a map
Function.iterate_succ_apply	∀ (f : Obj → Obj) (n : ℕ) (x : Obj), f^[succ n] x = f^[n] (f x) — the second clause: f^[succ n] is f^[n] ∘ f
Function.iterate_add_apply	∀ (f : Obj → Obj) (m n : ℕ) (x : Obj), f^[m + n] x = f^[m] (f^[n] x) — the first law of exponents
Nat.factorial_zero	0 ! = succ 0 — the first clause of the factorial
Nat.factorial_succ	∀ (n : ℕ), (succ n) ! = succ n * n ! — the second clause of the factorial
Nat.choose_zero_right	∀ (n : ℕ), choose n 0 = succ 0 — the empty set is the one 0-subset
Nat.choose_eq_zero_of_lt	∀ {n k : ℕ}, n < k → choose n k = 0 — no subset is larger than the whole
Nat.choose_succ_succ	∀ (n k : ℕ), choose (succ n) (succ k) = choose n k + choose n (succ k) — Pascal's identity

From the logical core

And.intro	∀ {a b : Prop}, a → b → a ∧ b
And.left	∀ {a b : Prop}, a ∧ b → a
And.right	∀ {a b : Prop}, a ∧ b → b
And.symm	∀ {a b : Prop}, a ∧ b → b ∧ a
Or.inl	∀ {a b : Prop}, a → a ∨ b
Or.inr	∀ {a b : Prop}, b → a ∨ b
Or.elim	∀ {a b c : Prop}, a ∨ b → (a → c) → (b → c) → c
Or.symm	∀ {a b : Prop}, a ∨ b → b ∨ a
Iff.intro	∀ {a b : Prop}, (a → b) → (b → a) → (a ↔ b)
Iff.mp	∀ {a b : Prop}, (a ↔ b) → a → b
Iff.mpr	∀ {a b : Prop}, (a ↔ b) → b → a
Iff.symm	∀ {a b : Prop}, (a ↔ b) → (b ↔ a)
Iff.rfl	∀ {a : Prop}, a ↔ a
Iff.trans	∀ {a b c : Prop}, (a ↔ b) → (b ↔ c) → (a ↔ c)
True.intro	True
False.elim	∀ {a : Prop}, False → a
absurd	∀ {a b : Prop}, a → ¬a → b
id	∀ {a : Prop}, a → a
mt	∀ {a b : Prop}, (a → b) → ¬b → ¬a
Eq.refl	∀ {α : Type} (a : α), a = a
Eq.symm	∀ {α : Type} {a b : α}, a = b → b = a
Eq.trans	∀ {α : Type} {a b c : α}, a = b → b = c → a = c
congrArg	∀ {α : Type} {β : Type} {a b : α} (f : α → β), a = b → f a = f b
Exists.intro	∀ {α : Type} {p : α → Prop} (w : α), p w → ∃ x : α, p x
Exists.elim	∀ {α : Type} {p : α → Prop} {b : Prop}, (∃ x : α, p x) → (∀ y : α, p y → b) → b

Lesson 9

Groups

Taught 22 Aug 2026

Operations and Semigroups

The last chapter produced a set with a multiplication on it. The permutations of a set compose, the composite is again a permutation, the composition is associative, there is an identity, and every element has an inverse. None of those five facts mentions what a permutation is. This chapter keeps the five facts and drops the permutations. Many other sets with an operation satisfy the same five facts, so a result proved from them holds for all of these at once.

Operations

Definition 9.1 (Operation).

Let $S$ be a non-empty set. An operation on $S$ is a function

* \, : \, S \times S \to S.

One writes $a * b$ for the value of $*$ at the ordered pair $(a, b)$ . The pair $(S, *)$ is a set equipped with an operation.

Two things are asked for. The first is that a value is named for every ordered pair of elements, which is what makes $*$ a function on all of $S \times S$ . The second is that the value lands back in $S$ , and one says that $S$ is closed under $*$ . Writing $a * b$ instead of $*\bigl((a,b)\bigr)$ is only notation; the object underneath is still a function, and two operations are equal exactly when they agree at every pair.

Example 9.2 (Operations already in hand).

$(\mathbb{N}_0, +)$ and $(\mathbb{N}_0, \cdot)$ , with addition as built in the chapter on the natural numbers and multiplication as defined in its problems. So are $(\mathbb{N}, +)$ and $(\mathbb{N}, \cdot)$ , since a sum or product of positive numbers is positive.
For a non-empty set $A$ , the Cartesian power $A^A$ , the set of all functions $A \to A$ , carries composition, $(f, g) \mapsto f \circ g$ . Its subset $S_A$ of permutations is closed under the same operation, since composition sends bijections to bijections; that closure is what let us write $\sigma\lambda$ at all.
For any set $X$ , the power set $\mathcal{P}(X)$ carries both $\cup$ and $\cap$ .
On a totally ordered set $M$ , the rule sending $(a, b)$ to whichever of $a$ and $b$ is the larger is an operation, since comparability names one of them and antisymmetry makes the choice unambiguous when they are equal.
On any non-empty $S$ , the rule $a * b \defeq a$ (keep the left factor) is an operation.

Example 9.3 (Two failures of closure).

Call $n \in \mathbb{N}$ prime if $n > 1$ and the only $k \in \mathbb{N}$ dividing $n$ are $1$ and $n$ , in the sense of divisibility. The primes are closed under neither addition nor multiplication: $3 + 5 = 8$ and $3 \cdot 5 = 15$ , and neither is prime. Restricting $+$ or $\cdot$ to the primes therefore does not produce an operation on them, and the restricted function is not a function into the primes at all.

The odd numbers show that the two can differ. They are closed under multiplication, since the remainder classes write two odd numbers as $2j + 1$ and $2k + 1$ , whose product is $2(2jk + j + k) + 1$ ; they are not closed under addition, since $1 + 1 = 2$ .

Definition 9.4 (Closed subsets).

Let $*$ be an operation on $S$ and let $S' \subset S$ be non-empty. Then $S'$ is closed under $*$ if $a * b \in S'$ for all $a, b \in S'$ . In that case the restriction of $*$ to $S' \times S'$ is an operation on $S'$ , and $(S', *)$ is a substructure of $(S, *)$ .

Closure is all that must be checked. Every named structure below (semigroup, group, subgroup) is this definition with further conditions added, and each further condition is either always inherited or always needs checking.

Example 9.5 (A substructure of the natural numbers).

Fix $\ell \in \mathbb{N}$ and let $\ell\mathbb{N}_0 \defeq \{n \in \mathbb{N}_0 \mid n = \ell k \text{ for some } k \in \mathbb{N}_0\}$ be the multiples of $\ell$ . If $n = \ell k$ and $m = \ell j$ then $n + m = \ell(k + j)$ by distributivity and $n \cdot m = \ell(k \cdot \ell \cdot j)$ , so $\ell\mathbb{N}_0$ is closed under both operations and is a substructure of $(\mathbb{N}_0, +)$ and of $(\mathbb{N}_0, \cdot)$ .

Associativity and Commutativity

Definition 9.6 (Associative and commutative operations).

Let $*$ be an operation on $S$ .

The operation is associative if $(a * b) * c = a * (b * c)$ for all $a, b, c \in S$ . In that case $(S, *)$ is a semigroup.
The operation is commutative if $a * b = b * a$ for all $a, b \in S$ . A commutative semigroup is also called abelian.

Both conditions are equalities required to hold identically, so both are inherited by every substructure: a law true of all elements of $S$ is in particular true of all elements of a subset. We note this once here rather than re-checking it in each example.

Example 9.7 (Which of them associate).

Addition and multiplication on $\mathbb{N}_0$ are associative and commutative by the laws of addition and the corresponding laws of multiplication. Composition on $A^A$ is associative by the theorem on associativity of composition, so $A^A$ and $S_A$ are semigroups under $\circ$ . Union and intersection on $\mathcal{P}(X)$ are associative and commutative by the basic properties of unions and by the algebra of intersection. Taking the larger of two elements of a totally ordered set is associative and commutative, both sides of each law naming the largest of the elements involved.

Keeping the left factor, $a * b \defeq a$ , is associative, since both $(a * b) * c$ and $a * (b * c)$ are $a$ , and is not commutative as soon as $S$ has two distinct points.

Composition shows that the two conditions are independent. It is associative on every $A^A$ , and already in $S_3$ there are permutations with $\sigma\lambda \neq \lambda\sigma$ , as the worked product of the last chapter showed in $S_4$ . Associativity is the more useful of the two, because of the next theorem.

Products of Many Factors

Definition 9.8 (Left-associated product).

Let $(S, *)$ be a set with an operation and let $a_1, \ldots, a_n \in S$ with $n \in \mathbb{N}$ . The left-associated product is defined by the recursion

p(a_1) \defeq a_1, \qquad p(a_1, \ldots, a_{k+1}) \defeq p(a_1, \ldots, a_k) * a_{k+1} \quad (1 \leqslant k < n).

A full bracketing of $a_1 * \cdots * a_n$ is either the single term $a_1$ , when $n = 1$ , or an expression $L * R$ in which $L$ is a full bracketing of $a_1 * \cdots * a_k$ and $R$ is a full bracketing of $a_{k+1} * \cdots * a_n$ , for some $1 \leqslant k < n$ .

Theorem 9.9 (General associativity).

Let $(S, *)$ be a semigroup, let $n \in \mathbb{N}$ and let $a_1, \ldots, a_n \in S$ .

For every $1 \leqslant k < n$ ,

p(a_1, \ldots, a_n) = p(a_1, \ldots, a_k) * p(a_{k+1}, \ldots, a_n).

Every full bracketing of $a_1 * \cdots * a_n$ equals $p(a_1, \ldots, a_n)$ .
If $*$ is commutative and $(b_1, \ldots, b_n)$ is obtained from $(a_1, \ldots, a_n)$ by exchanging two neighbouring entries, then $p(b_1, \ldots, b_n) = p(a_1, \ldots, a_n)$ .

Discussion.

This is the theorem on general associativity and neighbour swaps with $+$ replaced by $*$ , and the earlier proof never mentioned $\mathbb{N}_0$ : it used the associativity of $+$ for the first two parts and its commutativity for the third, and nothing else. So the argument transfers unchanged. We repeat the main step, which is the first part, the splitting identity, which says a left-associated product may be cut anywhere; it is an induction on the length of the second block, with the recursion clause as the base and a single application of associativity in the step. The second part is then an induction on $n$ in which a bracketing $L * R$ is split at its own top-level cut and the two halves identified by the inductive hypothesis. The third puts the exchanged pair in the middle by two applications of the first part, exchanges it by commutativity, and reassembles.

Proof.

For the first part, fix $k$ and induct on the length $m$ of the second block, so that $n = k + m$ . If $m = 1$ the assertion is $p(a_1, \ldots, a_{k+1}) = p(a_1, \ldots, a_k) * a_{k+1}$ , which is the recursion clause. Suppose it holds at $m$ . Then

\begin{aligned} p(a_1, \ldots, a_{n+1}) &= p(a_1, \ldots, a_n) * a_{n+1} \\ &= \bigl(p(a_1, \ldots, a_k) * p(a_{k+1}, \ldots, a_n)\bigr) * a_{n+1} \\ &= p(a_1, \ldots, a_k) * \bigl(p(a_{k+1}, \ldots, a_n) * a_{n+1}\bigr) \\ &= p(a_1, \ldots, a_k) * p(a_{k+1}, \ldots, a_{n+1}), \end{aligned}

by the recursion clause, the inductive hypothesis, associativity, and the recursion clause again.

The second and third parts are the two parts of the earlier theorem, whose proof used only the binary laws just used here; substituting $*$ for $+$ throughout gives the argument verbatim, with the first part standing in for the splitting identity recorded there.

The second part allows us to write $a_1 * a_2 * \cdots * a_n$ with no brackets at all in a semigroup, and we do so from here on. The third part allows exchanging neighbours when the operation commutes, and neighbours are enough, because the last chapter proved that every rearrangement is a string of neighbour exchanges.

Corollary 9.10 (Rearranging a product).

Let $(S, *)$ be an abelian semigroup, let $a_1, \ldots, a_n \in S$ and let $\sigma \in S_n$ . Then

a_{\sigma(1)} * a_{\sigma(2)} * \cdots * a_{\sigma(n)} = a_1 * a_2 * \cdots * a_n.

Proof.

The proposition that neighbour swaps suffice writes $\sigma$ as a product $\pi_1 \pi_2 \cdots \pi_m$ of transpositions of neighbouring points. Applying the third part of the theorem once for each $\pi_i$ carries the list $(a_{\sigma(1)}, \ldots, a_{\sigma(n)})$ to $(a_1, \ldots, a_n)$ without changing the left-associated product, and the second part identifies that product with the bracket-free expression on either side.

Remark.

The corollary uses results from three chapters. The bracketing was settled for sums in the chapter on sequences, the reduction of an arbitrary rearrangement to neighbour exchanges was set as a problem there and proved in the chapter on permutations, and the assembly is one line. Nothing about $\mathbb{N}_0$ was used, so the statement holds for unions, for intersections, for the larger-of-two operation, and for every abelian semigroup yet to be met.

Problem 9.1.

On $S = \{a, b, c\}$ define an operation by the table

\begin{array}{c|ccc} * & a & b & c \\ \hline a & a & b & c \\ b & b & c & a \\ c & c & a & b \end{array}

the entry in row $x$ and column $y$ being $x * y$ . Decide whether $*$ is associative and whether it is commutative.

Problem 9.2.

Let $S$ be non-empty and define $a * b \defeq b$ , keeping the right factor. Decide whether $*$ is associative and whether it is commutative, and compare with the left-factor rule.

Problem 9.3.

Let $(S, *)$ be an abelian semigroup and let $a_{j,k} \in S$ for $j \in \langle m \rangle$ and $k \in \langle n \rangle$ . Show that forming the products along the rows and then combining, and forming them along the columns and then combining, give the same element of $S$ .

Identity, Inverses and Groups

A semigroup asks one law of its operation. Two further conditions, an identity element and an inverse for each element, give a group. We look at each condition separately first.

The Identity

Definition 9.11 (Identity element).

Let $(S, *)$ be a set with an operation. An element $e \in S$ is an identity for $*$ if

e * a = a * e = a \qquad \text{for every } a \in S.

A semigroup with an identity is called a monoid.

An operation written additively usually has its identity written $0$ , and one written multiplicatively usually has it written $1$ . The choice is only notation. The definition asks for the law on both sides, and the next section shows why one side alone is weaker.

Proposition 9.12 (Uniqueness of the identity).

An operation has at most one identity.

Discussion.

The claim is a uniqueness statement, so we suppose two identities $e$ and $e'$ and derive $e = e'$ . We evaluate one product in two ways: reading $e * e'$ with the left factor as the identity gives $e'$ , reading it with the right factor as the identity gives $e$ , so the two are equal. No associativity is needed, which is why the proposition is stated for an operation and not for a semigroup.

Proof.

Let $e$ and $e'$ be identities for $*$ . Since $e$ is an identity, $e * e' = e'$ . Since $e'$ is an identity, $e * e' = e$ . Hence $e = e'$ .

Example 9.13 (Identities in the running examples).

The identity for $+$ on $\mathbb{N}_0$ is $0$ , and the identity for $\cdot$ on $\mathbb{N}_0$ and on $\mathbb{N}$ is $1$ . Composition on $A^A$ has the identity map $\mathrm{id}_A$ as its identity, by the identity laws, and the same map serves for $S_A$ . On $\mathcal{P}(X)$ the identity for $\cup$ is $\emptyset$ and the identity for $\cap$ is $X$ . Taking the larger of two elements of a totally ordered set has an identity exactly when the set has a minimum, which is then the identity.

Two of the examples have none. Keeping the left factor has no identity once $S$ has two distinct points: an identity $e$ would satisfy $e * a = a$ for every $a$ , whereas $e * a = e$ , forcing every $a$ to equal $e$ . And $(\mathbb{N}, +)$ has no identity, not because no element behaves like one but because the element that would, namely $0$ , is not in the set.

Inverses

Definition 9.14 (Inverse element).

Let $(S, *)$ be a set with an operation and with identity $e$ , and let $a \in S$ . An element $b \in S$ is an inverse of $a$ if

a * b = b * a = e.

If $a$ has an inverse, $a$ is invertible.

Multiplicative notation writes $a^{-1}$ for the inverse and additive notation writes $-a$ . The two-sided demand is the same shape as the one made of inverse functions, and for the same reason: a one-sided inverse is weaker unless associativity lets us combine the two sides, as in the next proposition.

Proposition 9.15 (Uniqueness of inverses).

Let $(S, *)$ be a monoid and let $a \in S$ be invertible. Then $a$ has exactly one inverse.

Discussion.

Uniqueness again, so we take two inverses $b$ and $c$ of the same $a$ and show they agree. We form one expression that can be read as either: put $a$ between them, as $b * a * c$ , and bracket it the two available ways. Bracketing to the right uses that $c$ is an inverse and leaves $b$ ; bracketing to the left uses that $b$ is one and leaves $c$ . The re-bracketing uses associativity, and without it the argument fails.

Proof.

Let $b$ and $c$ be inverses of $a$ . Then

\begin{aligned} b &= b * e && \text{($e$ is the identity)} \\ &= b * (a * c) && \text{($c$ is an inverse of $a$)} \\ &= (b * a) * c && \text{(associativity)} \\ &= e * c && \text{($b$ is an inverse of $a$)} \\ &= c && \text{($e$ is the identity)}. \end{aligned}

So one may speak of the inverse of an invertible element, and write $a^{-1}$ for it. Two consequences follow at once. First, $e$ is invertible with $e^{-1} = e$ , since $e * e = e$ . Second, $(a^{-1})^{-1} = a$ : the defining equations $a * a^{-1} = a^{-1} * a = e$ are unchanged when $a$ and $a^{-1}$ are exchanged, so $a$ is an inverse of $a^{-1}$ , and uniqueness makes it the inverse.

Proposition 9.16 (Inverse of a product).

Let $(S, *)$ be a monoid and let $a, b \in S$ be invertible. Then $a * b$ is invertible and

(a * b)^{-1} = b^{-1} * a^{-1}.

Discussion.

There are two assertions, an existence claim and a formula, and one calculation settles both: if the element named on the right multiplies with $a * b$ to give $e$ on both sides, then $a * b$ is invertible and, by the uniqueness just proved, that element is the inverse. The check is a product of four factors, which general associativity lets us write without brackets; cancelling the inner pair $a^{-1} * a$ leaves $b^{-1} * b$ , and cancelling that leaves $e$ . The order is reversed so that the inner pair is adjacent.

Proof.

Using general associativity to drop brackets,

(b^{-1} * a^{-1}) * (a * b) = b^{-1} * (a^{-1} * a) * b = b^{-1} * e * b = b^{-1} * b = e,

and in the same way $(a * b) * (b^{-1} * a^{-1}) = a * (b * b^{-1}) * a^{-1} = a * a^{-1} = e$ . So $b^{-1} * a^{-1}$ is an inverse of $a * b$ , and by uniqueness it is the inverse.

The reversal is not new. It is the inverse of a composition met for functions, and the rule $(\sigma\lambda)^{-1} = \lambda^{-1}\sigma^{-1}$ used without comment for permutations. The proof here covers both, and every other monoid as well.

Proposition 9.17 (The invertible elements).

Let $(S, *)$ be a monoid with identity $e$ and let

G \defeq \{a \in S \mid a \text{ is invertible}\}.

Then $e \in G$ ; if $a, b \in G$ then $a * b \in G$ ; and if $a \in G$ then $a^{-1} \in G$ . So $(G, *)$ is a monoid with identity $e$ , in which every element is invertible.

Discussion.

Three closure claims and then a summary that follows from them. That $e \in G$ was noted above, $e$ being its own inverse. That $G$ is closed under $*$ is the previous proposition, which not only says $a * b$ is invertible but names the inverse, so nothing further is needed. That $G$ is closed under inversion is the remark that $(a^{-1})^{-1} = a$ , which exhibits an inverse for $a^{-1}$ . The summary then needs only that associativity and the identity laws are inherited by a subset, which they are, being laws required of all elements.

Proof.

From $e * e = e$ we get $e \in G$ with $e^{-1} = e$ . If $a, b \in G$ then $a * b$ is invertible by the previous proposition, so $a * b \in G$ , and $G$ is closed under $*$ . If $a \in G$ then $(a^{-1})^{-1} = a$ exhibits $a$ as an inverse of $a^{-1}$ , so $a^{-1} \in G$ .

Hence $G$ is a closed subset containing $e$ , and it inherits associativity and the identity laws from $S$ . Each of its elements is invertible, and its inverse lies in $G$ , so $(G, *)$ is a monoid in which every element is invertible.

Example 9.18 (Who is invertible).

In $(\mathbb{N}_0, +)$ only $0$ is invertible, since $a + b = 0$ forces $a = b = 0$ by the corollary on sums that vanish. In $(\mathbb{N}_0, \cdot)$ only $1$ is invertible. In $\mathcal{P}(X)$ under $\cup$ only $\emptyset$ is invertible, and under $\cap$ only $X$ .

The main example is composition. In the monoid $A^A$ the invertible elements are exactly the bijections $A \to A$ , by the theorem on invertibility and bijections; that is to say, they are exactly the permutations, and the set they form is $S_A$ . The previous proposition, applied to $A^A$ , is therefore the statement that $S_A$ is closed under composition and under inversion. We proved this twice already, once for functions and once for permutations; here it follows from a proposition about monoids.

Problem 9.4.

Let $*$ be an operation on $S$ with a left identity $\ell$ , meaning $\ell * a = a$ for every $a$ , and a right identity $r$ , meaning $a * r = a$ for every $a$ . Show that $\ell = r$ , and that it is then an identity. No associativity is needed.

Problem 9.5.

Show that under the left-factor rule $a * b \defeq a$ every element of $S$ is a right identity, and that under the right-factor rule $a * b \defeq b$ every element is a left identity. So a one-sided identity need be neither unique nor accompanied by one on the other side, and neither rule has an identity once $S$ has two distinct points.

Problem 9.6.

Let $(S, *)$ be a monoid and let $a, b, c \in S$ satisfy $a * b = e$ and $b * c = e$ . Show that $a = c$ , so that $b$ is a two-sided inverse of $a$ . Which hypothesis fails if $S$ is only a set with an operation?

Groups

Definition 9.19 (Group).

A group is a monoid in which every element is invertible: a set $G$ with an associative operation, an identity $e$ , and an inverse for each of its elements. A group whose operation is commutative is abelian, in the sense already fixed for semigroups.

The proposition on invertible elements now reads: the invertible elements of any monoid form a group. Applied to $A^A$ under composition it says that $S_A$ is a group, without using anything about permutations.

Example 9.20 (Groups already met).

For any non-empty set $A$ , the symmetric group $S_A$ under composition. For $A = \langle n \rangle$ this is $S_n$ , with $\#S_n = n!$ by the count of bijections.
The dihedral group $D_n \subset S_n$ . That it is closed under products and inverses was a problem of the last chapter, and the identity permutation is a symmetry of every polygon, so $D_n$ is a group with $\#D_n = 2n$ .
The even permutations in $S_n$ , closed under products and inverses by another problem there. For $n \geqslant 2$ this group is written $A_n$ and called the alternating group on $n$ letters.
The one-element set $\{e\}$ with the only operation available to it. It is a group, called the trivial group, and it is the smallest group, since a group must contain an identity.
$S_2 = \{\mathrm{id}, \tau_{1,2}\}$ , the smallest group with more than one element. Its whole multiplication is $\tau_{1,2} * \tau_{1,2} = \mathrm{id}$ , which is the proposition that a transposition is its own inverse.

None of $(\mathbb{N}_0, +)$ , $(\mathbb{N}, +)$ and $(\mathbb{N}_0, \cdot)$ is a group. The first is a monoid in which only $0$ is invertible; the second has no identity at all; the third is a monoid in which only $1$ is invertible.

Remark (The groups not yet available).

The familiar first examples of groups are the integers under addition and the non-zero rationals or reals under multiplication. We do not have them: the integers are the subject of the next chapter, and they are built by adding to $(\mathbb{N}_0, +)$ the inverses it lacks. The examples above are the ones already constructed, and they are enough to prove everything in this chapter; the arithmetic ones become further examples once they are built.

The definition asks for a two-sided identity and two-sided inverses, four conditions in all. The two left-handed ones already imply the others.

Theorem 9.21 (A left identity with left inverses is enough).

Let $(G, *)$ be a semigroup. Suppose there is $\ell \in G$ with $\ell * a = a$ for every $a \in G$ , and suppose that for every $a \in G$ there is $b \in G$ with $b * a = \ell$ . Then $\ell$ is an identity, every such $b$ is an inverse of $a$ , and $(G, *)$ is a group.

Discussion.

The statement is an implication from a shortened list of axioms to the full one, so we assume the two left-handed conditions and must produce the two right-handed ones. We apply the hypotheses repeatedly and re-bracket, in a particular order: the right-inverse law must come first, because the right-identity law is proved from it.

To get $a * b = \ell$ , take a left inverse $b$ of $a$ and then a left inverse $c$ of $b$ . Insert $\ell$ in front of $a * b$ using the left identity, replace that $\ell$ by $c * b$ , and re-bracket so that $b * a$ becomes adjacent; that inner pair becomes $\ell$ , which the left identity removes, leaving $c * b = \ell$ . Once $a * b = \ell$ is known, $a * \ell$ can be written as $a * (b * a)$ , re-bracketed to $(a * b) * a = \ell * a = a$ , which is the right-identity law. No appeal to uniqueness is needed anywhere; the calculation gives the two-sided laws directly.

Proof.

Fix $a \in G$ , choose $b \in G$ with $b * a = \ell$ , and choose $c \in G$ with $c * b = \ell$ . Then

\begin{aligned} a * b &= \ell * (a * b) = (c * b) * (a * b) \\ &= c * \bigl((b * a) * b\bigr) = c * (\ell * b) = c * b = \ell, \end{aligned}

using the left identity, the choice of $c$ , associativity, the choice of $b$ , and the left identity again. So $b$ is a right inverse of $a$ as well as a left one.

Consequently

a * \ell = a * (b * a) = (a * b) * a = \ell * a = a,

so $\ell$ is a right identity as well as a left one. Hence $\ell$ is an identity, each $a$ has the two-sided inverse $b$ , and $(G, *)$ is a group.

Remark (Right axioms, mixed axioms).

The same argument with left and right exchanged shows that a right identity together with right inverses is equally sufficient. A left identity paired with right inverses is not enough, and a counterexample with two elements exists; producing one is a problem below. So the two halves of the definition may be dropped in matching pairs but not in mismatched ones.

Proposition 9.22 (Cancellation in a group).

Let $(G, *)$ be a group and let $a, x_1, x_2, y_1, y_2 \in G$ . If $a * x_1 = a * x_2$ then $x_1 = x_2$ ; and if $y_1 * a = y_2 * a$ then $y_1 = y_2$ .

Discussion.

Two implications of the same shape, and the second is the mirror image of the first, so only one need be written out. In each the hypothesis is an equation with a common factor on one side, and the way to remove a factor is to introduce its inverse on that same side and re-bracket: the inverse then meets $a$ and the pair becomes $e$ , which the identity law removes. Both the associativity and the existence of $a^{-1}$ are used, and each once.

Proof.

Suppose $a * x_1 = a * x_2$ . Then

\begin{aligned} x_1 &= e * x_1 = (a^{-1} * a) * x_1 = a^{-1} * (a * x_1) \\ &= a^{-1} * (a * x_2) = (a^{-1} * a) * x_2 = e * x_2 = x_2. \end{aligned}

The second implication is the same computation with $a^{-1}$ applied on the right.

Cancellation says exactly that, for each fixed $a$ , the maps $x \mapsto a * x$ and $y \mapsto y * a$ are injective. When $G$ is finite they are therefore bijective, so each of the equations $a * x = b$ and $y * a = b$ has exactly one solution. In fact they are bijective whether or not $G$ is finite, since $x \mapsto a^{-1} * x$ undoes the first; the finite case is worth noting separately, because it uses nothing about inverses and so applies to semigroups that are not groups.

Definition 9.23 (Regular semigroup).

A monoid in which cancellation holds on both sides (that is, $a * x_1 = a * x_2$ implies $x_1 = x_2$ , and $y_1 * a = y_2 * a$ implies $y_1 = y_2$ ) is called regular.

Every group is regular, by the proposition. The converse fails, and our standard examples show it.

Example 9.24 (The natural numbers cancel).

$(\mathbb{N}_0, +)$ is regular, since $x + z = y + z$ implies $x = y$ by the cancellation problem of the chapter on the natural numbers. It is not a group. Multiplication on $\mathbb{N}$ is regular too: if $a \cdot x_1 = a \cdot x_2$ with $a$ positive and, say, $x_1 \leqslant x_2$ , write $x_2 = x_1 + s$ and cancel to get $a \cdot s = 0$ , whence $s = 0$ because a product of positive numbers is positive. It is not a group either.

So regularity is strictly weaker than being a group, and it is the condition we need of a monoid we want to enlarge into a group. That is what the next chapter does to $(\mathbb{N}_0, +)$ .

Problem 9.7.

Construct a semigroup with a left identity in which every element has a right inverse, but which is not a group. Two elements suffice.

Problem 9.8.

Let $(G, *)$ be a finite group and write out its multiplication table, the entry in row $x$ and column $y$ being $x * y$ . Show that every element of $G$ occurs exactly once in each row and exactly once in each column.

Problem 9.9.

Show that $(\mathbb{N}_0, \cdot)$ is not regular, and that $\mathcal{P}(X)$ under $\cup$ is not regular once $X$ is non-empty.

Problem 9.10.

The symmetric difference $A \oplus B \defeq (A \setminus B) \cup (B \setminus A)$ , the set of objects lying in exactly one of $A$ and $B$ , was met in the exercises of the chapter on sets. Show that $\bigl(\mathcal{P}(X), \oplus\bigr)$ is an abelian group, and identify its identity and the inverse of each element.

Subgroups

Definition 9.25 (Subgroup).

Let $(G, *)$ be a group with identity $e$ and let $H \subset G$ be non-empty. Then $H$ is a subgroup of $G$ if $e \in H$ , if $a * b \in H$ for all $a, b \in H$ , and if $a^{-1} \in H$ for all $a \in H$ . A subgroup other than $\{e\}$ and $G$ is called proper.

The three conditions say precisely that $(H, *)$ is itself a group under the restricted operation, and nothing extra is assumed. Suppose $H$ is a closed subset that happens to be a group in its own right, with identity $f \in H$ . Then $f * f = f$ holds in $G$ , and cancelling $f$ there gives $f = e$ : a subset cannot have an identity of its own. Likewise the inverse of $a \in H$ computed inside $H$ is an inverse of $a$ inside $G$ , so uniqueness of inverses makes it $a^{-1}$ . The conditions listed are therefore necessary as well as sufficient, and they may be checked by computing in $G$ .

Proposition 9.26 (One-step subgroup criterion).

Let $(G, *)$ be a group and let $H \subset G$ be non-empty. Then $H$ is a subgroup of $G$ if and only if $a * b^{-1} \in H$ for all $a, b \in H$ .

Discussion.

A biconditional between the three conditions of the definition and a single one, and one direction is much easier than the other. Forwards is immediate: a subgroup contains $b^{-1}$ and is closed, so it contains $a * b^{-1}$ . Backwards, we get the three conditions from the one by choosing $a$ and $b$ suitably. Non-emptiness supplies some $a \in H$ to start from; taking $b = a$ produces $e$ ; taking $a = e$ then produces $b^{-1}$ for every $b \in H$ ; and once inverses are available, the product $a * b$ is $a * (b^{-1})^{-1}$ , which the hypothesis covers. Each step uses the hypothesis once, and the steps must come in this order.

Proof.

Suppose $H$ is a subgroup and let $a, b \in H$ . Then $b^{-1} \in H$ , so $a * b^{-1} \in H$ .

Conversely suppose $a * b^{-1} \in H$ whenever $a, b \in H$ . Since $H$ is non-empty, choose $a \in H$ ; taking $b = a$ gives $e = a * a^{-1} \in H$ . Now let $b \in H$ ; taking the pair $(e, b)$ gives $b^{-1} = e * b^{-1} \in H$ , so $H$ is closed under inverses. Finally let $a, b \in H$ ; then $b^{-1} \in H$ , so taking the pair $(a, b^{-1})$ gives $a * (b^{-1})^{-1} = a * b \in H$ . Hence $H$ satisfies the three conditions.

Example 9.27 (Subgroups already met).

$D_n$ is a subgroup of $S_n$ , and for $n \geqslant 2$ the alternating group $A_n$ is a subgroup of $S_n$ . Inside $S_A$ , the permutations fixing a chosen point $a \in A$ form a subgroup: the identity fixes $a$ , a composite of two maps fixing $a$ fixes $a$ , and if $\sigma(a) = a$ then $\sigma^{-1}(a) = a$ . Every group has the two subgroups $\{e\}$ and $G$ .

For each $\ell \in \mathbb{N}$ the multiples of $\ell$ form a subgroup of the integers under addition, an example we can state properly once the integers are built.

Problem 9.11.

Show that a non-empty finite subset $H$ of a group $G$ that is closed under the operation is already a subgroup. For fixed $a \in H$ the map $x \mapsto a * x$ sends $H$ into $H$ ; use finiteness to make it surjective, and read off $e \in H$ and $a^{-1} \in H$ .

Problem 9.12.

Show that an intersection of subgroups of $G$ is a subgroup of $G$ , for any non-empty indexed family of them. Show by example that a union of two subgroups need not be one.

Problem 9.13.

List every subgroup of $S_3$ , and check each against the one-step criterion.

Homomorphisms and Quotients

A subgroup cuts a group into pieces of equal size, and sometimes the pieces form a group themselves. The last notion of this chapter is a map that respects the operation.

Cosets

Definition 9.28 (Cosets).

Let $H$ be a subgroup of a group $(G, *)$ and let $b \in G$ . The right coset and the left coset of $H$ by $b$ are

H * b \defeq \{h * b \mid h \in H\}, \qquad b * H \defeq \{b * h \mid h \in H\}.

Proposition 9.29 (The right cosets partition the group).

Let $H$ be a subgroup of a group $(G, *)$ and define $a_1 \sim a_2$ to mean $a_1 * a_2^{-1} \in H$ . Then $\sim$ is an equivalence relation on $G$ whose classes are exactly the right cosets of $H$ . Moreover $\#(H * b) = \#H$ for every $b$ , when $H$ is finite.

Discussion.

Three assertions. The first asks for the three defining properties of an equivalence relation, and each comes from one of the three subgroup conditions: reflexivity is $e \in H$ , symmetry is closure under inverses applied to $a_1 * a_2^{-1}$ , and transitivity is closure under the operation applied to a product that telescopes. The second is an equality of sets, so it is two inclusions, and both come from writing an element of one side in the form the other side demands. The third is a counting claim, and since cardinality classifies finite sets it is enough to produce a bijection $H \to H * b$ ; the map $h \mapsto h * b$ is one, being surjective by the definition of the coset and injective by cancellation.

Proof.

For reflexivity, $a * a^{-1} = e \in H$ , so $a \sim a$ . For symmetry, if $a_1 * a_2^{-1} \in H$ then its inverse $(a_1 * a_2^{-1})^{-1} = a_2 * a_1^{-1}$ lies in $H$ , so $a_2 \sim a_1$ . For transitivity, if $a_1 * a_2^{-1} \in H$ and $a_2 * a_3^{-1} \in H$ then their product

(a_1 * a_2^{-1}) * (a_2 * a_3^{-1}) = a_1 * a_3^{-1}

lies in $H$ , so $a_1 \sim a_3$ .

For the classes, fix $b$ . If $a \sim b$ then $h \defeq a * b^{-1} \in H$ and $a = h * b \in H * b$ . Conversely if $a = h * b$ with $h \in H$ then $a * b^{-1} = h \in H$ , so $a \sim b$ . Hence $[b] = H * b$ .

For the count, the map $h \mapsto h * b$ from $H$ to $H * b$ is surjective by the definition of $H * b$ and injective by cancellation, hence a bijection, so the two sets have the same cardinality.

Since the classes of an equivalence relation partition the set, the right cosets partition $G$ into pieces all of the same size as $H$ . The left cosets do the same, by the mirror argument with $a_1^{-1} * a_2 \in H$ .

Problem 9.14.

Let $G$ be a finite group and $H$ a subgroup. Show that $\#G$ is the product of $\#H$ with the number of right cosets of $H$ . Deduce that $\#H$ divides $\#G$ .

Problem 9.15.

Show that the left cosets of $H$ partition $G$ as the right cosets do, and that the two partitions have the same number of pieces, by checking that $H * b \mapsto b^{-1} * H$ is a well-defined bijection between them.

Homomorphisms

Definition 9.30 (Homomorphism).

Let $(G_1, *)$ and $(G_2, *)$ be groups. A function $\alpha : G_1 \to G_2$ is a homomorphism if

\alpha(a * b) = \alpha(a) * \alpha(b) \qquad \text{for all } a, b \in G_1.

Its kernel is $\ker \alpha \defeq \alpha^{-1}(\{e_2\})$ , the preimage of the identity of $G_2$ , and its image is $\alpha(G_1)$ . A bijective homomorphism is an isomorphism; groups with an isomorphism between them are isomorphic, written $G_1 \cong G_2$ . An isomorphism from a group to itself is an automorphism. Throughout, $e_1$ and $e_2$ are the identities of $G_1$ and of $G_2$ .

The same symbol $*$ denotes two different operations, one on each side of the equation, and that is deliberate: the condition says that computing in $G_1$ and then transporting gives the same answer as transporting and then computing in $G_2$ .

Proposition 9.31 (Homomorphisms respect the identity and inverses).

Let $\alpha : G_1 \to G_2$ be a homomorphism. Then $\alpha(e_1) = e_2$ and $\alpha(a^{-1}) = \alpha(a)^{-1}$ for every $a \in G_1$ .

Discussion.

Neither conclusion is assumed by the definition, which mentions only products, so both must be derived from the product condition. For the first, $e_1$ is characterised inside $G_1$ by $e_1 * e_1 = e_1$ ; applying $\alpha$ turns that into an equation about $\alpha(e_1)$ alone, and cancellation in $G_2$ finishes it. For the second, apply $\alpha$ to the defining equation $a * a^{-1} = e_1$ and use the first part: the result says $\alpha(a^{-1})$ is an inverse of $\alpha(a)$ , and inverses are unique.

Proof.

From $e_1 * e_1 = e_1$ we get $\alpha(e_1) * \alpha(e_1) = \alpha(e_1) = \alpha(e_1) * e_2$ , and cancelling $\alpha(e_1)$ on the left gives $\alpha(e_1) = e_2$ .

Then $\alpha(a) * \alpha(a^{-1}) = \alpha(a * a^{-1}) = \alpha(e_1) = e_2$ , and likewise $\alpha(a^{-1}) * \alpha(a) = e_2$ . So $\alpha(a^{-1})$ is an inverse of $\alpha(a)$ , hence is $\alpha(a)^{-1}$ by uniqueness.

Proposition 9.32 (Kernel and image).

Let $\alpha : G_1 \to G_2$ be a homomorphism.

$\ker \alpha$ is a subgroup of $G_1$ and $\alpha(G_1)$ is a subgroup of $G_2$ .
$\alpha$ is injective if and only if $\ker \alpha = \{e_1\}$ .
If $\alpha$ is an isomorphism then so is $\alpha^{-1}$ .

Discussion.

Three separate claims. The first two subgroup checks are the one-step criterion, and each is a short computation using the previous proposition to convert $\alpha$ of an inverse into an inverse of $\alpha$ ; non-emptiness comes from $\alpha(e_1) = e_2$ in both cases. The second claim is a biconditional: forwards, an injective map sends at most one point to $e_2$ and $e_1$ is already one; backwards, the way to turn an equality $\alpha(a_1) = \alpha(a_2)$ into a statement about the kernel is to move everything to one side, forming $a_1 * a_2^{-1}$ , which the hypothesis then forces to be $e_1$ . The third is again a matter of transporting a condition: given $a_2, b_2 \in G_2$ , name their preimages, apply the homomorphism condition to those, and read the resulting equation backwards through $\alpha^{-1}$ ; that $\alpha^{-1}$ is bijective is the theorem that the inverse of a bijection is a bijection.

Proof.

For the first, $\alpha(e_1) = e_2$ puts $e_1 \in \ker \alpha$ and $e_2 \in \alpha(G_1)$ , so both are non-empty. If $a, b \in \ker \alpha$ then

\alpha(a * b^{-1}) = \alpha(a) * \alpha(b)^{-1} = e_2 * e_2^{-1} = e_2,

so $a * b^{-1} \in \ker\alpha$ and the one-step criterion applies. If $\alpha(a), \alpha(b) \in \alpha(G_1)$ then $\alpha(a) * \alpha(b)^{-1} = \alpha(a * b^{-1}) \in \alpha(G_1)$ , and the criterion applies again.

For the second, suppose $\alpha$ is injective and $a \in \ker\alpha$ . Then $\alpha(a) = e_2 = \alpha(e_1)$ , so $a = e_1$ . Conversely suppose $\ker\alpha = \{e_1\}$ and $\alpha(a_1) = \alpha(a_2)$ . Then

\alpha(a_1 * a_2^{-1}) = \alpha(a_1) * \alpha(a_2)^{-1} = e_2,

so $a_1 * a_2^{-1} = e_1$ and hence $a_1 = a_2$ .

For the third, let $\alpha$ be an isomorphism and let $a_2, b_2 \in G_2$ have preimages $a_1, b_1$ . Then $\alpha(a_1 * b_1) = a_2 * b_2$ , so $\alpha^{-1}(a_2 * b_2) = a_1 * b_1 = \alpha^{-1}(a_2) * \alpha^{-1}(b_2)$ . And $\alpha^{-1}$ is a bijection, so it is an isomorphism.

Example 9.33 (The sign is a homomorphism).

The sign of a permutation takes two values, and the group with two elements is $S_2 = \{\mathrm{id}, \tau_{1,2}\}$ . Define $\mathrm{sg} : S_n \to S_2$ by sending an even permutation to $\mathrm{id}$ and an odd one to $\tau_{1,2}$ . The theorem on the properties of the sign says the sign of a product is the product of the signs, which is exactly the statement that $\mathrm{sg}$ is a homomorphism: the four cases of “even or odd times even or odd” are the four products in $S_2$ .

Its kernel is the set of even permutations, which is $A_n$ . So the fact that the alternating group is a subgroup, checked by hand last chapter, now follows from the proposition on kernels. For $n \geqslant 2$ the map is surjective, since a transposition is odd, so its image is all of $S_2$ .

Theorem 9.34 (Cayley's theorem).

Every group is isomorphic to a subgroup of a symmetric group. Precisely: if $(G, *)$ is a group then $a \mapsto \lambda_a$ , where $\lambda_a(x) \defeq a * x$ , is an injective homomorphism $G \to S_G$ , so $G$ is isomorphic to a subgroup of $S_G$ .

Discussion.

There are three things to check. First, the proposed values must lie where the theorem says: $\lambda_a$ must be a permutation of $G$ , and it is, because $\lambda_{a^{-1}}$ undoes it on both sides, so it is invertible and hence a bijection. Second, the map must be a homomorphism, and since both sides are functions on $G$ the check is an equality of functions, settled by evaluating at an arbitrary $x$ ; associativity is the only thing used, and it is used once. Third, injectivity, which by the previous proposition is the triviality of the kernel: if $\lambda_a$ is the identity permutation then it fixes $e$ in particular, and $\lambda_a(e) = a$ . The isomorphism onto a subgroup is then the image, which the previous proposition already knows to be a subgroup.

Proof.

For each $a \in G$ the maps $\lambda_a$ and $\lambda_{a^{-1}}$ satisfy

\lambda_{a^{-1}}\bigl(\lambda_a(x)\bigr) = a^{-1} * (a * x) = x

and likewise in the other order, so $\lambda_a$ is invertible and hence a permutation of $G$ ; that is, $\lambda_a \in S_G$ .

Let $\Lambda : G \to S_G$ be $a \mapsto \lambda_a$ . For $a, b \in G$ and any $x \in G$ ,

\lambda_{a * b}(x) = (a * b) * x = a * (b * x) = \lambda_a\bigl(\lambda_b(x)\bigr) = (\lambda_a \lambda_b)(x),

so $\Lambda(a * b) = \Lambda(a)\Lambda(b)$ and $\Lambda$ is a homomorphism.

If $a \in \ker\Lambda$ then $\lambda_a = \mathrm{id}_G$ , so $a = a * e = \lambda_a(e) = e$ . Hence $\ker\Lambda = \{e\}$ and $\Lambda$ is injective. Its image is a subgroup of $S_G$ , and $\Lambda$ restricted to that image is a bijective homomorphism, hence an isomorphism.

Remark.

Cayley’s theorem says that every group is a group of permutations, so in that sense the last chapter already covered all groups. The statement is weak, though: $S_G$ has $\#G\,!$ elements when $G$ is finite, so the copy of $G$ inside it is a very small part of a very large group, and locating it is no easier than knowing $G$ already.

Problem 9.16.

Show that $\mathrm{sg} : S_n \to S_2$ of the example above is the only homomorphism $S_n \to S_2$ that is not constant, for $n \geqslant 2$ . Every transposition must go to the same place; say why.

Problem 9.17.

Show that a homomorphism carries a subgroup of $G_1$ to a subgroup of $G_2$ , and that the preimage of a subgroup of $G_2$ is a subgroup of $G_1$ .

Problem 9.18.

Show that $D_3$ and $S_3$ are isomorphic, and that $D_4$ and $S_4$ are not.

Normal Subgroups and Quotients

Kernels satisfy a condition that not every subgroup does, and it is exactly the condition needed to make the cosets into a group.

Definition 9.35 (Normal subgroup).

A subgroup $H$ of a group $(G, *)$ is normal if $b^{-1} * a * b \in H$ for all $a \in H$ and $b \in G$ .

In an abelian group $b^{-1} * a * b = a$ , so every subgroup is normal, and the condition only matters in the non-commutative case.

Proposition 9.36 (Kernels are normal).

The kernel of a homomorphism $\alpha : G_1 \to G_2$ is a normal subgroup of $G_1$ .

Discussion.

That the kernel is a subgroup is already proved, so only the normality condition is left, and it is a membership claim: we must show $\alpha$ sends $b^{-1} * a * b$ to $e_2$ . Applying the homomorphism condition twice turns that element into a product of three values of $\alpha$ , the middle one being $e_2$ because $a$ lies in the kernel; the outer two are inverse to each other by the proposition on how homomorphisms treat inverses, so the product is $e_2$ .

Proof.

Let $a \in \ker\alpha$ and $b \in G_1$ . Then

\begin{aligned} \alpha(b^{-1} * a * b) &= \alpha(b)^{-1} * \alpha(a) * \alpha(b) \\ &= \alpha(b)^{-1} * e_2 * \alpha(b) = \alpha(b)^{-1} * \alpha(b) = e_2, \end{aligned}

so $b^{-1} * a * b \in \ker\alpha$ . With the subgroup property already established, $\ker\alpha$ is normal.

Theorem 9.37 (Normal subgroups are those whose cosets agree).

Let $H$ be a subgroup of a group $(G, *)$ . Then $H$ is normal if and only if $H * b = b * H$ for every $b \in G$ .

Discussion.

A biconditional, and each direction converts one membership statement into another by moving a factor across. Forwards: an element of $H * b$ is $h * b$ , and to exhibit it in $b * H$ we must write it as $b$ times something in $H$ ; inserting $b * b^{-1}$ gives $b * (b^{-1} * h * b)$ , and normality is precisely the statement that the bracketed factor lies in $H$ . The reverse inclusion is the same move with $b^{-1}$ in place of $b$ , which normality also covers. Backwards: given the equality of the two cosets, an element $b^{-1} * a * b$ with $a \in H$ is a member of $b^{-1} * (H * b)$ , which the hypothesis rewrites as $b^{-1} * (b * H) = H$ .

Proof.

Suppose $H$ is normal and let $b \in G$ . If $h \in H$ then $h * b = b * (b^{-1} * h * b)$ and $b^{-1} * h * b \in H$ , so $H * b \subset b * H$ . If $h \in H$ then $b * h = \bigl(b * h * b^{-1}\bigr) * b$ , and $b * h * b^{-1} = (b^{-1})^{-1} * h * b^{-1} \in H$ by normality applied with $b^{-1}$ , so $b * H \subset H * b$ . Hence the two cosets are equal.

Conversely suppose $H * b = b * H$ for every $b$ , and let $a \in H$ and $b \in G$ . Then $a * b \in H * b = b * H$ , so $a * b = b * h$ for some $h \in H$ , and $b^{-1} * a * b = h \in H$ . So $H$ is normal.

When $H$ is normal the two partitions of $G$ coincide and there is just one collection of cosets, which we write $G/H$ — the quotient of $G$ by the equivalence relation of the coset proposition.

Theorem 9.38 (The quotient group).

Let $H$ be a normal subgroup of a group $(G, *)$ . Then

(H * a) * (H * b) \defeq H * (a * b)

is a well-defined operation on $G/H$ , making $G/H$ a group with identity $H = H * e$ and with $(H * a)^{-1} = H * a^{-1}$ . The map $q : G \to G/H$ sending $a$ to $H * a$ is a surjective homomorphism with $\ker q = H$ .

Discussion.

The main difficulty is showing that the operation is well defined. The proposed rule names the answer using $a$ and $b$ , but the arguments of the operation are the cosets, and a coset has many names; so we must show that replacing $a$ by any $a'$ with $H * a' = H * a$ , and $b$ likewise, leaves $H * (a * b)$ unchanged. Written in terms of the equivalence relation, that means $a' * a^{-1} \in H$ and $b' * b^{-1} \in H$ should force $(a' * b') * (a * b)^{-1} \in H$ , and this calculation uses normality: the two known members of $H$ are separated by $a$ , and normality lets us move it past.

The rest follows from the laws in $G$ . Associativity, the identity law and the inverse law each reduce, through the definition of the operation, to the corresponding law in $G$ applied to representatives. That $q$ is a homomorphism is the definition of the operation read backwards, surjectivity is the definition of $G/H$ , and the kernel is the set of $a$ with $H * a = H$ , which is $H$ .

Proof.

For well-definedness, suppose $H * a' = H * a$ and $H * b' = H * b$ , so that $h \defeq a' * a^{-1}$ and $k \defeq b' * b^{-1}$ lie in $H$ . Then $a' = h * a$ and $b' = k * b$ , so

(a' * b') * (a * b)^{-1} = h * a * k * b * b^{-1} * a^{-1} = h * (a * k * a^{-1}),

and $a * k * a^{-1} = (a^{-1})^{-1} * k * a^{-1} \in H$ by normality. So the displayed element is a product of two members of $H$ and lies in $H$ , giving $H * (a' * b') = H * (a * b)$ . The operation is therefore well defined.

Associativity follows from associativity in $G$ :

\begin{aligned} \bigl((H * a) * (H * b)\bigr) * (H * c) &= H * \bigl((a * b) * c\bigr) \\ &= H * \bigl(a * (b * c)\bigr) \\ &= (H * a) * \bigl((H * b) * (H * c)\bigr). \end{aligned}

Also $(H * e) * (H * a) = H * a = (H * a) * (H * e)$ , so $H * e = H$ is an identity, and $(H * a) * (H * a^{-1}) = H * e = (H * a^{-1}) * (H * a)$ , so $H * a^{-1}$ is an inverse of $H * a$ . Hence $G/H$ is a group.

Finally $q(a * b) = H * (a * b) = (H * a) * (H * b) = q(a) * q(b)$ , so $q$ is a homomorphism; it is surjective because every coset is $H * a$ for some $a$ ; and $q(a) = H$ holds exactly when $a * e^{-1} = a \in H$ , so $\ker q = H$ .

Remark.

So normal subgroups and kernels are the same thing: every kernel is normal, by the proposition, and every normal subgroup is the kernel of the map $q$ onto its own quotient. So the question “which subgroups can be quotiented by?” and the question “which subgroups arise as kernels?” have one answer.

Theorem 9.39 (First isomorphism theorem).

Let $\alpha : G_1 \to G_2$ be a homomorphism. Then

G_1/\ker\alpha \;\cong\; \alpha(G_1),

by the isomorphism sending the coset $\ker\alpha * a$ to $\alpha(a)$ .

Discussion.

The statement claims an isomorphism between two groups both of which are already known to exist: the quotient by the previous theorem, since the kernel is normal, and the image by the proposition on kernels and images. So we only need to check four things about the named map. That it is well defined is the same issue as before, since a coset has many names, and it holds because two representatives differ by a member of the kernel, which $\alpha$ sends to the identity. That it is a homomorphism is the definition of the quotient operation together with the homomorphism property of $\alpha$ . Surjectivity onto the image is immediate. Injectivity is the triviality of the kernel of the new map, and it comes out as the statement that $\alpha(a) = e_2$ puts $a$ in $\ker\alpha$ , so the coset is the identity of the quotient.

Proof.

Write $K \defeq \ker\alpha$ , a normal subgroup by the proposition, so that $G_1/K$ is a group. Define $\beta(K * a) \defeq \alpha(a)$ .

Well defined. If $K * a' = K * a$ then $a' * a^{-1} \in K$ , so $\alpha(a') * \alpha(a)^{-1} = \alpha(a' * a^{-1}) = e_2$ and hence $\alpha(a') = \alpha(a)$ .

Homomorphism. $\beta\bigl((K * a) * (K * b)\bigr) = \beta\bigl(K * (a * b)\bigr) = \alpha(a * b) = \alpha(a) * \alpha(b) = \beta(K * a) * \beta(K * b)$ .

Surjective onto $\alpha(G_1)$ . Every element of the image is $\alpha(a) = \beta(K * a)$ for some $a$ .

Injective. If $\beta(K * a) = e_2$ then $\alpha(a) = e_2$ , so $a \in K$ and $K * a = K$ , the identity of $G_1/K$ . So the kernel of $\beta$ is trivial and $\beta$ is injective.

Hence $\beta$ is a bijective homomorphism from $G_1/K$ onto $\alpha(G_1)$ .

Example 9.40 (The sign again).

Take $\alpha = \mathrm{sg} : S_n \to S_2$ with $n \geqslant 2$ . Its kernel is $A_n$ and its image is all of $S_2$ , so the theorem gives

S_n/A_n \;\cong\; S_2 .

The quotient has two elements, the even permutations and the odd ones, and its multiplication is the arithmetic of parity: even times even is even, even times odd is odd, odd times odd is even. Counting the cosets recovers the problem that $\#S_n$ is twice the number of even permutations, this time as a consequence rather than a construction.

Problem 9.19.

Show that a subgroup $H \subset G$ with exactly two right cosets is normal, without computing anything about $H$ itself.

Problem 9.20.

Show that $\{\mathrm{id}, \tau_{1,2}\}$ is a subgroup of $S_3$ that is not normal, and that $A_3$ is a normal subgroup of $S_3$ . Which of the two has a quotient group, and what is it?

Problem 9.21.

Let $\alpha : G_1 \to G_2$ be a homomorphism and let $H$ be a normal subgroup of $G_1$ contained in $\ker\alpha$ . Show that there is exactly one homomorphism $\gamma : G_1/H \to G_2$ with $\alpha = \gamma \circ q$ , where $q : G_1 \to G_1/H$ is the quotient map. The first isomorphism theorem is the case $H = \ker\alpha$ .

Exercises

Exercises in Lean

The proofs below are checked in your browser. Nothing is sent anywhere, and nothing is stored but your own work. Type \to for →, \and for ∧, \< and \> for ⟨ ⟩.

Every statement of this chapter is made of one group, so the checker is given one and the sheet works inside it. The objects are its elements, and three new pieces of notation carry the structure:

a ∗ b    the product, typed \ast
e        the identity
a⁻¹      the inverse of a, typed \-1

The operation binds like a product, so a ∗ b ∗ c is (a ∗ b) ∗ c and brackets are written where the other reading is wanted. A subgroup is a set, hence an object like any other, and a ∈ H is the membership of the sets chapter.

Five entries carry Definition 9.19 and nothing else is given away:

mul_assoc        (a ∗ b) ∗ c = a ∗ (b ∗ c)
e_mul            e ∗ a = a
mul_e            a ∗ e = a
inv_mul_cancel   a⁻¹ ∗ a = e
mul_inv_cancel   a ∗ a⁻¹ = e

Everything else in the chapter is a consequence, and the sheet asks for the consequences.

Multiplying an equation through

The axioms are equations, so the work is rewriting. What the paper proofs do in one line (multiply both sides by $a^{-1}$ ) the checker needs spelled out, and congrArg says it: from h : x = y it builds f x = f y for any f written out as fun t => …. Once the equation has the factor on it, rw [...] at h cancels.

Example.

The observation that settled the identity of a subgroup, in the discussion after Definition 9.25 . Multiplying by $a^{-1}$ on the left is the whole of it.

Example.

The left half of Proposition 9.22 . It is listed below as mul_left_cancel, so the exercises may lean on it.

Example.

Inverting twice returns the element, noted after Proposition 9.15 . It is listed below as inv_inv.

lean worked

1example (a : Obj) : (a⁻¹)⁻¹ = a := by

2  have h : a⁻¹ ∗ (a⁻¹)⁻¹ = e := mul_inv_cancel a⁻¹3  have h2 : a ∗ (a⁻¹ ∗ (a⁻¹)⁻¹) = a ∗ e := congrArg (fun t => a ∗ t) h4  rw [← mul_assoc, mul_inv_cancel, e_mul, mul_e] at h25  exact h2

verified

goalGoals accomplished.

Exercise 9.12.

The identity is its own inverse.

Exercise 9.13.

Proposition 9.12 , with only the left-hand law assumed.

Exercise 9.14.

The right half of Proposition 9.22 .

Exercise 9.15.

Proposition 9.15 , in the form Problem 9.6 puts it.

Exercise 9.16.

The equation $a * x = b$ has no second solution, as the remark after Proposition 9.22 observes.

Exercise 9.17.

Proposition 9.16 . Show first that $b^{-1} * a^{-1}$ multiplies with $a * b$ to give $e$ .

Exercise 9.18.

The rearrangement that carries a right coset onto a left one in Theorem 9.37 .

Exercise 9.19.

The first step of the backward direction of Proposition 9.26 .

Exercise 9.20.

Its second step.

Exercise 9.21.

And its third, which is where inv_inv earns its place.

Exercise 9.22.

The first half of Proposition 9.31 . A homomorphism is a map with the one property, so it arrives as an arrow together with the equation it satisfies.

Exercise 9.23.

Its second half, with the first half to hand.

Exercise 9.24.

The kernel is closed under the operation, part of Proposition 9.32 . The kernel arrives as a set together with its membership criterion.

Exercise 9.25.

Proposition 9.36 : the kernel is normal. (Harder.)

lean proof

1example (f : Obj → Obj) (K : Obj) (hf : ∀ a b : Obj, f (a ∗ b) = f a ∗ f b) (hfe : f e = e)2    (hinv : ∀ a : Obj, f a⁻¹ = (f a)⁻¹) (hK : ∀ x : Obj, x ∈ K ↔ f x = e)3    (a b : Obj) (ha : a ∈ K) : b⁻¹ ∗ a ∗ b ∈ K := by

4  sorry

goalf : Obj → Obj K : Obj hf : ∀ (a : Obj), ∀ (b : Obj), f (a ∗ b) = f a ∗ f b hfe : f e = e hinv : ∀ (a : Obj), f a⁻¹ = (f a)⁻¹ hK : ∀ (x : Obj), x ∈ K ↔ f x = e a b : Obj ha : a ∈ K ⊢ (b⁻¹ ∗ a) ∗ b ∈ K

What the checker understands

Tactics

intro h	assume the hypothesis of an implication, naming it h
exact e	give the proof outright
apply f	reduce the goal to the hypotheses of f
assumption	close the goal with a hypothesis already present
trivial	close the goal True
exfalso	replace the goal with False
by_contra h	assume the negation of the goal
constructor	split ∧ into both halves, or ↔ into both directions
left / right	choose which half of a ∨ to prove
rcases h with a \| b	argue by cases on a disjunction
obtain ⟨a, b⟩ := h	take a conjunction or an existential apart
cases h	as above, keeping the name
refine e	give the proof with holes left in it
have h : p := …	record an intermediate result
show p	restate the goal in an equal form
use w	give a witness for ∃
specialize h a	instantiate a ∀ hypothesis
rw [h]	rewrite with an equation, ← to go backwards
rfl	both sides compute to the same thing
decide / norm_num	settle a closed computation
tauto	close a goal that is true by pure logic
induction n with k ih	the fifth Peano condition: prove the goal at 0, then at succ k from ih

Results you may cite

Classical.em	∀ (a : Prop), a ∨ ¬a — the law of excluded middle
Classical.byContradiction	∀ {a : Prop}, (¬a → False) → a — proof by contradiction; the tactic by_contra does this for you
Classical.byCases	∀ {a b : Prop}, (a → b) → (¬a → b) → b — split on whether a holds
not_not	∀ {a : Prop}, ¬¬a ↔ a — double negation
not_and_or	∀ {a b : Prop}, ¬(a ∧ b) ↔ ¬a ∨ ¬b — De Morgan
not_or	∀ {a b : Prop}, ¬(a ∨ b) ↔ ¬a ∧ ¬b — De Morgan
not_imp	∀ {a b : Prop}, ¬(a → b) ↔ a ∧ ¬b
and_comm	∀ {a b : Prop}, a ∧ b ↔ b ∧ a
or_comm	∀ {a b : Prop}, a ∨ b ↔ b ∨ a
Set.ext	∀ {A B : Obj}, (∀ x : Obj, x ∈ A ↔ x ∈ B) → A = B — extensionality: sets with the same elements are equal
Set.ext_iff	∀ {A B : Obj}, A = B ↔ (∀ x : Obj, x ∈ A ↔ x ∈ B) — extensionality and substitution, in one biconditional
Set.subset_antisymm	∀ {A B : Obj}, A ⊆ B → B ⊆ A → A = B — mutual inclusion is equality
Set.empty_subset	∀ {A : Obj}, ∅ ⊆ A — the empty set is a subset of every set
Set.pair_eq	∀ {a b c d : Obj}, ((a, b) = (c, d)) ↔ (a = c ∧ b = d) — two ordered pairs are equal exactly when their coordinates are
Nat.succ_inj	∀ {m n : ℕ}, succ m = succ n → m = n — the third Peano condition: the successor is injective
Nat.succ_ne_zero	∀ (n : ℕ), succ n ≠ 0 — the fourth: zero is nobody's successor
Nat.pred	∀ {n : ℕ}, n ≠ 0 → ∃ m : ℕ, n = succ m — predecessors: everything but zero is a successor
Nat.add_zero	∀ (m : ℕ), m + 0 = m — the first clause of addition
Nat.add_succ	∀ (m n : ℕ), m + succ n = succ (m + n) — the second clause of addition
Nat.zero_add	∀ (n : ℕ), 0 + n = n — addition from the left
Nat.succ_add	∀ (m n : ℕ), succ m + n = succ (m + n) — addition from the left, at a successor
Nat.add_assoc	∀ (m n p : ℕ), (m + n) + p = m + (n + p) — addition is associative
Nat.add_comm	∀ (m n : ℕ), m + n = n + m — addition is commutative
Nat.add_ne_zero	∀ {a : ℕ} (b : ℕ), a ≠ 0 → a + b ≠ 0 — positivity is absorbing
Nat.mul_zero	∀ (m : ℕ), m * 0 = 0 — the first clause of multiplication
Nat.mul_succ	∀ (m n : ℕ), m * succ n = m * n + m — the second clause of multiplication
Nat.zero_mul	∀ (m : ℕ), 0 * m = 0 — multiplication from the left
Nat.succ_mul	∀ (m n : ℕ), succ m * n = m * n + n — multiplication from the left, at a successor
Nat.add_right_cancel	∀ {m n k : ℕ}, m + k = n + k → m = n — cancellation, from the last sheet
Nat.add_eq_zero	∀ {m n : ℕ}, m + n = 0 → m = 0 ∧ n = 0 — a sum is zero only when both parts are, from the last sheet
Nat.mul_comm	∀ (m n : ℕ), m * n = n * m — multiplication is commutative, from the last sheet
Nat.mul_add	∀ (m n p : ℕ), m * (n + p) = m * n + m * p — multiplication distributes over addition, from the last sheet
Nat.mul_assoc	∀ (m n p : ℕ), (m * n) * p = m * (n * p) — multiplication associates, from the problems of the last chapter
Nat.add_mul	∀ (m n p : ℕ), (m + n) * p = m * p + n * p — distributivity on the other side
Nat.add_left_cancel	∀ {a m n : ℕ}, a + m = a + n → m = n — uniqueness of differences
Nat.lt_trichotomy	∀ (m n : ℕ), m < n ∨ m = n ∨ n < m — trichotomy, from the theorem that ℕ is strictly ordered
Nat.lt_irrefl	∀ (n : ℕ), ¬(n < n) — anti-reflexivity, from the last sheet
Nat.lt_trans	∀ {m n p : ℕ}, m < n → n < p → m < p — transitivity of the strict order, from the last sheet
Nat.lt_succ_self	∀ (n : ℕ), n < succ n — every number is below its successor, from the last sheet
Nat.not_lt_zero	∀ {n : ℕ}, ¬(n < 0) — nothing lies below zero
Nat.lt_succ_iff	∀ {m n : ℕ}, m < succ n ↔ m < n ∨ m = n — nothing lies strictly between n and succ n
Num.inj	∀ {m n : ℕ}, ↑m = ↑n → m = n — distinct numbers name distinct objects of ω
swap_apply_left	∀ (a b : Obj), swap a b a = b — the transposition sends a to b
swap_apply_right	∀ (a b : Obj), swap a b b = a — and b to a
swap_apply_of_ne_of_ne	∀ {a b x : Obj}, x ≠ a → x ≠ b → swap a b x = x — and fixes every other point
swap_swap	∀ (a b x : Obj), swap a b (swap a b x) = x — a transposition is its own inverse
Function.iterate_zero_apply	∀ (f : Obj → Obj) (x : Obj), f^[0] x = x — the first clause of the powers of a map
Function.iterate_succ_apply	∀ (f : Obj → Obj) (n : ℕ) (x : Obj), f^[succ n] x = f^[n] (f x) — the second clause: f^[succ n] is f^[n] ∘ f
Function.iterate_add_apply	∀ (f : Obj → Obj) (m n : ℕ) (x : Obj), f^[m + n] x = f^[m] (f^[n] x) — the first law of exponents
Nat.factorial_zero	0 ! = succ 0 — the first clause of the factorial
Nat.factorial_succ	∀ (n : ℕ), (succ n) ! = succ n * n ! — the second clause of the factorial
Nat.choose_zero_right	∀ (n : ℕ), choose n 0 = succ 0 — the empty set is the one 0-subset
Nat.choose_eq_zero_of_lt	∀ {n k : ℕ}, n < k → choose n k = 0 — no subset is larger than the whole
Nat.choose_succ_succ	∀ (n k : ℕ), choose (succ n) (succ k) = choose n k + choose n (succ k) — Pascal's identity
mul_assoc	∀ (a b c : Obj), (a ∗ b) ∗ c = a ∗ (b ∗ c) — the operation associates
e_mul	∀ (a : Obj), e ∗ a = a — the identity on the left
mul_e	∀ (a : Obj), a ∗ e = a — and on the right
inv_mul_cancel	∀ (a : Obj), a⁻¹ ∗ a = e — the inverse on the left
mul_inv_cancel	∀ (a : Obj), a ∗ a⁻¹ = e — and on the right
inv_inv	∀ (a : Obj), (a⁻¹)⁻¹ = a — worked above: inverting twice gives the element back
mul_left_cancel	∀ {a x y : Obj}, a ∗ x = a ∗ y → x = y — worked above: the left half of prop-9-5

From the logical core

And.intro	∀ {a b : Prop}, a → b → a ∧ b
And.left	∀ {a b : Prop}, a ∧ b → a
And.right	∀ {a b : Prop}, a ∧ b → b
And.symm	∀ {a b : Prop}, a ∧ b → b ∧ a
Or.inl	∀ {a b : Prop}, a → a ∨ b
Or.inr	∀ {a b : Prop}, b → a ∨ b
Or.elim	∀ {a b c : Prop}, a ∨ b → (a → c) → (b → c) → c
Or.symm	∀ {a b : Prop}, a ∨ b → b ∨ a
Iff.intro	∀ {a b : Prop}, (a → b) → (b → a) → (a ↔ b)
Iff.mp	∀ {a b : Prop}, (a ↔ b) → a → b
Iff.mpr	∀ {a b : Prop}, (a ↔ b) → b → a
Iff.symm	∀ {a b : Prop}, (a ↔ b) → (b ↔ a)
Iff.rfl	∀ {a : Prop}, a ↔ a
Iff.trans	∀ {a b c : Prop}, (a ↔ b) → (b ↔ c) → (a ↔ c)
True.intro	True
False.elim	∀ {a : Prop}, False → a
absurd	∀ {a b : Prop}, a → ¬a → b
id	∀ {a : Prop}, a → a
mt	∀ {a b : Prop}, (a → b) → ¬b → ¬a
Eq.refl	∀ {α : Type} (a : α), a = a
Eq.symm	∀ {α : Type} {a b : α}, a = b → b = a
Eq.trans	∀ {α : Type} {a b c : α}, a = b → b = c → a = c
congrArg	∀ {α : Type} {β : Type} {a b : α} (f : α → β), a = b → f a = f b
Exists.intro	∀ {α : Type} {p : α → Prop} (w : α), p w → ∃ x : α, p x
Exists.elim	∀ {α : Type} {p : α → Prop} {b : Prop}, (∃ x : α, p x) → (∀ y : α, p y → b) → b

Lesson 10

Integers

Taught 22 Aug 2026

Formal Differences

At the end of the last chapter $(\mathbb{N}_0, +)$ was a monoid in which cancellation holds and in which nothing but $0$ has an inverse. It is regular and it is not a group, and this chapter closes the gap between those two facts. We do not add an axiom and we do not assume the integers exist. We build them, out of the natural numbers we already have, by writing down what a difference ought to be and then declaring two of them equal when they ought to be equal.

Pairs and the Sum Criterion

Every integer we want is a difference of two natural numbers. We want $3$ to be $5 - 2$ and $-3$ to be $5 - 8$ , and we want $3$ to be $6 - 3$ and $7 - 4$ as well: the pair recording the difference is not unique, and any construction must say when two pairs record the same thing.

The condition we want is $m - n = m' - n'$ , and as it stands that is not a statement about $\mathbb{N}_0$ , since subtraction is available only when the answer stays in $\mathbb{N}_0$ . But adding $n + n'$ to both sides of it turns it into

m + n' = n + m',

which mentions nothing but addition and is a perfectly good statement about four natural numbers. That equation is the relation we put on pairs, and the rest of the chapter is built on it.

Definition 10.1 (Net-difference equivalence).

On the set $\mathbb{N}_0 \times \mathbb{N}_0$ declare

(m, n) \sim (m', n') \qquad \text{if and only if} \qquad m + n' = n + m'.

The relation $\sim$ is called net-difference equivalence, and the displayed equation is the sum criterion.

Read the pair $(m, n)$ as a formal stand-in for the difference $m - n$ , one that makes sense whether or not that difference exists in $\mathbb{N}_0$ . Two pairs are related exactly when the differences they stand for ought to agree. The symbol $\sim$ is used for equivalence relations throughout mathematics and throughout this book; in this chapter it means only the relation just defined.

Proposition 10.2 (Net-difference equivalence is an equivalence relation).

The relation $\sim$ is an equivalence relation on $\mathbb{N}_0 \times \mathbb{N}_0$ .

Discussion.

Three properties to check, and each unfolds the sum criterion into a statement about addition in $\mathbb{N}_0$ . Reflexivity asks for $m + n = n + m$ , which is commutativity. Symmetry asks that $m + n' = n + m'$ give $m' + n = n' + m$ , which is the same equation read backwards with the two sides commuted. Transitivity is the only part that needs work: two sum criteria are given and a third must be produced, and the obstacle is that the middle pair $(m', n')$ appears in both hypotheses and in neither conclusion. The way to remove it is to add $n''$ to the first equation so that the second becomes substitutable, and then to strip the surviving $n'$ off both sides by cancellation, the property that made $(\mathbb{N}_0, +)$ regular.

Proof.

For reflexivity, $m + n = n + m$ by the laws of addition, so $(m, n) \sim (m, n)$ .

For symmetry, suppose $(m, n) \sim (m', n')$ , so $m + n' = n + m'$ . Commuting each side gives $m' + n = n' + m$ , which is the sum criterion for $(m', n') \sim (m, n)$ .

For transitivity, suppose $(m, n) \sim (m', n')$ and $(m', n') \sim (m'', n'')$ , so

m + n' = n + m' \qquad \text{and} \qquad m' + n'' = n' + m''.

Adding $n''$ to the first equation and rearranging by associativity and commutativity gives $m + n' + n'' = n + m' + n''$ . The second equation replaces $m' + n''$ on the right by $n' + m''$ , so

m + n' + n'' = n + n' + m''.

Cancelling $n'$ leaves $m + n'' = n + m''$ , which is $(m, n) \sim (m'', n'')$ .

The Net Difference

The sum criterion is easy to check but does not describe what a class looks like. A second description of the same relation gives each pair a standard representative: compare the two coordinates using trichotomy and subtract the smaller from the larger.

Definition 10.3 (Net difference).

The net-difference function

\Delta : \mathbb{N}_0 \times \mathbb{N}_0 \to \mathbb{N}_0 \times \mathbb{N}_0

is defined by

\Delta(m, n) \defeq \begin{cases} (m - n, \, 0) & \text{if } n < m, \\ (0, \, n - m) & \text{if } m < n, \\ (0, \, 0) & \text{if } m = n, \end{cases}

where $m - n$ is the difference of the order chapter: the unique element with $n + (m - n) = m$ , which exists because $n \leqslant m$ .

By trichotomy the three clauses cover every pair exactly once, so $\Delta$ is a function. Informally, $(x, 0)$ with $x \in \mathbb{N}$ is a positive difference, $(0, x)$ a negative one, and $(0,0)$ is zero; the next few propositions make that reading precise and tie $\Delta$ back to $\sim$ .

Proposition 10.4 (Coordinates of a net difference).

Let $m, n \in \mathbb{N}_0$ . Then at least one coordinate of $\Delta(m, n)$ is $0$ , and

if $\Delta(m, n) = (x, 0)$ with $x \in \mathbb{N}$ , then $n < m$ ;
if $\Delta(m, n) = (0, x)$ with $x \in \mathbb{N}$ , then $m < n$ ;
if $\Delta(m, n) = (0, 0)$ , then $m = n$ .

Discussion.

The opening claim is read straight off the definition, every clause of which puts a $0$ in one coordinate. The three numbered parts run the definition backwards, which is legitimate because the three cases are mutually exclusive: given the value of $\Delta(m, n)$ , we ask which clause could have produced it and find that only one can. For the first two the argument is that the other two clauses put $0$ in the coordinate that is here positive. The third is the leftover case and is best argued by contradiction: if $m \neq n$ then trichotomy puts one of them strictly below the other, and either way one coordinate of $\Delta(m, n)$ is a positive natural number, so the value is not $(0,0)$ .

Proof.

Trichotomy gives exactly one of $n < m$ , $m < n$ , $m = n$ , and each of the three clauses of the definition places $0$ in at least one coordinate.

Suppose $\Delta(m, n) = (x, 0)$ with $x \in \mathbb{N}$ . The second and third clauses put $0$ in the first coordinate, so neither produced this value, and the first clause applies; hence $n < m$ . The same argument with the coordinates exchanged gives the second part.

For the third, suppose $m \neq n$ . Trichotomy gives $n < m$ or $m < n$ . In the first case $\Delta(m, n) = (m - n, 0)$ and $m - n$ is positive, since $m - n = 0$ would give $m = n$ ; in the second case the second coordinate is positive for the same reason. Either way $\Delta(m, n) \neq (0, 0)$ .

Proposition 10.5 (Shifting both coordinates).

For all $m, n, x \in \mathbb{N}_0$ ,

\Delta(m, n) = \Delta(m + x, \, n + x).

Discussion.

An equality of values of $\Delta$ , so we split on the trichotomy comparison of $m$ with $n$ and check the three cases separately. Shifting does not change the comparison: adding a fixed element preserves the order, so if $n < m$ then $n + x < m + x$ and both sides of the claimed equality are produced by the same clause. Once that is known the two sides are $(m - n, 0)$ and $\bigl((m+x) - (n+x), 0\bigr)$ , and what remains is that shifting both arguments leaves the difference alone, which is uniqueness of differences applied to the defining equation. The case $m < n$ is the same with the coordinates exchanged, and $m = n$ is immediate.

Proof.

Suppose $n < m$ . Then $n + x < m + x$ , so

\Delta(m, n) = (m - n, \, 0), \qquad \Delta(m + x, n + x) = \bigl((m + x) - (n + x), \, 0\bigr).

Write $d \defeq m - n$ , so $n + d = m$ and hence $(n + x) + d = m + x$ by associativity and commutativity. Uniqueness of differences therefore gives $(m + x) - (n + x) = d$ , and the two values agree.

If $m < n$ the same argument applies with the coordinates exchanged. If $m = n$ then $m + x = n + x$ , and both sides are $(0, 0)$ .

Proposition 10.6 (Equal net differences satisfy the sum criterion).

If $\Delta(m, n) = \Delta(m', n')$ , then $m + n' = n + m'$ .

Discussion.

The hypothesis is an equality of pairs and the conclusion an equality of natural numbers, and we must convert one into the other, and the conversion depends on which of the three shapes the common value has. We therefore split on the comparison of $m$ with $n$ , and in each case the previous proposition on coordinates transfers the same comparison to $m'$ and $n'$ . Once both pairs are known to be of the same shape with the same entry $x$ , the defining property of the difference rewrites the hypothesis as $m = n + x$ and $m' = n' + x$ , two equations with no subtraction in them, and substituting both into $m + n'$ produces $n + m'$ after rearrangement.

Proof.

Suppose first that $n < m$ . Then $\Delta(m, n) = (x, 0)$ with $x \defeq m - n \in \mathbb{N}$ , so $\Delta(m', n') = (x, 0)$ as well; by the proposition on coordinates $n' < m'$ , and $m' - n' = x$ . Hence $m = n + x$ and $m' = n' + x$ , and

m + n' = (n + x) + n' = n + (x + n') = n + (n' + x) = n + m',

by associativity and commutativity.

The case $m < n$ is the same argument with the coordinates exchanged. If $m = n$ then $\Delta(m, n) = (0,0)$ , so $\Delta(m', n') = (0,0)$ and $m' = n'$ ; then $m + n' = n + m'$ reads $m + m' = m + m'$ .

Proposition 10.7 (The sum criterion yields a common shift).

Let $(m, n)$ and $(m', n')$ satisfy $m + n' = n + m'$ with $n \leqslant n'$ . Then there is $x \in \mathbb{N}_0$ with

n' = n + x \qquad \text{and} \qquad m' = m + x.

Discussion.

An existence claim, and the hypothesis $n \leqslant n'$ gives the witness directly: the description of the associated order says that $n \leqslant n'$ is exactly the existence of an $x$ with $n' = n + x$ . It remains to check that the same $x$ works in the other coordinate. Substituting $n + x$ for $n'$ in the sum criterion and rearranging leaves $n$ in front on both sides, and cancelling it gives $m' = m + x$ .

Proof.

Since $n \leqslant n'$ , there is $x \in \mathbb{N}_0$ with $n' = n + x$ . Then

n + m' = m + n' = m + (n + x) = n + (m + x),

the first equality being the sum criterion and the last associativity and commutativity. Cancelling $n$ gives $m' = m + x$ .

Three readings of “the same formal difference” are now available, and they agree.

Theorem 10.8 (Characterisations of net-difference equivalence).

Let $(m, n)$ and $(m', n')$ lie in $\mathbb{N}_0 \times \mathbb{N}_0$ . The following are equivalent.

$\Delta(m, n) = \Delta(m', n')$ .
$m + n' = n + m'$ .
Either $n' = n + x$ and $m' = m + x$ for some $x \in \mathbb{N}_0$ , or $n = n' + x$ and $m = m' + x$ for some $x \in \mathbb{N}_0$ .

Discussion.

Three conditions asserted to be equivalent, so rather than six implications we run a cycle through them, in the manner already used for equivalence classes. The first arrow, from the net difference to the sum criterion, is the proposition just proved. The second, from the sum criterion to the shift, is the other proposition together with a case split: the shift condition offers two alternatives because the previous proposition needed $n \leqslant n'$ , and trichotomy guarantees that one of $n \leqslant n'$ and $n' \leqslant n$ holds, each giving one alternative once the roles of the two pairs are exchanged. The third arrow, from the shift back to the net difference, is the proposition on shifting both coordinates, read in one direction for the first alternative and in the other for the second.

Proof.

That the first implies the second is the proposition on equal net differences.

Suppose the second holds. If $n \leqslant n'$ , the proposition on common shifts gives $x$ with $n' = n + x$ and $m' = m + x$ , which is the first alternative of the third condition. Otherwise $n' \leqslant n$ by trichotomy, and the same proposition applied to the pairs in the other order — the sum criterion $m' + n = n' + m$ being the given equation read backwards — gives $x$ with $n = n' + x$ and $m = m' + x$ , which is the second alternative.

Suppose the third holds. Under the first alternative, the proposition on shifting both coordinates gives

\Delta(m, n) = \Delta(m + x, \, n + x) = \Delta(m', n').

Under the second it gives $\Delta(m', n') = \Delta(m' + x, \, n' + x) = \Delta(m, n)$ . Either way the first condition holds, and the cycle is closed.

Remark (Which characterisation to use).

Each of the three is the convenient one somewhere below. The sum criterion is easiest to use in algebra, since it is an equation in $\mathbb{N}_0$ and nothing else, and every well-definedness check in this chapter uses it. The shift criterion says that two pairs agree when one is the other with the same amount added to both coordinates. The net difference gives each class a single named representative, and that is what lets us say what the integers are rather than only when two of them are equal.

Problem 10.1.

Compute $\Delta(19, 15)$ and $\Delta(14, 16)$ .

Problem 10.2.

Show that $m - 0 = m$ for every $m \in \mathbb{N}_0$ , from the defining property of the difference. Deduce that $\Delta(m, 0) = (m, 0)$ and $\Delta(0, n) = (0, n)$ for all $m, n \in \mathbb{N}_0$ , and hence that $\Delta \circ \Delta = \Delta$ .

Problem 10.3.

Prove the proposition that $\sim$ is an equivalence relation a second time, taking the first condition of the theorem as the definition of $\sim$ and checking the three properties directly from properties of $\Delta$ .

The Integers

Integers as Classes

Definition 10.9 (The integers).

The set of integers is the quotient

\mathbb{Z} \defeq \bigl(\mathbb{N}_0 \times \mathbb{N}_0\bigr)\big/{\sim}

of $\mathbb{N}_0 \times \mathbb{N}_0$ by net-difference equivalence. We write $[(m, n)]$ for the class of the pair $(m, n)$ .

No axiom has been added. The set is carved out of a product of two copies of $\mathbb{N}_0$ by a relation already checked to be an equivalence, and both of those operations have been available since the chapters on sets and on relations. What remains is to define the arithmetic.

By the theorem and the proposition on coordinates, every class has exactly one representative among the pairs

(x, 0) \text{ with } x \in \mathbb{N}_0, \qquad (0, x) \text{ with } x \in \mathbb{N},

these being the values $\Delta$ takes, and $(0,0)$ appearing once in the first list. We call it the normal form of the class. In particular $[(m,n)] = [(m - n, 0)]$ when $n \leqslant m$ , and $[(m, n)] = [(0, n - m)]$ when $m \leqslant n$ .

Proposition 10.10 (Embedding the natural numbers).

The map

\iota : \mathbb{N}_0 \to \mathbb{Z}, \qquad \iota(n) \defeq [(n, 0)]

is injective.

Discussion.

Injectivity is the statement that equal values force equal arguments, so we assume $\iota(n) = \iota(n')$ and must reach $n = n'$ . Equal classes mean related representatives, by the characterisation of equivalence classes, and the sum criterion for those two particular pairs is $n + 0 = 0 + n'$ , which is the conclusion up to two zeros.

Proof.

Suppose $[(n, 0)] = [(n', 0)]$ . Then $(n, 0) \sim (n', 0)$ , so $n + 0 = 0 + n'$ by the sum criterion, and $n = n'$ by the identity law for addition.

So $\iota$ embeds $\mathbb{N}_0$ in $\mathbb{Z}$ , and the normal form says every class is either $\iota(x)$ for a unique $x \in \mathbb{N}_0$ , or $[(0, x)]$ for a unique $x \in \mathbb{N}$ . Once addition is in place the second family will turn out to be the additive inverses of the first.

Problem 10.4.

List five distinct pairs in the class $[(3, 5)]$ , and give the normal form of that class.

Problem 10.5.

Show that $[(m, n)] = [(0, 0)]$ if and only if $m = n$ .

Addition and Negation

The arithmetic must be defined on classes, but the only formulae available are written with pairs. We follow the same pattern three times: define the operation on pairs, prove that equivalent inputs give equivalent outputs, and only then pass to classes. The middle step ensures that the result does not depend on which representatives were picked.

Pretend for a moment that differences already exist. Then $-(m - n) = n - m$ and

(m - n) + (m' - n') = (m + m') - (n + n'),

and those two identities suggest the definitions on pairs. Nothing below uses this; it only motivates the definitions.

Definition 10.11 (Pre-negation).

The pre-negation of a pair $(m, n) \in \mathbb{N}_0 \times \mathbb{N}_0$ is the pair $(n, m)$ .

Proposition 10.12 (Pre-negation respects equivalence).

If $(m, n) \sim (m', n')$ , then $(n, m) \sim (n', m')$ .

Discussion.

Hypothesis and conclusion are both sum criteria, so both unfold into equations in $\mathbb{N}_0$ . The hypothesis is $m + n' = n + m'$ . The conclusion, written out for the pairs $(n, m)$ and $(n', m')$ , is $n + m' = m + n'$ . Those are the same equation with the two sides exchanged, so symmetry of equality is the whole proof.

Proof.

The hypothesis is $m + n' = n + m'$ . Reading it backwards gives $n + m' = m + n'$ , which is the sum criterion for $(n, m) \sim (n', m')$ .

Definition 10.13 (Negation).

Define negation on $\mathbb{Z}$ by

-[(m, n)] \defeq [(n, m)].

By the proposition just proved, the right-hand side depends only on the class of $(m, n)$ and not on the representative chosen for it.

Definition 10.14 (Pre-addition).

Define pre-addition on pairs by

(m, n) + (m', n') \defeq (m + m', \; n + n').

Proposition 10.15 (Pre-addition respects equivalence).

If $(m, n) \sim (a, b)$ and $(m', n') \sim (a', b')$ , then

(m + m', \; n + n') \sim (a + a', \; b + b').

Discussion.

Two sum criteria are given and one is wanted, and since all three are equations between sums in $\mathbb{N}_0$ we add the two hypotheses together. What comes out is an equation whose two sides are the four summands in some order, and the general associativity and rearrangement available in any abelian semigroup lets us collect them into the grouping the conclusion asks for. No case split and no cancellation are needed.

Proof.

The hypotheses are $m + b = n + a$ and $m' + b' = n' + a'$ . Adding the two equations and rearranging the summands on each side gives

(m + m') + (b + b') = (n + n') + (a + a'),

which is exactly the sum criterion for $(m + m', n + n') \sim (a + a', b + b')$ .

Definition 10.16 (Addition and subtraction).

Define addition on $\mathbb{Z}$ by

[(m, n)] + [(m', n')] \defeq [(m + m', \; n + n')],

which the previous proposition makes independent of the representatives; and for $x, y \in \mathbb{Z}$ define $x - y \defeq x + (-y)$ .

Theorem 10.17 (The integers form an abelian group).

With the addition just defined, $(\mathbb{Z}, +)$ is an abelian group. Its identity is $[(0, 0)]$ and the inverse of $[(m, n)]$ is $[(n, m)]$ . Moreover

\iota(m + n) = \iota(m) + \iota(n) \qquad \text{for all } m, n \in \mathbb{N}_0,

and $\iota(0) = [(0, 0)]$ .

Discussion.

The definition of a group asks for four things and the last claim adds a fifth, but every one of them is a law in $\mathbb{N}_0$ applied in each coordinate, so the proof is a sequence of short computations. Associativity and commutativity of the new addition are associativity and commutativity of the old one, used once in each coordinate. The class $[(0,0)]$ is an identity because $0$ is one in $\mathbb{N}_0$ . The only step that needs thought is the inverse: adding $[(m,n)]$ to $[(n,m)]$ gives the class of $(m + n, \, n + m)$ , and this is $[(0,0)]$ not because the two coordinates are zero (they are not) but by the sum criterion, which asks only that $(m+n) + 0 = (n+m) + 0$ . That is commutativity. The claim about $\iota$ is the definition of addition read on pairs whose second coordinate is $0$ .

Proof.

Let $x = [(m, n)]$ , $y = [(m', n')]$ and $z = [(m'', n'')]$ . Then

(x + y) + z = \bigl[\bigl((m + m') + m'', \; (n + n') + n''\bigr)\bigr] = x + (y + z)

by associativity of addition in $\mathbb{N}_0$ in each coordinate, so the operation is associative; and $x + y = y + x$ by commutativity in each coordinate. So $(\mathbb{Z}, +)$ is an abelian semigroup.

Next,

x + [(0, 0)] = [(m + 0, \; n + 0)] = [(m, n)] = x,

so $[(0,0)]$ is an identity, and it is the only one by uniqueness of the identity. Finally

x + [(n, m)] = [(m + n, \; n + m)] = [(0, 0)],

the last equality by the sum criterion, which asks for $(m + n) + 0 = (n + m) + 0$ and gets it from commutativity. So every element is invertible and $(\mathbb{Z}, +)$ is an abelian group.

For the last claim, $\iota(m) + \iota(n) = [(m, 0)] + [(n, 0)] = [(m + n, \, 0)] = \iota(m + n)$ , and $\iota(0) = [(0,0)]$ by definition.

In particular $-\iota(n) = -[(n, 0)] = [(0, n)]$ , so the classes with normal form $(0, n)$ are precisely the additive inverses of the embedded positive naturals. Every integer is therefore $\iota(x)$ for a unique $x \in \mathbb{N}_0$ or $-\iota(x)$ for a unique $x \in \mathbb{N}$ , which is the normal form restated using addition. Writing $n$ for $\iota(n)$ and $-n$ for $-\iota(n)$ , the familiar list $\ldots, -2, -1, 0, 1, 2, \ldots$ names every integer exactly once.

Remark (What was gained).

The chapter on groups observed that $(\mathbb{N}_0, +)$ is regular but not a group, and said that regularity is the condition needed to enlarge a monoid into a group. This theorem does that for $(\mathbb{N}_0, +)$ . The regularity was used once, in the proof that $\sim$ is transitive, and without it the relation would not be an equivalence and there would be no quotient.

Problem 10.6.

Compute $[(5, 2)] + [(1, 4)]$ and $-[(5, 2)]$ , and give each answer in normal form.

Problem 10.7.

Prove directly from the definitions that $-(x + y) = (-x) + (-y)$ and $-(-x) = x$ for all $x, y \in \mathbb{Z}$ . Which of the two also follows from a proposition of the chapter on groups, and why does the other not?

Problem 10.8.

Show that $\iota(m) - \iota(n) = [(m, n)]$ for all $m, n \in \mathbb{N}_0$ , so that every class really is a difference of two embedded natural numbers.

Multiplication

Pretend again that differences exist. Then

(m - n)(m' - n') = (mm' + nn') - (mn' + nm'),

and the rule on pairs is forced.

Definition 10.18 (Pre-multiplication).

Define pre-multiplication on pairs by

(m, n) \cdot (m', n') \defeq (mm' + nn', \; mn' + nm').

Proposition 10.19 (Pre-multiplication respects equivalence).

Let $(m, n) \sim (a, b)$ . Then for every pair $(m', n')$ ,

(m, n) \cdot (m', n') \sim (a, b) \cdot (m', n').

Likewise, if $(m', n') \sim (a', b')$ then $(m, n) \cdot (m', n') \sim (m, n) \cdot (a', b')$ for every $(m, n)$ .

Discussion.

Two statements, one for each factor, and together they are what a definition on classes needs: changing the representatives one at a time changes them both. Each is again an implication between sum criteria, but unlike the additive case the hypothesis cannot simply be added to something, because the conclusion involves products. So we write down the sum criterion the conclusion asks for, expand both of its sides using distributivity in $\mathbb{N}_0$ , and regroup the terms until the hypothesis becomes visible. The left side collects into $(m + b)m' + (n + a)n'$ and the right into $(m + b)n' + (n + a)m'$ ; the hypothesis says the two bracketed factors are equal, so the two sides are the same sum written twice. The second statement is the first with the roles of the factors exchanged, which is legitimate because pre-multiplication is symmetric in its two arguments.

Proof.

Assume $m + b = n + a$ . The two pre-products are

(mm' + nn', \; mn' + nm') \qquad \text{and} \qquad (am' + bn', \; an' + bm'),

and the sum criterion for their equivalence asks that

(mm' + nn') + (an' + bm') = (mn' + nm') + (am' + bn').

Regrouping the left side by distributivity, associativity and commutativity gives $(m + b)m' + (n + a)n'$ , and the right side gives $(m + b)n' + (n + a)m'$ . Writing $k$ for the common value of $m + b$ and $n + a$ , both sides are $km' + kn'$ , so they agree.

For the second statement, pre-multiplication is unchanged when its two arguments are exchanged, since $mm' + nn'$ and $mn' + nm'$ are symmetric under exchanging $(m,n)$ with $(m',n')$ ; so the first statement applied to the exchanged pairs gives it.

Definition 10.20 (Multiplication).

Define multiplication on $\mathbb{Z}$ by

[(m, n)] \cdot [(m', n')] \defeq [(mm' + nn', \; mn' + nm')],

which the previous proposition makes independent of both choices of representative.

Theorem 10.21 (Distributivity in the integers).

For all $x, y, z \in \mathbb{Z}$ ,

x(y + z) = xy + xz.

Discussion.

An identity between two integers, so we name representatives for the three of them and compute both sides down to a single class each. The left side expands by the definition of addition and then of multiplication, the right side by multiplication twice and then addition, and each expansion is an application of distributivity in $\mathbb{N}_0$ inside a coordinate. The two resulting pairs then have the same four terms in each coordinate, in a different order, so commutativity of addition finishes. We write the computation with the pairs visible, since the identity holds coordinatewise.

Proof.

Let $x = [(a, b)]$ , $y = [(c, d)]$ and $z = [(e, f)]$ . Then $y + z = [(c + e, \; d + f)]$ , so

\begin{aligned} x(y + z) &= \bigl[\bigl(a(c + e) + b(d + f), \; a(d + f) + b(c + e)\bigr)\bigr] \\ &= \bigl[\bigl(ac + ae + bd + bf, \; ad + af + bc + be\bigr)\bigr], \end{aligned}

while

\begin{aligned} xy + xz &= [(ac + bd, \; ad + bc)] + [(ae + bf, \; af + be)] \\ &= \bigl[\bigl(ac + bd + ae + bf, \; ad + bc + af + be\bigr)\bigr]. \end{aligned}

The two pairs have the same terms in each coordinate, so they are equal by commutativity of addition in $\mathbb{N}_0$ , and the two classes agree.

Proposition 10.22 (The embedding preserves multiplication).

For all $m, n \in \mathbb{N}_0$ we have $\iota(mn) = \iota(m) \cdot \iota(n)$ . Moreover $[(1, 0)]$ is an identity for multiplication on all of $\mathbb{Z}$ .

Discussion.

Two computations, each one line. For the first, expand $\iota(m) \cdot \iota(n)$ by the definition: both second coordinates are $0$ , so three of the four products in the formula vanish and what is left is $(mn, 0)$ , which is $\iota(mn)$ . The second claim is not the case $m = 1$ of the first, since that only says $\iota(1)$ acts as an identity on the image of $\iota$ , and the image is not all of $\mathbb{Z}$ ; so it needs its own computation, which is the same expansion carried out against a general class. Only one side need be checked, because pre-multiplication is symmetric in its arguments.

Proof.

For the first claim,

\begin{aligned} \iota(m) \cdot \iota(n) &= [(m, 0)] \cdot [(n, 0)] \\ &= [(mn + 0 \cdot 0, \; m \cdot 0 + 0 \cdot n)] = [(mn, 0)] = \iota(mn). \end{aligned}

For the second, let $[(a, b)]$ be any class. Then

[(1, 0)] \cdot [(a, b)] = [(1 \cdot a + 0 \cdot b, \; 1 \cdot b + 0 \cdot a)] = [(a, b)],

and the other side is the same computation, pre-multiplication being symmetric in its two arguments.

Remark (What is left to check).

The remaining algebra (associativity and commutativity of multiplication, the sign rules $(-x)y = x(-y) = -(xy)$ and $(-x)(-y) = xy$ , and the absence of zero divisors) follows the pattern of this section without exception: name representatives, expand, and quote the corresponding law in $\mathbb{N}_0$ . None of it needs a new idea, so it is set as problems. What is still missing is the order on $\mathbb{Z}$ .

Problem 10.9.

Prove that multiplication on $\mathbb{Z}$ is commutative and associative.

Problem 10.10.

Prove that $(-x)y = x(-y) = -(xy)$ and $(-x)(-y) = xy$ for all $x, y \in \mathbb{Z}$ .

Problem 10.11.

Prove that if $x, y \in \mathbb{Z}$ and $xy = [(0, 0)]$ , then $x = [(0,0)]$ or $y = [(0,0)]$ . Reduce to normal form first, and then use that a product of positive natural numbers is positive.

Problem 10.12.

Using the problem that $\iota(m) - \iota(n) = [(m, n)]$ , show that $[(m, n)] = [(m', n')]$ if and only if $\iota(m) - \iota(n) = \iota(m') - \iota(n')$ in $\mathbb{Z}$ .

The Order on the Integers

So far nothing distinguishes $[(5,2)]$ from $[(2,5)]$ except the labels; the arithmetic treats them alike. The order does distinguish them, and we define it from the normal form, which sorts every class into one of three shapes.

Definition 10.23 (Positive integers).

An integer $x$ is positive if $x = \iota(n)$ for some $n \in \mathbb{N}$ .

Proposition 10.24 (Every integer has exactly one sign).

Let $x \in \mathbb{Z}$ . Then exactly one of the following holds: $x$ is positive; $x = \iota(0)$ ; $x = -y$ for a positive $y$ .

Discussion.

The claim is that three cases cover $\mathbb{Z}$ and no two of them overlap, so there are two halves to it. Coverage is the normal form: every class is $[(a, 0)]$ with $a \in \mathbb{N}_0$ or $[(0, a)]$ with $a \in \mathbb{N}$ , and the first of those is $\iota(a)$ , positive when $a \in \mathbb{N}$ and $\iota(0)$ when $a = 0$ , while the second is $-\iota(a)$ by the remark following the group theorem. Exclusivity is the uniqueness half of the normal form, together with the injectivity of $\iota$ : two of the cases could only collide if two distinct normal forms named the same class, or if $\iota$ identified $0$ with something in $\mathbb{N}$ .

Proof.

Every class has exactly one normal form. If it is $(a, 0)$ with $a \in \mathbb{N}$ , then $x = \iota(a)$ is positive. If it is $(0, 0)$ , then $x = \iota(0)$ . If it is $(0, a)$ with $a \in \mathbb{N}$ , then $x = -\iota(a)$ with $\iota(a)$ positive. So the three cases cover $\mathbb{Z}$ .

For exclusivity, suppose $x$ is positive and also $x = \iota(0)$ . Then $\iota(n) = \iota(0)$ for some $n \in \mathbb{N}$ , so $n = 0$ by injectivity of $\iota$ , contradicting $0 \notin \mathbb{N}$ . Suppose $x$ is positive and also $x = -\iota(n')$ with $n' \in \mathbb{N}$ . Then $x$ has normal forms $(n, 0)$ and $(0, n')$ with $n, n' \in \mathbb{N}$ , and those are different pairs, contradicting uniqueness. The same argument rules out $\iota(0) = -\iota(n')$ .

Proposition 10.25 (Positive integers are closed under the operations).

If $x$ and $y$ are positive, then so are $x + y$ and $xy$ .

Discussion.

Both claims say that a value built from two positive integers is again of the form $\iota(n)$ with $n \in \mathbb{N}$ , and the two facts that produce such a form are already proved: $\iota$ carries sums to sums and carries products to products. So each claim reduces immediately to the corresponding closure in $\mathbb{N}$ , which is the proposition that the positive natural numbers are closed under addition and multiplication. Nothing about $\mathbb{Z}$ is used beyond the two formulas for $\iota$ .

Proof.

Write $x = \iota(m)$ and $y = \iota(n)$ with $m, n \in \mathbb{N}$ . Then $x + y = \iota(m + n)$ and $xy = \iota(mn)$ , and both $m + n$ and $mn$ lie in $\mathbb{N}$ . So both values are positive.

Definition 10.26 (The order on the integers).

For $x, y \in \mathbb{Z}$ write

x < y \qquad \text{if and only if} \qquad y - x \text{ is positive},

and write $x \leqslant y$ for ” $x < y$ or $x = y$ ”, as in the associated order of the chapter on relations.

Theorem 10.27 (The integers are strictly ordered).

The relation $<$ is a strict linear order on $\mathbb{Z}$ .

Discussion.

Two conditions are asked for. Transitivity is a computation: from $y - x$ and $z - y$ positive we must produce $z - x$ positive, and the way to reach $z - x$ from those two is to add them, since the middle terms cancel in the group $(\mathbb{Z}, +)$ ; the previous proposition then keeps the sum positive. Trichotomy is the proposition on signs, applied not to $x$ or $y$ but to the single integer $y - x$ : its three cases say that $y - x$ is positive, or zero, or the negative of a positive, and those are exactly $x < y$ , $x = y$ and $y < x$ since $-(y - x) = x - y$ . That last identity is the rule for the inverse of a product read additively, together with $-(-x) = x$ .

Proof.

For transitivity, suppose $x < y$ and $y < z$ , so $y - x$ and $z - y$ are positive. Their sum is positive by the previous proposition, and

(z - y) + (y - x) = z + \bigl((-y) + y\bigr) + (-x) = z - x

by associativity and the inverse law in $(\mathbb{Z}, +)$ . So $x < z$ .

For trichotomy, let $x, y \in \mathbb{Z}$ and apply the proposition on signs to $y - x$ . If $y - x$ is positive then $x < y$ . If $y - x = \iota(0)$ then, adding $x$ to both sides, $y = x$ . If $y - x = -w$ with $w$ positive, then $x - y = -(y - x) = w$ is positive, so $y < x$ . Exactly one of the three cases holds, and they are the three alternatives trichotomy demands.

Proposition 10.28 (The order and the operations).

Let $x, y, z \in \mathbb{Z}$ .

If $x < y$ then $x + z < y + z$ .
If $x < y$ and $z$ is positive, then $xz < yz$ .
$\iota(m) < \iota(n)$ if and only if $m < n$ , for all $m, n \in \mathbb{N}_0$ .

Discussion.

Each part unfolds the definition of $<$ into a statement about a difference being positive, so each is settled by computing that difference. In the first the difference is $(y + z) - (x + z)$ , and the two copies of $z$ cancel in the group, leaving $y - x$ unchanged: so the hypothesis is the conclusion. In the second the difference is $yz - xz$ , which factors as $(y - x)z$ by distributivity together with the sign rule $(-x)z = -(xz)$ , and the previous proposition then multiplies two positives. Both of those facts about multiplication were set as problems above, and we quote them. The third is the definition of $<$ on each side: $\iota(n) - \iota(m)$ is $\iota(n - m)$ when $m \leqslant n$ , so positivity of the difference in $\mathbb{Z}$ says exactly what $m < n$ says in $\mathbb{N}_0$ .

Proof.

For the first, $(y + z) - (x + z) = y - x$ by associativity, commutativity and the inverse law, so one difference is positive exactly when the other is.

For the second, the problems on multiplication give $(-x)z = -(xz)$ , and distributivity gives

(y - x)z = yz + (-x)z = yz - xz.

If $x < y$ then $y - x$ is positive, and $z$ is positive, so the product is positive by the previous proposition; hence $xz < yz$ .

For the third, suppose $m < n$ in $\mathbb{N}_0$ . Then $n = m + b$ with $b \in \mathbb{N}$ , so $\iota(n) - \iota(m) = \iota(m + b) - \iota(m) = \iota(b)$ , which is positive; hence $\iota(m) < \iota(n)$ . Conversely, if $\iota(m) < \iota(n)$ then $\iota(n) - \iota(m) = \iota(b)$ for some $b \in \mathbb{N}$ , so $\iota(n) = \iota(m + b)$ and $n = m + b$ by injectivity, giving $m < n$ .

Corollary 10.29 (One is the least positive integer).

If $x \in \mathbb{Z}$ is positive, then $\iota(1) \leqslant x$ .

Proof.

Write $x = \iota(n)$ with $n \in \mathbb{N}$ . Then $1 \leqslant n$ , since $n$ is positive and $\mathbb{N}$ has least element $1$ among its own members, so $\iota(1) \leqslant \iota(n)$ by the third part of the proposition.

Remark (Dropping the embedding from the notation).

From here on we write $n$ for $\iota(n)$ and $-n$ for $-\iota(n)$ , and treat $\mathbb{N}_0$ as a subset of $\mathbb{Z}$ . This is harmless, by the proposition on the order: $\iota$ carries sums to sums, products to products, and the order to the order, and it is injective, so every statement about $\mathbb{N}_0$ transfers unchanged to its image and back. The positive integers are then exactly $\mathbb{N}$ , the list

\ldots, \; -2, \; -1, \; 0, \; 1, \; 2, \; \ldots

names every integer exactly once, and $0 < 1$ by the corollary.

Problem 10.13.

Show that $x < y$ if and only if $-y < -x$ , and that $x$ is positive if and only if $-x < 0$ .

Problem 10.14.

Show that if $x < y$ and $z < 0$ then $yz < xz$ , and that $x^2 \geqslant 0$ for every $x \in \mathbb{Z}$ , with equality only at $x = 0$ .

Problem 10.15.

Show that no integer lies strictly between $n$ and $n + 1$ , and deduce that $\mathbb{Z}$ has no maximum and no minimum.

Sequences, Sums and Products

Many of our notions are already functions. Addition is a map $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ , iteration is repeated composition, and counting a finite set is a bijection $\langle n \rangle \to S$ . A sequence is the same idea read in the other direction: a list of values is a function whose domain is a set of indices. That notion is not new: finite and infinite sequences were defined in the chapter on relations, and the summation symbol in the chapter on the natural numbers. What is new is that the indices may now run over an interval of integers rather than an opening stretch of $\mathbb{N}_0$ , which makes it easier to write a sum starting at $0$ or at $-3$ , or to shift one along.

Indexing by an Integer Interval

Definition 10.30 (Integer intervals).

For $m, n \in \mathbb{Z}$ with $m \leqslant n$ , write

\{m, \ldots, n\} \defeq \{x \in \mathbb{Z} \mid m \leqslant x \leqslant n\},

and for $n \in \mathbb{Z}$ write

\{n, n+1, \ldots\} \defeq \{x \in \mathbb{Z} \mid n \leqslant x\}.

The first is finite and the second is not, as the next paragraph records. When $m = 1$ and $n \in \mathbb{N}$ the interval $\{1, \ldots, n\}$ is exactly the block $\langle n \rangle$ .

Proposition 10.31 (Intervals are finite and are counted by their length).

Let $m \leqslant n$ in $\mathbb{Z}$ . Then $j \mapsto m + j$ is a bijection from the cut $L_{n - m + 1}$ onto $\{m, \ldots, n\}$ , so $\{m, \ldots, n\}$ is finite with $\#\{m, \ldots, n\} = n - m + 1$ . The ray $\{n, n+1, \ldots\}$ is infinite.

Discussion.

The first claim is an existence statement about a bijection, and the map comes from the shape of the interval: an element of $\{m, \ldots, n\}$ is $m$ plus something, and the something is bounded by the length. So we only check the two halves of bijectivity, and both come from the order: injectivity is cancellation for addition in $\mathbb{Z}$ , and surjectivity is the observation that $m \leqslant x \leqslant n$ makes $x - m$ a natural number below $n - m + 1$ . The cardinality is then read off, a cut $L_k$ having $k$ elements. The last claim is a contradiction: a finite subset of a totally ordered set has a maximum, and the ray has none, since $x + 1$ is always a larger member.

Proof.

Write $k \defeq n - m + 1$ , a positive integer since $m \leqslant n$ . If $j \in L_k$ then $0 \leqslant j \leqslant n - m$ , so $m \leqslant m + j \leqslant n$ and the map lands in the interval. It is injective because $m + j = m + j'$ gives $j = j'$ by cancellation, and surjective because $m \leqslant x \leqslant n$ makes $x - m$ a natural number with $x - m \leqslant n - m$ , so $x - m \in L_k$ and $x = m + (x - m)$ .

Hence $\{m, \ldots, n\} \approx L_k$ and $\#\{m, \ldots, n\} = k$ .

Were the ray finite it would be a non-empty finite subset of the totally ordered set $\mathbb{Z}$ , hence would have a maximum $M$ ; but $M + 1$ lies in the ray and $M < M + 1$ , which no upper bound permits.

Definition 10.32 (Sequences on an interval).

Let $S$ be a set. A finite sequence in $S$ indexed by $\{m, \ldots, n\}$ is a function $c : \{m, \ldots, n\} \to S$ , and an infinite sequence indexed by $\{n, n+1, \ldots\}$ is a function on that ray. One writes $c_i$ for $c(i)$ , calls $c_i$ the $i$ -th term and $i$ the index, and calls the domain the index set.

Composing with the bijection of the proposition turns such a sequence into one indexed by a cut, and back again, so this is the notion of the chapter on relations with the indices relabelled and not a second notion. The gain is convenience: one may now write a rule on a generic index, as in ” $c_i = 2i + 1$ for $i \in \{0, \ldots, 5\}$ ”, and start wherever the problem starts. The index is a bound variable and may be renamed, so the same sequence is equally $c_j = 2j + 1$ for $j \in \{0, \ldots, 5\}$ .

Remark (Shifts are different functions).

A finite sequence is commonly written $(c_i)_{i = m, \ldots, n}$ , or $(c_i)$ when the index set is understood, and sequences may be built from sequences: $(3a_i + 4)_{i = m, \ldots, n}$ is a sequence once $(a_i)$ is one. But

(b_{i+k})_{i = m-k, \ldots, n-k} \qquad \text{and} \qquad (b_i)_{i = m, \ldots, n}

are not equal, even though they list the same values in the same order. Their domains differ, and functions with different domains are different functions. One is a shift of the other, and the theorem on index shifts below is what relates their sums.

Problem 10.16.

Take the values $5, 10, 17, 26, 37$ . Give a finite sequence with exactly those terms, and name its index set. Write a shifted sequence with the same values in the same order, and extend the original to an infinite sequence.

Problem 10.17.

Let $(b_i)_{i = m, \ldots, n}$ be a finite sequence and let $k \in \mathbb{Z}$ . Show that $(b_{i+k})_{i = m-k, \ldots, n-k}$ and $(b_i)_{i = m, \ldots, n}$ have the same image, but are equal as functions if and only if $k = 0$ .

Sums over an Interval

Definition 10.33 (Summation over an interval).

Let $(b_i)$ be a sequence with values in a set $U$ carrying an operation written $+$ , and let $m$ lie in its index set. For every $n \geqslant m$ in that index set define

\sum_{i=m}^{m} b_i \defeq b_m, \qquad \sum_{i=m}^{k+1} b_i \defeq \left(\sum_{i=m}^{k} b_i\right) + b_{k+1} \quad (k \geqslant m).

That the clauses determine one value at each $n$ is the recursion theorem, applied after the running length $n - m$ has been shifted into $\mathbb{N}_0$ ; it is the same argument that produced the summation symbol in the chapter on the natural numbers, and taking $m = 1$ with values in a Peano system recovers that symbol exactly. What the present definition adds is an arbitrary lower limit and an arbitrary value set: $U$ needs an operation and nothing else. If every term lies in a subset of $U$ closed under $+$ , the sum lies there too, since the recursion never leaves that subset.

Proposition 10.34 (An interval sum is a left-associated sum).

Let $(U, +)$ be a set with an operation and let $m \leqslant n$ . Then

\sum_{i=m}^{n} b_i = p(b_m, b_{m+1}, \ldots, b_n),

where $p$ is the left-associated product of the chapter on groups, written additively.

Discussion.

Both sides are defined by a recursion, and the claim is that the two recursions are the same one in different notation. So the proof is an induction on the number of terms in which each step compares the two recursive clauses: the base cases are the single-term clauses, which both give $b_m$ , and the step appends $b_{k+1}$ on the right in both definitions. Nothing about $+$ is used. We record it because it lets us quote the results of the chapter on groups here instead of proving them again.

Proof.

Induct on $t \in \mathbb{N}_0$ with $n = m + t$ . At $t = 0$ both sides are $b_m$ . If the two agree at $t$ , then

\sum_{i=m}^{m+t+1} b_i = \left(\sum_{i=m}^{m+t} b_i\right) + b_{m+t+1} = p(b_m, \ldots, b_{m+t}) + b_{m+t+1} = p(b_m, \ldots, b_{m+t+1}),

the outer equalities being the recursive clauses of the two definitions and the middle one the inductive hypothesis.

Theorem 10.35 (Splitting a sum).

Let $(U, +)$ be a semigroup and let $\ell \leqslant m - 1$ and $m \leqslant n$ in $\mathbb{Z}$ , with $(b_i)$ defined on $\{\ell, \ldots, n\}$ . Then

\sum_{i=\ell}^{n} b_i = \sum_{i=\ell}^{m-1} b_i \; + \sum_{i=m}^{n} b_i .

Discussion.

This is the splitting identity of general associativity, which says a left-associated product may be cut anywhere, and the previous proposition has just identified an interval sum with such a product. So we only need to match the two statements: the cut after the $(m-1)$ -st term of the interval corresponds to the cut after the $(m - \ell)$ -th factor of the list, and the two blocks are the two sub-intervals.

Proof.

Write the list $b_\ell, b_{\ell+1}, \ldots, b_n$ , of length $n - \ell + 1$ , and cut it after position $k \defeq m - \ell$ , which satisfies $1 \leqslant k < n - \ell + 1$ . The first part of the theorem on general associativity gives

p(b_\ell, \ldots, b_n) = p(b_\ell, \ldots, b_{m-1}) + p(b_m, \ldots, b_n),

and the previous proposition rewrites each of the three left-associated sums as the corresponding interval sum.

Remark (Moving parentheses).

Once $+$ is associative the outer bracketing of a finite sum is irrelevant, and the theorem is the interval form of that fact. The reading fixed by the definition is the left-associated one, so $a_1 + a_2 + a_3 + a_4$ means $\bigl((a_1 + a_2) + a_3\bigr) + a_4$ , and the theorem equates it with $(a_1 + a_2) + (a_3 + a_4)$ and with $a_1 + (a_2 + a_3 + a_4)$ . It does not reorder the terms. Reordering needs commutativity, which the rearrangement corollary supplies for a list and the last section of this chapter supplies for an unordered index set.

Theorem 10.36 (General distributivity).

Let $U$ carry operations $+$ and $\cdot$ with $c \cdot (x + y) = (c \cdot x) + (c \cdot y)$ for all $c, x, y \in U$ . Then for $c \in U$ and $m \leqslant n$ ,

c \cdot \sum_{i=m}^{n} b_i = \sum_{i=m}^{n} (c \cdot b_i).

Discussion.

An identity over an interval whose length is not fixed, so it is an induction on that length, with the upper limit climbing from $m$ to $n$ . The base is the single-term clause, where both sides are $c \cdot b_m$ . The step uses the hypothesis on $U$ once: expand the left side by the recursive clause, apply the binary distributive law to split $c$ across the two summands, replace the shorter sum by the inductive hypothesis, and reassemble by the recursive clause on the right. Associativity is not needed anywhere, so the theorem is stated for any pair of operations.

Proof.

Induct on $t \in \{0, \ldots, n - m\}$ , writing $k = m + t$ . At $t = 0$ both sides are $c \cdot b_m$ . Suppose the identity holds at $t$ and $t < n - m$ . Then

\begin{aligned} c \cdot \sum_{i=m}^{k+1} b_i &= c \cdot \left(\left(\sum_{i=m}^{k} b_i\right) + b_{k+1}\right) \\ &= \left(c \cdot \sum_{i=m}^{k} b_i\right) + (c \cdot b_{k+1}) \\ &= \left(\sum_{i=m}^{k} (c \cdot b_i)\right) + (c \cdot b_{k+1}) = \sum_{i=m}^{k+1} (c \cdot b_i), \end{aligned}

by the recursive clause, distributivity, the inductive hypothesis, and the recursive clause again. Taking $t = n - m$ gives the theorem.

If the multiplication is not commutative the matching right-hand law, $\bigl(\sum b_i\bigr) \cdot c = \sum (b_i \cdot c)$ , is proved by the same induction from the right distributive law.

Theorem 10.37 (Termwise sums, zeros and negatives).

Let $m \leqslant n$ and let $(b_i)$ and $(c_i)$ be defined on $\{m, \ldots, n\}$ .

If $(U, +)$ is an abelian semigroup, then $\displaystyle\sum_{i=m}^{n} (b_i + c_i) = \sum_{i=m}^{n} b_i + \sum_{i=m}^{n} c_i$ .
If $U$ has an identity $0$ for $+$ , then $\displaystyle\sum_{i=m}^{n} 0 = 0$ .
If $(U, +)$ is an abelian group, then $\displaystyle -\sum_{i=m}^{n} b_i = \sum_{i=m}^{n} (-b_i)$ .

Discussion.

The first two are inductions on the length, of the same shape as the one just done. In the first the step produces a four-term expression on the left and must match a four-term expression on the right, and the matching is a rearrangement, so this uses commutativity as well as associativity. In the second the step adds one more $0$ to a running total already equal to $0$ , and $0 + 0 = 0$ . The third needs no induction: by the first two parts the two sums add to $0$ , so each is the inverse of the other by uniqueness of inverses.

Proof.

For the first, induct on $t$ with $k = m + t$ . At $t = 0$ both sides are $b_m + c_m$ . At the step,

\sum_{i=m}^{k+1} (b_i + c_i) = \left(\sum_{i=m}^{k} b_i + \sum_{i=m}^{k} c_i\right) + (b_{k+1} + c_{k+1}),

and rearranging the four summands by associativity and commutativity gives $\bigl(\sum_{i=m}^{k+1} b_i\bigr) + \bigl(\sum_{i=m}^{k+1} c_i\bigr)$ .

For the second, induct likewise: the single-term sum is $0$ , and $0 + 0 = 0$ .

For the third, the first two parts give

\sum_{i=m}^{n} b_i + \sum_{i=m}^{n} (-b_i) = \sum_{i=m}^{n} \bigl(b_i + (-b_i)\bigr) = \sum_{i=m}^{n} 0 = 0,

and the same computation in the other order, so $\sum (-b_i)$ is an inverse of $\sum b_i$ and hence is $-\sum b_i$ .

Theorem 10.38 (Index shift).

Let $(U, +)$ be a set with an operation, let $m \leqslant n$ , and let $k \in \mathbb{Z}$ . Then

\sum_{i=m}^{n} b_i = \sum_{i=m+k}^{n+k} b_{i-k}.

Discussion.

First, the right-hand sum makes sense, because $m + k \leqslant i \leqslant n + k$ gives $m \leqslant i - k \leqslant n$ , so every term named is a term of $(b_i)$ . The identity itself is an induction on the length: at the base both sides are $b_m$ , and at each step the term appended on the left is $b_{k'+1}$ while the term appended on the right is $b_{(k'+1+k)-k}$ , and those are the same element. This is the relation between shifted sequences mentioned earlier: two different functions with the same sum.

Proof.

Induct on $t \in \{0, \ldots, n - m\}$ , writing $l = m + t$ . At $t = 0$ both sides are $b_m$ . If the identity holds at $t$ , then

\sum_{i=m}^{l+1} b_i = \left(\sum_{i=m}^{l} b_i\right) + b_{l+1} = \left(\sum_{i=m+k}^{l+k} b_{i-k}\right) + b_{(l+1+k)-k} = \sum_{i=m+k}^{(l+1)+k} b_{i-k},

by the recursive clause, the inductive hypothesis, and the recursive clause again.

Proposition 10.39 (Agreement on an interval).

Let $(a_i)$ and $(b_i)$ be sequences in $U$ whose index sets both contain $\{m, \ldots, n\}$ , and suppose $a_i = b_i$ for every $i$ in that interval. Then their sums over it agree.

Discussion.

We need this because a sequence may be defined on a larger index set than the one being summed over, so the two sequences need not be equal as functions. The sum depends only on the terms named: equal single terms give the base, and the recursive clause appends equal terms at each step, so equality is preserved all the way up.

Proof.

Induct on $t \in \{0, \ldots, n - m\}$ . At $t = 0$ both sums are $a_m = b_m$ . If the sums to $m + t$ agree and $t < n - m$ , the recursive clause appends $a_{m+t+1} = b_{m+t+1}$ to each.

Problem 10.18.

Using only the definition, show that $\sum_{i=1}^{1} a_i = a_1$ , that $\sum_{i=1}^{2} a_i = a_1 + a_2$ , that $\sum_{i=1}^{3} a_i = (a_1 + a_2) + a_3$ , and that $\sum_{i=1}^{4} a_i = \bigl((a_1 + a_2) + a_3\bigr) + a_4$ .

Problem 10.19.

Show that the binary distributive law in $U$ is the case $m = 1$ , $n = 2$ of general distributivity, and that the case $\ell = 1$ , $m = 2$ , $n = 3$ of the splitting theorem is the ordinary associative law.

Problem 10.20.

Show that $\sum_{i=m}^{n} b_i$ agrees with the summation symbol of the chapter on the natural numbers when $m = 1$ and the values lie in a Peano system with addition.

Products over an Interval

Definition 10.40 (Products over an interval).

Let $(b_i)$ have values in a set $U$ carrying an operation written multiplicatively, and let $m$ lie in its index set. For $n \geqslant m$ define

\prod_{i=m}^{m} b_i \defeq b_m, \qquad \prod_{i=m}^{k+1} b_i \defeq \left(\prod_{i=m}^{k} b_i\right) \cdot b_{k+1} \quad (k \geqslant m).

Everything proved for sums holds for products, because nothing proved for sums used anything about $+$ beyond the recursion and the laws named in each hypothesis. Replacing $+$ by $\cdot$ and $0$ by $1$ throughout turns each statement above into its multiplicative twin, and each proof into a proof of it: the splitting theorem, general distributivity read the other way, termwise products, the product of ones, the index shift, and agreement on an interval. We use them under those names without restating them, and taking $m = 1$ recovers the product symbol of the chapter on the natural numbers.

One consequence is used often when taking products apart.

Proposition 10.41 (Splitting off the first factor).

Let $U$ have a multiplication with identity $1$ , and let $m \leqslant n$ . Then there is $b \in U$ with

\prod_{j=m}^{n} a_j = a_m \cdot b.

Discussion.

An existence claim, and there are two cases according to whether the interval has one term or more. If $n = m$ the product is $a_m$ itself, and $b = 1$ serves, which is the only place the identity element is needed. If $m < n$ the splitting theorem cuts the product after the first factor, and the second block is the required $b$ ; no computation is involved beyond naming it.

Proof.

If $n = m$ then $\prod_{j=m}^{m} a_j = a_m = a_m \cdot 1$ , so $b = 1$ serves. If $m < n$ then the splitting theorem with the cut after the first factor gives

\prod_{j=m}^{n} a_j = \left(\prod_{j=m}^{m} a_j\right) \cdot \left(\prod_{j=m+1}^{n} a_j\right) = a_m \cdot \left(\prod_{j=m+1}^{n} a_j\right),

so the second block serves as $b$ .

Remark (Left association as the default).

An unbracketed $a + b + c$ means the sum of the three-term sequence, hence $(a + b) + c$ , and longer unbracketed sums and products are read the same way. That matches the convention already fixed for the left-associated product and for composition of functions. The factorial may now be written $n! = \prod_{k=1}^{n} k$ for $n \in \mathbb{N}$ , with the empty product $0! = 1$ already recorded there.

Problem 10.21.

Using only the definition, expand $\prod_{i=1}^{k} a_i$ for $k = 1, 2, 3, 4$ and check that the reading is left-associated. State the identity obtained from the splitting theorem for products with $\ell = 1$ , $m = 3$ , $n = 5$ .

Sums over an Unordered Index Set

An interval carries an order, so its terms arrive in a fixed sequence. When $+$ is commutative as well as associative the order should not matter, and then the index set need not be ordered at all: one wants $\sum_{i \in I} b_i$ for an arbitrary finite $I$ . Throughout this section $(U, +)$ is a commutative monoid (associative, commutative, with an identity $0$ ), and $(b_i)_{i \in I}$ means a function $b : I \to U$ .

The plan is to sum along an ordering and then prove the answer independent of it.

Definition 10.42 (Sum along an ordering).

Let $I$ be finite with $\#I = n$ , let $(b_i)_{i \in I}$ have values in $U$ , and let $f : \langle n \rangle \to I$ be a bijection, with $\langle 0 \rangle = \emptyset$ . Define

\sum_{i \in I}^{(f)} b_i \defeq \begin{cases} 0 & \text{if } n = 0, \\ \displaystyle\sum_{j=1}^{n} b_{f(j)} & \text{if } n \geqslant 1. \end{cases}

Proposition 10.43 (Peeling the last index).

Let $\#I = n \geqslant 1$ , let $f : \langle n \rangle \to I$ be a bijection, put $x \defeq f(n)$ , and let $f_1$ be the restriction of $f$ to a bijection $\langle n-1 \rangle \to I \setminus \{x\}$ . Then

\sum_{i \in I}^{(f)} b_i = \left(\sum_{i \in I \setminus \{x\}}^{(f_1)} b_i\right) + b_x.

Discussion.

First, the restriction is a bijection onto $I \setminus \{x\}$ because $f$ is injective and $f(n) = x$ , so nothing below $n$ is sent to $x$ . The identity itself splits into two cases. When $n = 1$ the left side is $b_x$ and the right side is $0 + b_x$ , so the identity law applies; this is where the convention for the empty sum is used. When $n > 1$ the recursive clause peels the last term off the interval sum, and what is left is the interval sum of $j \mapsto b_{f(j)}$ over $\langle n - 1 \rangle$ ; that agrees with the sum along $f_1$ because the two sequences take equal values there, which is the proposition on agreement.

Proof.

If $n = 1$ then $I = \{x\}$ , the left side is $b_{f(1)} = b_x$ , and the right side is $0 + b_x = b_x$ .

If $n > 1$ the recursive clause gives

\sum_{j=1}^{n} b_{f(j)} = \left(\sum_{j=1}^{n-1} b_{f(j)}\right) + b_{f(n)},

and $b_{f(n)} = b_x$ . The sequences $j \mapsto b_{f(j)}$ and $j \mapsto b_{f_1(j)}$ agree on $\langle n-1 \rangle$ , so their sums there agree by the proposition on agreement, and that sum is the sum along $f_1$ .

Proposition 10.44 (The ordering does not matter).

Let $I$ be finite with $\#I = n$ and let $f, g : \langle n \rangle \to I$ be bijections. Then

\sum_{i \in I}^{(f)} b_i = \sum_{i \in I}^{(g)} b_i.

Discussion.

The claim is universal in $n$ and we use strong induction, since the step will need the hypothesis at $n - 2$ as well as at $n - 1$ . At $n = 0$ and $n = 1$ there is only one bijection available, so there is nothing to compare. For $n \geqslant 2$ we look at where the two orderings put their last element, $x = f(n)$ and $y = g(n)$ . If they agree, peeling both by the previous proposition leaves two sums over the same smaller set, which the hypothesis identifies. If they differ, we cannot peel to a common set in one step, so we take two: choose an ordering of $I \setminus \{x\}$ ending at $y$ and an ordering of $I \setminus \{y\}$ ending at $x$ (these exist because a transposition rearranges any ordering to end where we like), and peel each side twice. Both sides then reduce to the same sum over $I \setminus \{x, y\}$ with $b_x$ and $b_y$ attached, in opposite orders, and commutativity finishes.

Proof.

We first record that for any $z \in I$ there is a bijection $\langle n \rangle \to I$ sending $n$ to $z$ : take any bijection $h$ and compose it with the transposition of $h^{-1}(z)$ and $n$ in $\langle n \rangle$ , which is a bijection.

Now induct strongly on $n$ . If $n \leqslant 1$ the two bijections coincide, since $\langle 0 \rangle = \emptyset$ has one map to $\emptyset$ and $\langle 1 \rangle$ has one map to a singleton.

Let $n \geqslant 2$ and assume the claim for all index sets of smaller cardinality. Put $x = f(n)$ and $y = g(n)$ , with restrictions $f_1, g_1$ as in the previous proposition.

If $x = y$ , peeling both sides gives sums over $I \setminus \{x\}$ along $f_1$ and along $g_1$ , which agree by the inductive hypothesis, so the two sides agree.

If $x \neq y$ , choose a bijection $f_2 : \langle n-1 \rangle \to I \setminus \{x\}$ with $f_2(n-1) = y$ and a bijection $g_2 : \langle n-1 \rangle \to I \setminus \{y\}$ with $g_2(n-1) = x$ , and let $f_3, g_3$ be their restrictions to bijections onto $I \setminus \{x, y\}$ . Write $T$ for the common value of the sums over $I \setminus \{x,y\}$ along $f_3$ and along $g_3$ , equal by the inductive hypothesis. Peeling twice on each side,

\sum_{i \in I}^{(f)} b_i = (T + b_y) + b_x, \qquad \sum_{i \in I}^{(g)} b_i = (T + b_x) + b_y,

where the inductive hypothesis at $n-1$ was used to replace $f_1$ by $f_2$ and $g_1$ by $g_2$ . Associativity and commutativity make the two right-hand sides equal.

Definition 10.45 (Sum over a finite index set).

Let $I$ be finite and let $(b_i)_{i \in I}$ have values in $U$ . Define

\sum_{i \in I} b_i \defeq \sum_{i \in I}^{(f)} b_i

for any bijection $f : \langle \#I \rangle \to I$ , which the previous proposition makes independent of the choice.

Proposition 10.46 (Agreement with interval sums).

If $m \leqslant n$ in $\mathbb{Z}$ , then $\displaystyle\sum_{i \in \{m, \ldots, n\}} b_i = \sum_{i=m}^{n} b_i$ .

Discussion.

The two sides are sums of the same terms under two definitions, and to compare them we need a bijection $\langle N \rangle \to \{m, \ldots, n\}$ , which the proposition on intervals gives: $j \mapsto j + (m-1)$ is one, with $N = n - m + 1$ . Summing along it produces an interval sum from $1$ to $N$ of the shifted sequence, and the index-shift theorem carries that back to the sum from $m$ to $n$ . So the work is in the index shift.

Proof.

Put $N \defeq n - m + 1 = \#\{m, \ldots, n\}$ . The map $f : \langle N \rangle \to \{m, \ldots, n\}$ with $f(j) \defeq j + (m-1)$ is a bijection, with inverse $i \mapsto i - m + 1$ . Hence

\sum_{i \in \{m, \ldots, n\}} b_i = \sum_{j=1}^{N} b_{f(j)} = \sum_{j=1}^{N} b_{j + (m-1)} = \sum_{i=m}^{n} b_i,

the last step being the index shift with $k = m - 1$ .

Theorem 10.47 (Peeling and splitting).

Let $I$ be finite and let $(b_i)_{i \in I}$ have values in $U$ .

If $I = \emptyset$ then $\sum_{i \in I} b_i = 0$ , and if $I = \{x\}$ then $\sum_{i \in I} b_i = b_x$ .
If $x \in I$ then $\displaystyle\sum_{i \in I} b_i = \left(\sum_{i \in I \setminus \{x\}} b_i\right) + b_x$ .
If $I = I_1 \cup I_2$ with $I_1 \cap I_2 = \emptyset$ , then $\displaystyle\sum_{i \in I} b_i = \sum_{i \in I_1} b_i + \sum_{i \in I_2} b_i$ .

Discussion.

The first part is the definition, read at the two smallest index sets. The second is the proposition on peeling for a chosen element: that proposition peeled whichever element the ordering put last, and since the sum no longer depends on the ordering we may choose one that puts the element we want there, which the transposition trick supplies. The third is the main one and is proved by building an ordering of $I$ out of orderings of the two pieces, laid end to end; it is a bijection because the pieces are disjoint, and the cardinality of a disjoint union says its domain has the right size. Summing along it and cutting after the first block is the splitting theorem, and the second block needs an index shift to be recognised as a sum over $I_2$ . The empty cases are handled first, since an ordering of an empty piece is not available to concatenate.

Proof.

The first part is the definition together with the case $n = 1$ of peeling.

For the second, choose a bijection $f : \langle n \rangle \to I$ with $f(n) = x$ , as in the proof of the previous proposition, and apply peeling; the two sums are independent of the orderings by that proposition.

For the third, put $n_1 \defeq \#I_1$ and $n_2 \defeq \#I_2$ , so $\#I = n_1 + n_2$ . If $n_1 = 0$ then $I_1 = \emptyset$ and $I_2 = I$ , and the claim reads $s = 0 + s$ ; the case $n_2 = 0$ is the same. Suppose both are positive, choose bijections $f_1 : \langle n_1 \rangle \to I_1$ and $f_2 : \langle n_2 \rangle \to I_2$ , and define $f : \langle n_1 + n_2 \rangle \to I$ by

f(j) \defeq \begin{cases} f_1(j) & \text{if } j \leqslant n_1, \\ f_2(j - n_1) & \text{if } n_1 < j. \end{cases}

Disjointness of $I_1$ and $I_2$ makes $f$ injective, and it is surjective because $I$ is their union. Then

\sum_{i \in I} b_i = \sum_{j=1}^{n_1 + n_2} b_{f(j)} = \sum_{j=1}^{n_1} b_{f(j)} + \sum_{j = n_1 + 1}^{n_1 + n_2} b_{f(j)}

by the splitting theorem. The first block is $\sum_{i \in I_1} b_i$ . In the second, the index shift by $k = -n_1$ turns it into $\sum_{j=1}^{n_2} b_{f(j + n_1)} = \sum_{j=1}^{n_2} b_{f_2(j)}$ , which is $\sum_{i \in I_2} b_i$ .

Theorem 10.48 (Reindexing).

Let $I$ and $I'$ be finite, let $(b_i)_{i \in I}$ have values in $U$ , and let $\sigma : I' \to I$ be a bijection. Then

\sum_{i \in I'} b_{\sigma(i)} = \sum_{i \in I} b_i.

Discussion.

Both sides are defined by choosing an ordering, and the sum does not depend on which, so we are free to choose orderings that make the two sides identical term by term. An ordering $f$ of $I'$ produces one of $I$ by composing with $\sigma$ , since a composite of bijections is a bijection; and the $j$ -th term of the left sum along $f$ is $b_{\sigma(f(j))}$ , which is the $j$ -th term of the right sum along $\sigma \circ f$ . So the two interval sums are the same sum.

Proof.

If $I = \emptyset$ then $I' = \emptyset$ and both sides are $0$ . Otherwise put $n \defeq \#I = \#I'$ and choose a bijection $f : \langle n \rangle \to I'$ . Then $\sigma \circ f : \langle n \rangle \to I$ is a bijection, and

\sum_{i \in I'} b_{\sigma(i)} = \sum_{j=1}^{n} b_{\sigma(f(j))} = \sum_{j=1}^{n} b_{(\sigma \circ f)(j)} = \sum_{i \in I} b_i,

each outer equality being the definition of the sum over an index set.

Corollary 10.49 (Rearranging a sum).

Let $I$ be finite and let $\sigma$ be a permutation of $I$ . Then $\sum_{i \in I} b_i = \sum_{i \in I} b_{\sigma(i)}$ .

Proof.

Reindexing with $I' = I$ .

That is the general commutative law for sums, and the rearrangement corollary of the chapter on groups is the case where $I$ is an interval and $\sigma$ is written as a permutation of positions. Termwise addition carries over to unordered index sets the same way: choose an ordering and quote the interval statement.

Proposition 10.50 (Termwise sums, unordered).

Let $I$ be finite and let $(b_i)_{i \in I}$ and $(c_i)_{i \in I}$ have values in $U$ . Then

\sum_{i \in I} (b_i + c_i) = \sum_{i \in I} b_i + \sum_{i \in I} c_i.

Discussion.

The empty case is the identity law, $0 = 0 + 0$ . Otherwise choose any ordering; each of the three sums becomes an interval sum along it, and the interval version of termwise addition applies to the two ordered sequences. The choice of ordering does not matter, by the proposition on independence.

Proof.

If $I = \emptyset$ both sides are $0$ . Otherwise let $n = \#I$ and let $f : \langle n \rangle \to I$ be a bijection. Then

\begin{aligned} \sum_{i \in I} (b_i + c_i) &= \sum_{j=1}^{n} \bigl(b_{f(j)} + c_{f(j)}\bigr) \\ &= \sum_{j=1}^{n} b_{f(j)} + \sum_{j=1}^{n} c_{f(j)} = \sum_{i \in I} b_i + \sum_{i \in I} c_i, \end{aligned}

by the interval theorem on termwise sums.

Everything in this section holds for products over a commutative monoid written multiplicatively, with $1$ in place of $0$ ; the definitions, the independence of the ordering, peeling, splitting, reindexing and the commutative law all carry over under that substitution, and their proofs with them.

Problem 10.22.

State the product analogues of the definition of $\prod_{i \in I} b_i$ and of the peeling and splitting theorem, and say which line of each proof changes.

Problem 10.23.

Let $I = \{1, 2, 3\}$ and let $\sigma$ be the permutation $1 \mapsto 2 \mapsto 3 \mapsto 1$ . Expand both sides of the rearrangement corollary for a general $(b_i)_{i \in I}$ in $\mathbb{Z}$ .

Problem 10.24.

Let $I$ and $J$ be finite and let $a_{i,j} \in U$ for $i \in I$ and $j \in J$ . Show that summing along $I$ first and then along $J$ gives the same element as summing along $J$ first and then along $I$ .

Divisibility

$(\mathbb{Z}, +)$ is a group and $(\mathbb{Z}, \cdot)$ is not, since only $1$ and $-1$ have multiplicative inverses. So exact division inside $\mathbb{Z}$ is rare. The study of when one integer divides another is number theory, and this section is its beginning.

Divisors

Proposition 10.51 (The only invertible integers).

Let $a, b \in \mathbb{Z}$ with $ab = 1$ . Then either $a = b = 1$ or $a = b = -1$ .

Discussion.

The hypothesis pins a product to a particular value, and we use the order to get information about the factors. So we work through the signs. Neither factor can be $0$ , since a product with a zero factor is $0$ and $0 \neq 1$ . They cannot have opposite signs, since a positive times a negative is negative while $1$ is positive. That leaves both positive or both negative. If both are positive, each is at least $1$ by the corollary on the least positive integer, and if $a$ were strictly larger than $1$ then multiplying that inequality by the positive $b$ would push $ab$ above $b$ and so above $1$ ; so $a = 1$ , and then $b = 1$ . The negative case is the positive one applied to $-a$ and $-b$ , whose product is again $1$ .

Proof.

Neither factor is $0$ , since $0 \cdot b = 0 \neq 1$ . If one is positive and the other negative, their product is negative, while $1$ is positive; so both are positive or both are negative.

Suppose both are positive. Then $1 \leqslant a$ and $1 \leqslant b$ . If $1 < a$ , multiplying by the positive $b$ gives $b < ab = 1$ , contradicting $1 \leqslant b$ . So $a = 1$ , and $b = 1 \cdot b = ab = 1$ .

If both are negative then $-a$ and $-b$ are positive and $(-a)(-b) = ab = 1$ , so $-a = -b = 1$ and $a = b = -1$ .

Definition 10.52 (Divisibility).

Let $a, b \in \mathbb{Z}$ with $a \neq 0$ . Then $a$ divides $b$ , written $a \mid b$ , if $b = ac$ for some $c \in \mathbb{Z}$ ; in that case $a$ is a divisor of $b$ and $b$ a multiple of $a$ . When $a$ does not divide $b$ we write $a \nmid b$ .

The condition $a \neq 0$ is part of the notation: wherever $a \mid b$ appears as a hypothesis, $a \neq 0$ is part of it. Usage keeps divisor for the general statement and factor for a number appearing in a particular product, so $1, 2, 3, 4, 6$ and $12$ are the positive divisors of $12$ , while $3$ and $4$ are the factors of $12$ in the expression $12 = 3 \cdot 4$ . This is the notion the problem on divisibility introduced for natural numbers, now stated where negative multipliers are available.

Definition 10.53 (Prime).

An integer $p > 1$ whose only positive divisors are $1$ and $p$ is prime.

This makes precise a word we have used informally since the first chapter, and it restates two old results: that $n$ is even says exactly that $2 \mid n$ , and the proposition that odd squares are odd now reads $2 \mid n$ if and only if $2 \mid n^2$ . The four rules below are the basic arithmetic of divisibility.

Proposition 10.54 (The arithmetic of divisibility).

Let $a, b, c, d \in \mathbb{Z}$ with the divisors below non-zero.

If $a \mid b$ and $b \mid c$ , then $a \mid c$ .
If $a \mid c$ and $b \mid d$ , then $ab \mid cd$ .
If $a \mid b$ and $a \mid c$ , then $a \mid bx + cy$ for all $x, y \in \mathbb{Z}$ .
If $a \mid b$ or $a \mid c$ , then $a \mid bc$ .

Discussion.

All four are proved the same way. Each hypothesis $a \mid b$ unpacks by the definition into an equation $b = ak$ with $k$ an integer, and each conclusion asks for one integer of the same shape; so in every case the work is to substitute the hypotheses into the expression named in the conclusion, regroup by associativity, commutativity and distributivity until a single factor of the intended divisor stands in front, and observe that what remains in the bracket is an integer because $\mathbb{Z}$ is closed under the operations. The first substitutes one equation into the other and takes the product of the two multipliers. The second multiplies the two equations and regroups. The third is the one where the shared divisor matters: both unpackings carry the same $a$ , so distributivity pulls it out of the combination. The fourth is a disjunction, hence two cases, each of which is the first line of the others. Nothing needs the order, and nothing needs a case split on sign.

Proof.

For the first, write $b = ak_1$ and $c = bk_2$ . Then $c = (ak_1)k_2 = a(k_1k_2)$ , and $k_1k_2 \in \mathbb{Z}$ , so $a \mid c$ .

For the second, write $c = ak_1$ and $d = bk_2$ . Then $cd = (ak_1)(bk_2) = (ab)(k_1k_2)$ by associativity and commutativity, so $ab \mid cd$ .

For the third, write $b = ak_1$ and $c = ak_2$ . For any $x, y \in \mathbb{Z}$ ,

bx + cy = (ak_1)x + (ak_2)y = a(k_1x + k_2y)

by distributivity, and $k_1x + k_2y \in \mathbb{Z}$ , so $a \mid bx + cy$ .

For the fourth, suppose $a \mid b$ , so $b = ak$ . Then $bc = (ak)c = a(kc)$ and $kc \in \mathbb{Z}$ , so $a \mid bc$ . If instead $a \mid c$ , the same argument applies with $b$ and $c$ exchanged.

Corollary 10.55 (Sums and differences).

If $a \mid b$ and $a \mid c$ , then $a \mid b + c$ and $a \mid b - c$ .

Proof.

Take $x = y = 1$ , and then $x = 1$ , $y = -1$ , in the third part of the proposition.

Theorem 10.56 (Each factor divides the product).

Let $(a_j)$ be a sequence in $\mathbb{Z}$ defined on $\{m, \ldots, n\}$ with $m \leqslant n$ , and let $i \in \{m, \ldots, n\}$ with $a_i \neq 0$ . Then

a_i \; \Bigg| \; \prod_{j=m}^{n} a_j .

Discussion.

The conclusion asks for the product to be written with $a_i$ standing in front, so we move $a_i$ there. If $i$ is the first index, the proposition on splitting off the first factor has already done it. Otherwise cut the product just before $i$ by the splitting theorem for products; the second block now begins at $i$ , so splitting off its first factor exposes $a_i$ , and what is left is the first block times whatever followed. Commutativity and associativity of multiplication in $\mathbb{Z}$ then move $a_i$ to the front of the whole, and the remaining bracket is an integer because a product of integers is one. The hypothesis $a_i \neq 0$ is needed only because the divisibility symbol demands it.

Proof.

If $i = m$ , the proposition on splitting off the first factor gives $\prod_{j=m}^{n} a_j = a_m \cdot b$ for some $b \in \mathbb{Z}$ , so $a_m$ divides the product.

Suppose $m < i$ . The splitting theorem for products, cutting before $i$ , gives

\prod_{j=m}^{n} a_j = \left(\prod_{j=m}^{i-1} a_j\right) \cdot \left(\prod_{j=i}^{n} a_j\right),

and splitting off the first factor of the second block writes it as $a_i \cdot b$ for some $b \in \mathbb{Z}$ . Writing $A$ for the first block,

\prod_{j=m}^{n} a_j = A \cdot (a_i \cdot b) = a_i \cdot (b \cdot A)

by commutativity and associativity, and $b \cdot A \in \mathbb{Z}$ . So $a_i$ divides the product.

Division with Remainder

Non-divisibility is harder to use than divisibility, because it gives no equation. Instead we use the remainder, and every integer has exactly one. That was proved for $\mathbb{N}_0$ in the chapter on order; it extends to $\mathbb{Z}$ with one extra step.

Theorem 10.57 (Division with remainder in the integers).

Let $a \in \mathbb{Z}$ and $m \in \mathbb{N}$ . Then there is exactly one pair $q, r \in \mathbb{Z}$ with

a = qm + r, \qquad 0 \leqslant r < m.

Discussion.

An existence-and-uniqueness claim, and the two halves need different arguments, as they did over $\mathbb{N}_0$ . For existence we look for the remainder rather than the quotient: the numbers $a - qm$ that land in $\mathbb{N}_0$ form a set to which well-ordering applies, provided it is non-empty. That is the one new step, since a negative $a$ leaves nothing when $q = 0$ ; taking $q = a$ works, because multiplying a negative $a$ by $m \geqslant 1$ only makes it more negative. Minimality of the least member then forces it below $m$ , since otherwise one more $m$ could be subtracted. Uniqueness is the argument from the earlier theorem word for word: two decompositions give a multiple of $m$ equal to a difference of remainders, which is trapped strictly between $-m$ and $m$ and so is $0$ .

Proof.

For existence, let $R \defeq \{a - qm \mid q \in \mathbb{Z}\} \cap \mathbb{N}_0$ . It is non-empty: if $0 \leqslant a$ then $q = 0$ gives $a \in R$ ; and if $a < 0$ then $q = a$ gives $a - am = a(1 - m) \geqslant 0$ , since $a < 0$ and $1 - m \leqslant 0$ .

Well-ordering supplies $r \defeq \min R$ , with $a = qm + r$ for some $q \in \mathbb{Z}$ and $0 \leqslant r$ . If $m \leqslant r$ then $r - m \in \mathbb{N}_0$ and $a - (q+1)m = r - m$ , so $r - m \in R$ ; but $r - m < r$ because $m$ is positive, contradicting minimality. Hence $r < m$ .

For uniqueness, suppose $q_1m + r_1 = q_2m + r_2$ with $0 \leqslant r_1, r_2 < m$ . Then $(q_1 - q_2)m = r_2 - r_1$ , and $-m < r_2 - r_1 < m$ . If $q_1 - q_2$ were non-zero its absolute size would be at least $1$ , so $(q_1 - q_2)m$ would be at least $m$ or at most $-m$ , which the bounds forbid. Hence $q_1 = q_2$ , and then $r_1 = r_2$ .

Corollary 10.58 (Remainder forms).

Let $m \in \mathbb{N}$ . Every $a \in \mathbb{Z}$ has exactly one of the $m$ forms $qm$ , $qm + 1$ , …, $qm + (m-1)$ with $q \in \mathbb{Z}$ . In particular $m \mid a$ if and only if the remainder is $0$ .

Proof.

The theorem gives exactly one pair $q, r$ with $a = qm + r$ and $r < m$ , and the available values of $r$ are $0, 1, \ldots, m-1$ . If $r = 0$ then $a = qm$ and $m \mid a$ ; conversely $a = qm$ is a decomposition with remainder $0$ , which uniqueness makes the only one.

So $m \nmid a$ says that $a$ takes one of the $m - 1$ forms $qm + 1, \ldots, qm + (m-1)$ , and a proof from a non-divisibility hypothesis is a proof by cases with that many cases. Taking $m = 2$ recovers the familiar reading: $2 \nmid a$ says $a = 2q + 1$ . The next proposition is the pattern at $m = 3$ .

Proposition 10.59 (Divisibility by three).

Let $a \in \mathbb{Z}$ . If $3 \nmid a^2 - 1$ , then $3 \mid a$ .

Discussion.

The statement is a conditional $p \to q$ in which $p$ is " $3 \nmid a^2 - 1$ " and $q$ is " $3 \mid a$ ". A direct proof would have to start from a non-divisibility fact about $a^2 - 1$ , which offers no equation to work with, so we take the contrapositive: if $3 \nmid a$ then $3 \mid a^2 - 1$ . That still opens with a non-divisibility, but now it is a hypothesis rather than a conclusion, and the corollary converts it into two cases, $a = 3k + 1$ and $a = 3k + 2$ . Each case is then ordinary algebra: expand $a^2 - 1$ , take a factor of $3$ out, and check that what remains in the bracket is an integer.

Proof.

We prove the contrapositive: if $3 \nmid a$ , then $3 \mid a^2 - 1$ . By the corollary, $a = 3k + 1$ or $a = 3k + 2$ for some $k \in \mathbb{Z}$ .

If $a = 3k + 1$ , then

a^2 - 1 = (3k+1)^2 - 1 = 9k^2 + 6k = 3(3k^2 + 2k),

and $3k^2 + 2k \in \mathbb{Z}$ , so $3 \mid a^2 - 1$ .

If $a = 3k + 2$ , then

a^2 - 1 = (3k+2)^2 - 1 = 9k^2 + 12k + 3 = 3(3k^2 + 4k + 1),

and $3k^2 + 4k + 1 \in \mathbb{Z}$ , so $3 \mid a^2 - 1$ .

Remark (Exhibiting a remainder is enough).

The corollary works in both directions. If $a = 3k + 1$ or $a = 3k + 2$ for some $k$ , then $3 \nmid a$ , because divisibility would give $a$ the form $3k'$ and each integer has exactly one of the three forms. So when the conclusion wanted is itself a non-divisibility, exhibiting a remainder settles it, and the same holds with any $m$ in place of $3$ .

Problem 10.25.

Show that $3 \nmid a^2 + 1$ for every $a \in \mathbb{Z}$ .

Problem 10.26.

Show that $a \mid b$ and $b \mid a$ together force $b = a$ or $b = -a$ , and that $a \mid b$ with $b \neq 0$ forces $a \leqslant b$ or $b \leqslant a$ according to sign — state the inequality carefully before proving it.

Problem 10.27.

Let $p_1, \ldots, p_k \in \mathbb{Z}$ be non-zero and let $\sigma \in S_k$ . Using the theorem on factors and the rearrangement corollary, show that each $p_i$ divides $p_{\sigma(1)} \cdots p_{\sigma(k)}$ as well as $p_1 \cdots p_k$ .

Problem 10.28.

Show that the square of an integer leaves remainder $0$ or $1$ on division by $4$ , and deduce that no integer of the form $4k + 3$ is a sum of two squares.

Rings and Fields

This last section names the structure formed by $+$ and $\cdot$ on $\mathbb{Z}$ , and shows that the construction of $\mathbb{Z}$ works for any monoid with the right properties.

The Construction Was Not Special

Look back at what the first half of the chapter used. It took $(\mathbb{N}_0, +)$ , a commutative monoid in which cancellation holds, and produced a group containing a copy of it. Cancellation entered once, in the proof of transitivity. Commutativity and the identity entered in the routine checks. Nothing else about $\mathbb{N}_0$ was used: not the order, not induction, not the successor.

Theorem 10.60 (Every regular commutative monoid sits inside a group).

Let $(H, *)$ be a commutative regular monoid with identity $f$ . Then there are a group $(G, *)$ with identity $e$ and an injective map $\alpha : H \to G$ such that

\alpha(f) = e, \qquad \alpha(a * b) = \alpha(a) * \alpha(b),

and such that every $c \in G$ has the form $c = \alpha(a) * \alpha(b)^{-1}$ for some $a, b \in H$ . Moreover the pair $(G, \alpha)$ is unique up to a unique isomorphism: if $(G', \alpha')$ is another such pair, there is exactly one isomorphism $\beta : G \to G'$ with $\alpha' = \beta \circ \alpha$ .

Discussion.

Two claims, one much harder than the other. The first is an existence claim, and it needs no new work: the construction of $\mathbb{Z}$ from $\mathbb{N}_0$ used only the hypotheses now assumed, so repeating it with $*$ in place of $+$ builds $G$ , and the proofs of this chapter become its proofs. We say which pieces correspond.

The second is the one to prove. Uniqueness of $\beta$ comes first and is immediate: the condition $\alpha' = \beta \circ \alpha$ fixes $\beta$ on the image of $\alpha$ , and since every element of $G$ is a quotient $\alpha(a) * \alpha(b)^{-1}$ , and a homomorphism must send an inverse to an inverse, $\beta$ is forced everywhere. That forced formula is then taken as a definition, and since an element of $G$ may be written as such a quotient in many ways, we check that two writings give the same value, which uses injectivity of $\alpha$ . That $\beta$ is a homomorphism is a short computation, and we get bijectivity by building the map the other way and observing that both composites satisfy the defining condition of the identity, which the uniqueness half then identifies them with.

Proof.

For existence, run the construction of the first half of this chapter with $H$ in place of $\mathbb{N}_0$ and $*$ in place of $+$ . Declare $(a, b) \sim (a', b')$ when $a * b' = a' * b$ ; this is an equivalence relation, the proof of transitivity using cancellation exactly where it used it for $\mathbb{N}_0$ . Let $G$ be the quotient, with $[(a,b)] * [(a',b')] \defeq [(a * a', b * b')]$ , well defined by the same computation as before. Then $(G, *)$ is an abelian group with identity $e \defeq [(f,f)]$ and $[(a,b)]^{-1} = [(b,a)]$ , and $\alpha(a) \defeq [(a, f)]$ is injective with $\alpha(f) = e$ and $\alpha(a * b) = \alpha(a) * \alpha(b)$ . Finally

\alpha(a) * \alpha(b)^{-1} = [(a,f)] * [(f,b)] = [(a * f, f * b)] = [(a, b)],

so every element of $G$ has the required form.

For uniqueness, let $(G', \alpha')$ be another such pair and suppose $\beta : G \to G'$ is a homomorphism with $\alpha' = \beta \circ \alpha$ . For $c = \alpha(a) * \alpha(b)^{-1}$ we get

\beta(c) = \beta(\alpha(a)) * \beta\bigl(\alpha(b)^{-1}\bigr) = \alpha'(a) * \alpha'(b)^{-1},

using that homomorphisms respect inverses. So $\beta$ is determined, and at most one such map exists.

That formula does define a map. Suppose $\alpha(a) * \alpha(b)^{-1} = \alpha(a') * \alpha(b')^{-1}$ . Multiplying by $\alpha(b) * \alpha(b')$ gives $\alpha(a) * \alpha(b') = \alpha(a') * \alpha(b)$ , that is $\alpha(a * b') = \alpha(a' * b)$ , so $a * b' = a' * b$ by injectivity of $\alpha$ . Applying $\alpha'$ to that equation and reversing the steps gives $\alpha'(a) * \alpha'(b)^{-1} = \alpha'(a') * \alpha'(b')^{-1}$ , so the value does not depend on the writing.

The map $\beta$ so defined is a homomorphism, since

\bigl(\alpha(a) * \alpha(b)^{-1}\bigr) * \bigl(\alpha(a') * \alpha(b')^{-1}\bigr) = \alpha(a * a') * \alpha(b * b')^{-1}

in $G$ and the same identity holds in $G'$ , both by commutativity and the inverse of a product. It satisfies $\alpha' = \beta \circ \alpha$ , by taking $b = f$ .

Exchanging the roles of $G$ and $G'$ produces a homomorphism $\beta' : G' \to G$ with $\alpha = \beta' \circ \alpha'$ . Then $\beta' \circ \beta$ is a homomorphism $G \to G$ with $(\beta' \circ \beta) \circ \alpha = \alpha$ , and so is $\mathrm{id}_G$ ; the uniqueness just proved, applied with $G' = G$ and $\alpha' = \alpha$ , forces $\beta' \circ \beta = \mathrm{id}_G$ . Likewise $\beta \circ \beta' = \mathrm{id}_{G'}$ . So $\beta$ is a bijection, hence an isomorphism.

Remark.

The theorem says the integers are the only possible answer: any group containing a copy of $(\mathbb{N}_0,+)$ in which every element is a difference of two copied elements is isomorphic to $\mathbb{Z}$ , by a unique isomorphism respecting the copy. It also saves work later. The same theorem, applied to a multiplicative monoid instead of an additive one, is what will build the rationals out of the integers, and we shall not have to write the pairs down again.

Rings, Integral Domains and Fields

Definition 10.61 (Ring).

A ring is a set $R$ with two operations $+$ and $\cdot$ such that $(R, +)$ is an abelian group, written with identity $0$ ; $(R, \cdot)$ is a semigroup; and both distributive laws hold,

a(b + c) = ab + ac, \qquad (b + c)a = ba + ca .

The ring is commutative if $\cdot$ is commutative. It is a ring with identity if there is $1 \in R$ with $1 \neq 0$ and $1 \cdot a = a \cdot 1 = a$ for every $a$ . It is free of zero divisors if $ab = 0$ forces $a = 0$ or $b = 0$ . A commutative ring with identity and free of zero divisors is an integral domain, and an integral domain in which every $a \neq 0$ is invertible under $\cdot$ is a field.

The identity of $+$ and the identity of $\cdot$ are each unique, by uniqueness of the identity applied to the two operations separately, so the notation $0$ and $1$ names something definite. Nothing yet says how the two operations interact beyond distributivity; the next proposition gives some consequences of distributivity.

Proposition 10.62 (Arithmetic in a ring).

Let $R$ be a ring and let $a, b \in R$ . Then

$0 \cdot a = a \cdot 0 = 0$ ;
$(-a)b = a(-b) = -(ab)$ ;
$(-a)(-b) = ab$ .

Discussion.

None of these is an axiom, and each has to be derived from distributivity, which is the only link between the two operations. For the first, write $0$ as $0 + 0$ and distribute: the result is an equation saying that $0 \cdot a$ added to itself is itself, and in a group only the identity does that, so cancellation finishes. The second uses the first as its target: $ab + (-a)b$ becomes, by distributivity, $\bigl(a + (-a)\bigr)b = 0 \cdot b = 0$ , which says $(-a)b$ is an additive inverse of $ab$ , and inverses are unique. The third is the second applied twice, or once with $b$ replaced by $-b$ , together with $-(-x) = x$ .

Proof.

For the first, $0 \cdot a = (0 + 0) \cdot a = 0 \cdot a + 0 \cdot a$ by the right distributive law. Cancelling $0 \cdot a$ in the group $(R, +)$ gives $0 = 0 \cdot a$ . The computation for $a \cdot 0$ is the same with the left law.

For the second, $ab + (-a)b = \bigl(a + (-a)\bigr)b = 0 \cdot b = 0$ , and likewise $(-a)b + ab = 0$ , so $(-a)b$ is an additive inverse of $ab$ and hence equals $-(ab)$ . The same argument on the other side gives $a(-b) = -(ab)$ .

For the third, replacing $b$ by $-b$ in the second part gives $(-a)(-b) = -\bigl(a(-b)\bigr) = -\bigl(-(ab)\bigr) = ab$ .

Theorem 10.63 (The integers are an integral domain).

$(\mathbb{Z}, +, \cdot)$ is an integral domain, and it is not a field.

Discussion.

Each clause of the definition has already been proved or set as a problem, so the proof mostly collects them: the additive group is the theorem of the second section, the multiplicative semigroup and its commutativity and identity are the problems of the third, and distributivity is its theorem. Two points need a comment. Freedom from zero divisors is the problem on products that vanish. That $\mathbb{Z}$ is not a field is the proposition on invertible integers, which leaves only $1$ and $-1$ invertible, so $2$ has no inverse and $2x = 1$ has no solution.

Proof.

$(\mathbb{Z}, +)$ is an abelian group by the theorem of the second section. Multiplication is associative and commutative by the problems there, with identity $1$ by the proposition on the embedding, and it distributes over addition by the theorem on distributivity; commutativity turns the one distributive law into both. The problem on vanishing products says $xy = 0$ forces $x = 0$ or $y = 0$ . So $\mathbb{Z}$ is an integral domain.

It is not a field: by the proposition on invertible integers, the only invertible elements are $1$ and $-1$ , so $2$ has no multiplicative inverse.

Example 10.64 (The smallest field).

Let $\mathbb{F}_2 \defeq \{0, 1\}$ with $0 \neq 1$ . The field axioms fix both tables. Multiplication is forced by the previous proposition and the identity law: $0 \cdot 0 = 0 \cdot 1 = 1 \cdot 0 = 0$ and $1 \cdot 1 = 1$ . For addition, only $1 + 1$ is not yet determined, and $1 + 1 = 1$ would give

1 = 1 + 0 = 1 + \bigl(1 + (-1)\bigr) = (1 + 1) + (-1) = 1 + (-1) = 0,

which is forbidden; so $1 + 1 = 0$ . The tables are

\begin{array}{c|cc} + & 0 & 1 \\ \hline 0 & 0 & 1 \\ 1 & 1 & 0 \end{array} \qquad\qquad \begin{array}{c|cc} \cdot & 0 & 1 \\ \hline 0 & 0 & 0 \\ 1 & 0 & 1 \end{array}

and one checks directly that they satisfy the axioms. A field needs $0 \neq 1$ , so no smaller field exists.

Corollary 10.65 (The binomial theorem in a commutative ring).

Let $R$ be a commutative ring with identity, let $a, b \in R$ and let $n \in \mathbb{N}_0$ . Then

(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k},

where $\binom{n}{k} x$ means the sum of $\binom{n}{k}$ copies of $x$ .

Proof.

The proof of the binomial theorem used only the commutative, associative and distributive laws and the recursion defining powers, as the remark following it recorded. Every one of those holds in a commutative ring with identity, so the argument applies word for word, with the coefficient $\binom{n}{k}$ read as an instruction to add $\binom{n}{k}$ copies.

Remark.

The coefficient cannot in general be read as an element of $R$ , since $R$ need not contain a copy of $\mathbb{N}_0$ ; what it names is a repeated sum, and repeated sums can be $0$ . In $\mathbb{F}_2$ , for instance, $2x = x + x = 0$ for every $x$ , so the middle term of $(a+b)^2$ disappears and $(a+b)^2 = a^2 + b^2$ .

Greatest Common Divisors

The next theorem, proved with division with remainder, is what we need to build fields from the integers.

Theorem 10.66 (Bézout's identity).

Let $a, b \in \mathbb{Z}$ be not both $0$ , and let

S \defeq \{ax + by \mid x, y \in \mathbb{Z}\}.

Then $S$ contains a positive element, its least positive element $d$ divides both $a$ and $b$ , and every common divisor of $a$ and $b$ divides $d$ . In particular $d = ax_0 + by_0$ for some $x_0, y_0 \in \mathbb{Z}$ .

Discussion.

Three assertions, and the first is needed for well-ordering to produce $d$ : taking $x = a$ and $y = b$ gives $a^2 + b^2$ , which is positive because a square is never negative and the two are not both $0$ . The second is the main one, and it uses division with remainder. Divide $a$ by $d$ ; the remainder is again of the form $ax + by$ , being $a$ minus a multiple of $ax_0 + by_0$ , and it is strictly below $d$ ; since $d$ was the least positive member, the remainder cannot be positive, so it is $0$ and $d$ divides $a$ . The same for $b$ . The third assertion is the easiest: a common divisor of $a$ and $b$ divides every combination $ax + by$ , by the arithmetic of divisibility, and $d$ is one of those.

Proof.

Taking $x = a$ and $y = b$ gives $a^2 + b^2 \in S$ , which is positive since squares are non-negative and $a$ and $b$ are not both $0$ . So the positive members of $S$ form a non-empty subset of $\mathbb{N}$ , and well-ordering supplies a least one, $d = ax_0 + by_0$ .

Divide: $a = qd + r$ with $0 \leqslant r < d$ . Then

r = a - qd = a - q(ax_0 + by_0) = a(1 - qx_0) + b(-qy_0),

so $r \in S$ . Were $r$ positive it would be a positive member of $S$ below $d$ , contrary to the choice of $d$ ; so $r = 0$ and $d \mid a$ . The same argument with $b$ in place of $a$ gives $d \mid b$ .

If $c$ divides both $a$ and $b$ , then $c$ divides $ax_0 + by_0 = d$ by the third part of the arithmetic of divisibility.

Definition 10.67 (Greatest common divisor and coprimality).

For $a, b \in \mathbb{Z}$ not both $0$ , the greatest common divisor $\gcd(a, b)$ is the $d$ of the theorem: the least positive integer of the form $ax + by$ . The integers $a$ and $b$ are coprime if $\gcd(a, b) = 1$ .

The name is justified by the theorem: $d$ is a common divisor, and every common divisor divides it, hence is at most $d$ . Coprimality says exactly that $1 = ax + by$ for some integers $x$ and $y$ , and the next section uses that equation.

The Integers Modulo $m$

Definition 10.68 (Congruence).

Let $m \in \mathbb{N}$ and $a, b \in \mathbb{Z}$ . Then $a$ is congruent to $b$ modulo $m$ , written $a \equiv b \pmod{m}$ , if $m \mid a - b$ .

Proposition 10.69 (Congruence is an equivalence relation).

For each $m \in \mathbb{N}$ , congruence modulo $m$ is an equivalence relation on $\mathbb{Z}$ , and its classes are in bijection with $\{0, 1, \ldots, m-1\}$ .

Discussion.

Each of the three properties follows from the definition. Reflexivity is $m \mid 0$ , which holds because $0 = m \cdot 0$ . Symmetry is the observation that $b - a$ is $-(a - b)$ , and a divisor of an integer divides its negative. Transitivity is the corollary on sums: $m$ divides $a - b$ and $b - c$ , hence their sum $a - c$ . For the count, division with remainder attaches to each integer exactly one remainder below $m$ , and two integers are congruent exactly when their remainders agree, because their difference is then a multiple of $m$ , and conversely a difference of two numbers both below $m$ cannot be a non-zero multiple of $m$ .

Proof.

Reflexivity: $a - a = 0 = m \cdot 0$ , so $m \mid a - a$ . Symmetry: if $a - b = mk$ then $b - a = m(-k)$ . Transitivity: if $m \mid a - b$ and $m \mid b - c$ then $m$ divides $(a-b) + (b-c) = a - c$ by the corollary on sums.

For the count, let $r(a)$ be the remainder of $a$ on division by $m$ . If $r(a) = r(b)$ then $a - b = (q_a - q_b)m$ , so $a \equiv b$ . Conversely if $a \equiv b$ then $r(a) - r(b) = (a - q_am) - (b - q_bm)$ is a multiple of $m$ lying strictly between $-m$ and $m$ , hence $0$ . So $a \mapsto r(a)$ induces a bijection from the classes onto $\{0, \ldots, m-1\}$ .

Definition 10.70 (The integers modulo $m$ ).

Write $[a]$ for the class of $a$ and

\mathbb{Z}/m\mathbb{Z} \defeq \mathbb{Z}\big/{\equiv} \pmod m

for the quotient, the integers modulo $m$ , with operations

[a] + [b] \defeq [a + b], \qquad [a] \cdot [b] \defeq [ab].

Theorem 10.71 (The integers modulo $m$ form a ring).

The two operations above are well defined, and $\mathbb{Z}/m\mathbb{Z}$ is a commutative ring with identity $[1]$ and zero $[0]$ , with $\#(\mathbb{Z}/m\mathbb{Z}) = m$ when $m \geqslant 2$ .

Discussion.

Well-definedness is the only thing that needs care, as in the first half of the chapter: the formulas name representatives, and a class has many. So we suppose $a \equiv a'$ and $b \equiv b'$ and must show $a + b \equiv a' + b'$ and $ab \equiv a'b'$ . The first is the corollary on sums applied to the two differences. The second needs one step more: $ab - a'b'$ does not obviously factor, so we insert and remove $a'b$ , splitting it into $b(a - a')$ and $a'(b - b')$ , each divisible by $m$ by the fourth part of the arithmetic of divisibility, and then add. Every ring axiom afterwards is the corresponding axiom of $\mathbb{Z}$ read inside the brackets, and the count is the bijection of the previous proposition.

Proof.

Suppose $m \mid a - a'$ and $m \mid b - b'$ . Then $m$ divides $(a - a') + (b - b') = (a + b) - (a' + b')$ , so the sum is well defined. And

ab - a'b' = b(a - a') + a'(b - b'),

in which $m$ divides each summand by the fourth part of the arithmetic of divisibility, hence divides the whole by the corollary on sums; so the product is well defined.

Associativity, commutativity and distributivity for the classes follow from the same laws in $\mathbb{Z}$ applied to representatives; $[0]$ is an additive identity with $-[a] = [-a]$ , and $[1]$ is a multiplicative identity. When $m \geqslant 2$ the classes $[0], \ldots, [m-1]$ are distinct and exhaust the quotient by the previous proposition, so there are $m$ of them and $[0] \neq [1]$ .

Theorem 10.72 (A prime modulus gives a field).

Let $m \geqslant 2$ . Then $\mathbb{Z}/m\mathbb{Z}$ is a field if and only if $m$ is prime. When $m = p$ is prime one writes $\mathbb{F}_p$ for it, a field with exactly $p$ elements.

Discussion.

A biconditional, and the two directions use opposite features of $m$ . Suppose $m$ is composite, say $m = m_1m_2$ with both factors strictly between $1$ and $m$ . Then $[m_1]$ and $[m_2]$ are non-zero while their product is $[m] = [0]$ , so the ring has zero divisors; and a zero divisor can never be invertible, since multiplying $[m_1][m_2] = [0]$ by a hypothetical inverse of $[m_1]$ would give $[m_2] = [0]$ . So no composite modulus works.

Suppose instead $m = p$ is prime and $[a] \neq [0]$ , that is $p \nmid a$ . The positive divisors of $p$ are $1$ and $p$ , and $p$ is not among the common divisors of $a$ and $p$ , so $\gcd(a, p) = 1$ . Bézout then writes $1 = ax + py$ , and reading that equation modulo $p$ leaves $[a][x] = [1]$ . Every non-zero class is therefore invertible, which with the previous theorem is the definition of a field.

Proof.

Suppose $m$ is not prime, so $m = m_1m_2$ with $1 < m_1, m_2 < m$ . Then $[m_1] \neq [0]$ and $[m_2] \neq [0]$ , since $m$ divides neither factor, while $[m_1][m_2] = [m] = [0]$ . If $[m_1]$ had an inverse $[c]$ then $[m_2] = [c][m_1][m_2] = [c][0] = [0]$ , a contradiction. So $\mathbb{Z}/m\mathbb{Z}$ is not a field.

Suppose $m = p$ is prime and let $[a] \neq [0]$ , so $p \nmid a$ . Any positive common divisor of $a$ and $p$ divides $p$ , hence is $1$ or $p$ ; it is not $p$ , since $p \nmid a$ . So $\gcd(a, p) = 1$ , and Bézout gives $x, y \in \mathbb{Z}$ with $ax + py = 1$ . Then $ax - 1 = p(-y)$ , so $[a][x] = [ax] = [1]$ and $[a]$ is invertible. With the previous theorem, $\mathbb{Z}/p\mathbb{Z}$ is a field, and it has $p$ elements.

Remark.

Taking $p = 2$ recovers $\mathbb{F}_2$ , whose tables were forced by the axioms a few pages ago; now it comes from the construction, which gives one field for every prime. These are not all the finite fields (there is one with $p^n$ elements for each prime $p$ and each $n \in \mathbb{N}$ ), but the others are not quotients of $\mathbb{Z}$ , and we do not build them here.

Problem 10.29.

Show that $[a] \in \mathbb{Z}/m\mathbb{Z}$ is invertible if and only if $\gcd(a, m) = 1$ , and that the invertible classes form an abelian group under multiplication.

Problem 10.30.

Write out the addition and multiplication tables of $\mathbb{Z}/m\mathbb{Z}$ for $m = 4$ and $m = 5$ , and say which classes are invertible in each.

Problem 10.31.

Show that a subset $R_0$ of a ring $R$ is itself a ring under the restricted operations if and only if it is non-empty and $a - b$ and $ab$ lie in $R_0$ whenever $a$ and $b$ do. Show that $2\mathbb{Z}$ is such a subset of $\mathbb{Z}$ but does not contain the identity of $\mathbb{Z}$ .

Problem 10.32.

Show that freedom from zero divisors is equivalent, in a commutative ring with identity, to cancellation: $ab = ac$ with $a \neq 0$ forces $b = c$ . Which of the two conditions was easier to check for $\mathbb{Z}$ ?

Subrings, Homomorphisms and Ideals

The chapter on groups asked, of every structure it defined, which subsets inherit it and which maps respect it. We ask the same two questions of rings. The answers are parallel, except that the kernel of a ring homomorphism has a strictly stronger property than a subring.

Definition 10.73 (Subring).

Let $R$ be a ring. A non-empty subset $R_0 \subset R$ is a subring if $a - b \in R_0$ and $ab \in R_0$ whenever $a, b \in R_0$ .

The two conditions say exactly that $R_0$ is a subgroup of $(R, +)$ , by the one-step criterion, and is closed under multiplication; associativity and distributivity are laws and so are inherited. Freedom from zero divisors is inherited too, being another law. What is not inherited is the identity: $2\mathbb{Z}$ is a subring of $\mathbb{Z}$ containing no multiplicative identity of $\mathbb{Z}$ , and a subring of a ring with identity may lack one.

Definition 10.74 (Ring homomorphism).

Let $R_1$ and $R_2$ be rings. A function $\alpha : R_1 \to R_2$ is a ring homomorphism if

\alpha(a + b) = \alpha(a) + \alpha(b) \qquad \text{and} \qquad \alpha(ab) = \alpha(a)\alpha(b)

for all $a, b \in R_1$ . A bijective ring homomorphism is a ring isomorphism. The kernel is $\ker\alpha \defeq \alpha^{-1}(\{0\})$ and the image is $\alpha(R_1)$ .

Proposition 10.75 (Kernel and image of a ring homomorphism).

Let $\alpha : R_1 \to R_2$ be a ring homomorphism. Then $\ker\alpha$ is a subring of $R_1$ and $\alpha(R_1)$ is a subring of $R_2$ ; and $\alpha$ is injective if and only if $\ker\alpha = \{0\}$ .

Discussion.

A ring homomorphism is in particular a homomorphism of the additive groups, so everything proved about kernels and images in the chapter on groups is available and settles the additive half of each claim, injectivity included. What remains is closure under multiplication, one line on each side: a product of two elements of the kernel maps to a product of zeros, and a product of two values is the value at a product. Nothing here needs the identity, and neither the kernel nor the image need contain one.

Proof.

Since $\alpha$ is a homomorphism of $(R_1, +)$ into $(R_2, +)$ , the proposition on kernels and images makes $\ker\alpha$ a subgroup of $(R_1, +)$ and $\alpha(R_1)$ a subgroup of $(R_2, +)$ , and makes $\alpha$ injective exactly when $\ker\alpha = \{0\}$ .

If $a, b \in \ker\alpha$ then $\alpha(ab) = \alpha(a)\alpha(b) = 0 \cdot 0 = 0$ , so $ab \in \ker\alpha$ . And $\alpha(a)\alpha(b) = \alpha(ab) \in \alpha(R_1)$ . So both are closed under multiplication and hence are subrings.

Remark (Neither identities nor freedom from zero divisors survive).

The image of a ring with identity need not contain the identity of the target, and the image of an integral domain need not be free of zero divisors: the map $\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}$ sending $a$ to $[a]$ is a ring homomorphism, and for composite $m$ its image is the whole of a ring with zero divisors. The kernel contains the identity only in one case: if $1 \in \ker\alpha$ then $\alpha(a) = \alpha(a \cdot 1) = \alpha(a) \cdot 0 = 0$ for every $a$ , so $\alpha$ is the zero map.

The kernel has a further property, the analogue of normality.

Proposition 10.76 (Kernels absorb multiplication).

Let $\alpha : R_1 \to R_2$ be a ring homomorphism, let $a \in \ker\alpha$ and let $b \in R_1$ . Then $ab \in \ker\alpha$ and $ba \in \ker\alpha$ .

Discussion.

A subring need only be closed under products of its own elements, but the kernel contains the product of any of its elements with anything at all, and the reason is the one-line computation that $\alpha$ sends $ab$ to $\alpha(a)\alpha(b) = 0 \cdot \alpha(b)$ , which is $0$ by the arithmetic of a ring. This property is strictly stronger, and like normality for subgroups it gets its own name.

Proof.

$\alpha(ab) = \alpha(a)\alpha(b) = 0 \cdot \alpha(b) = 0$ , so $ab \in \ker\alpha$ ; and $\alpha(ba) = \alpha(b) \cdot 0 = 0$ , so $ba \in \ker\alpha$ .

Definition 10.77 (Ideal).

Let $R$ be a commutative ring. A non-empty subset $\mathfrak{a} \subset R$ is an ideal if

$a_1 - a_2 \in \mathfrak{a}$ whenever $a_1, a_2 \in \mathfrak{a}$ ;
$ba \in \mathfrak{a}$ whenever $a \in \mathfrak{a}$ and $b \in R$ .

An ideal other than $\{0\}$ and $R$ is proper. For $a \in R$ the set $Ra \defeq \{ba \mid b \in R\}$ is the principal ideal generated by $a$ , written $(a)$ ; more generally $(a_1, \ldots, a_n) \defeq \{b_1a_1 + \cdots + b_na_n \mid b_i \in R\}$ .

The second clause makes the first look weak, and in a ring with identity it is: from $a \in \mathfrak{a}$ one gets $-a = (-1)a \in \mathfrak{a}$ , so closure under subtraction follows from closure under addition. Every ideal is a subring, and the kernel of every ring homomorphism is an ideal by the proposition above. The converse, that every ideal is a kernel, is proved below.

Example 10.78 (Ideals of the integers).

For $m \in \mathbb{N}$ the set $m\mathbb{Z}$ of multiples of $m$ is an ideal of $\mathbb{Z}$ , and it is the principal ideal $(m)$ . It is the kernel of $a \mapsto [a]$ , and congruence modulo $m$ is exactly the relation $a - b \in m\mathbb{Z}$ , so the general quotient construction below, applied to this ideal, reproduces the ring $\mathbb{Z}/m\mathbb{Z}$ built by hand earlier, which explains the notation. Bézout’s identity says more: $(a, b) = (\gcd(a,b))$ , so every ideal generated by finitely many integers is principal.

Theorem 10.79 (The quotient ring).

Let $R$ be a commutative ring and $\mathfrak{a}$ an ideal. Then $a \equiv b$ meaning $a - b \in \mathfrak{a}$ is an equivalence relation, and the operations

[a] + [b] \defeq [a + b], \qquad [a] \cdot [b] \defeq [ab]

are well defined on the quotient $R/\mathfrak{a}$ , making it a commutative ring. The map $q : R \to R/\mathfrak{a}$ sending $a$ to $[a]$ is a surjective ring homomorphism with $\ker q = \mathfrak{a}$ .

Discussion.

That the relation is an equivalence is the first clause of the definition, which makes $\mathfrak{a}$ a subgroup of $(R,+)$ : reflexivity is $0 \in \mathfrak{a}$ , symmetry is closure under negation, transitivity is closure under addition.

Well-definedness of the sum is the additive statement already proved for quotient groups, an abelian group having every subgroup normal. Well-definedness of the product uses the second clause of the definition: $ab - a'b'$ does not obviously lie in $\mathfrak{a}$ , so we insert and remove $a'b$ , splitting the difference into $b(a - a')$ and $a'(b - b')$ , each of which is a ring element times a member of $\mathfrak{a}$ , which lies in an ideal but need not lie in a subring. The ring axioms then follow from those of $R$ applied to representatives, and the statement about $q$ is the definition read backwards.

Proof.

The first clause makes $\mathfrak{a}$ a subgroup of the abelian group $(R,+)$ , so $a - b \in \mathfrak{a}$ is an equivalence relation whose classes are the cosets, and the theorem on quotient groups makes $[a] + [b] \defeq [a+b]$ well defined with $(R/\mathfrak{a}, +)$ an abelian group.

For the product, suppose $a - a' \in \mathfrak{a}$ and $b - b' \in \mathfrak{a}$ . Then

ab - a'b' = b(a - a') + a'(b - b'),

and the second clause puts both summands in $\mathfrak{a}$ , so their sum lies there and $[ab] = [a'b']$ .

Associativity, commutativity and distributivity for classes are the corresponding laws in $R$ applied to representatives. Finally $q(a + b) = [a+b] = q(a) + q(b)$ and $q(ab) = [ab] = q(a)q(b)$ , so $q$ is a ring homomorphism; it is surjective by construction; and $q(a) = [0]$ holds exactly when $a \in \mathfrak{a}$ .

Theorem 10.80 (First isomorphism theorem for rings).

Let $\alpha : R_1 \to R_2$ be a homomorphism of commutative rings. Then

R_1/\ker\alpha \;\cong\; \alpha(R_1),

by the ring isomorphism sending $[a]$ to $\alpha(a)$ .

Discussion.

Both sides exist already: the quotient because kernels are ideals, the image because it is a subring. So the work is in the named map, and it is the group version with one clause added. Well-definedness, additivity, surjectivity onto the image and injectivity are exactly as they were there, since $\alpha$ is in particular a homomorphism of additive groups. The one new thing to check is that the map respects multiplication, and that is the definition of the quotient product followed by the multiplicativity of $\alpha$ .

Proof.

Write $\mathfrak{k} \defeq \ker\alpha$ , an ideal by the proposition on kernels, and define $\beta([a]) \defeq \alpha(a)$ .

Everything about the additive structure — that $\beta$ is well defined, additive, surjective onto $\alpha(R_1)$ and injective — is the first isomorphism theorem for groups applied to $\alpha$ as a homomorphism of $(R_1,+)$ into $(R_2,+)$ .

For multiplication, $\beta([a][b]) = \beta([ab]) = \alpha(ab) = \alpha(a)\alpha(b) = \beta([a])\beta([b])$ . So $\beta$ is a bijective ring homomorphism onto $\alpha(R_1)$ .

Remark (Fields have no proper ideals).

If $\mathbb{K}$ is a field and $\mathfrak{a}$ an ideal containing some $a \neq 0$ , then $1 = a^{-1}a \in \mathfrak{a}$ , and then $b = b \cdot 1 \in \mathfrak{a}$ for every $b$ , so $\mathfrak{a} = \mathbb{K}$ . A field therefore has only the two trivial ideals, and consequently a ring homomorphism between fields is either the zero map or injective: its kernel, being an ideal not containing $1$ unless it is everything, must be $\{0\}$ .

Problem 10.33.

Show that an intersection of ideals of $R$ is an ideal, and that $(a_1, \ldots, a_n)$ is the smallest ideal containing $a_1, \ldots, a_n$ .

Problem 10.34.

Show that $a \mid b$ in a commutative ring with identity if and only if $(b) \subset (a)$ , that $u$ is a unit if and only if $(u) = R$ , and that $d$ is a greatest common divisor of $a$ and $b$ exactly when $(a, b) = (d)$ .

Problem 10.35.

Show that every ideal of $\mathbb{Z}$ is principal. Use well-ordering on the positive members, as in the proof of Bézout’s identity.

Polynomials

A polynomial is usually written as an expression, and an expression is not an object. We do what we did for ordered pairs and for the integers: say what the object is, in terms of things already built, and then recover the familiar notation as a theorem.

Definition 10.81 (Polynomials and formal power series).

Let $R$ be a commutative ring with identity. A formal power series over $R$ is a sequence $\alpha : \mathbb{N}_0 \to R$ , written $\alpha = (a_0, a_1, a_2, \ldots)$ with $a_n \defeq \alpha(n)$ . It is a polynomial if $a_n = 0$ for all but finitely many $n$ . Define

(\alpha + \beta)_n \defeq a_n + b_n, \qquad (\alpha\beta)_n \defeq \sum_{j = 0}^{n} a_j b_{n-j} .

Write $R[x]$ for the set of polynomials and $R[[x]]$ for the set of all formal power series, with these operations. The degree $\deg\alpha$ of a non-zero polynomial $\alpha$ is the largest index carrying a non-zero coefficient.

The product is the rule one would get by multiplying out two expressions and collecting the terms of each degree; here it is a definition, and the sum defining it is a finite sum over an interval, so it names an element of $R$ without any question of convergence. “Formal” means that the series is a sequence of coefficients and nothing more.

Theorem 10.82 (Polynomials form a ring).

Let $R$ be a commutative ring with identity. Then $R[[x]]$ is a commutative ring with identity under the operations above, and $R[x]$ is a subring containing that identity. The map sending $r \in R$ to $(r, 0, 0, \ldots)$ is an injective ring homomorphism, and writing $x \defeq (0, 1, 0, 0, \ldots)$ gives

x^n = (\underbrace{0, \ldots, 0}_{n}, 1, 0, \ldots), \qquad \alpha = \sum_{j=0}^{n} a_j x^j \quad \text{for every polynomial } \alpha \text{ with } a_j = 0 \text{ beyond } n .

Discussion.

Addition is coordinatewise, so the additive group is immediate. Commutativity of the product is the observation that reversing the order of summation in $\sum_j a_j b_{n-j}$ turns it into $\sum_j b_j a_{n-j}$ , which is reindexing a finite sum. Associativity is the corresponding statement for a double sum: both $(\alpha\beta)\gamma$ and $\alpha(\beta\gamma)$ have $n$ -th coefficient the sum of $a_ib_jc_k$ over all triples with $i + j + k = n$ , so the two agree once the sums are rearranged, which the results on unordered index sets allow. Distributivity is coordinatewise. That $R[x]$ is a subring is the observation that a sum or product of two sequences with finitely many non-zero terms has finitely many non-zero terms; for the product, because the $n$ -th coefficient vanishes once $n$ exceeds the sum of the two cut-offs. The claims about $x$ are a computation from the product rule, and the last display is then bookkeeping.

Proof.

Addition is coordinatewise, so $(R[[x]], +)$ is an abelian group with zero $(0,0,\ldots)$ and $(-\alpha)_n = -a_n$ .

For commutativity of the product, the substitution $j \mapsto n - j$ is a bijection of $\{0, \ldots, n\}$ with itself, so reindexing gives $\sum_{j=0}^{n} a_jb_{n-j} = \sum_{j=0}^{n} b_j a_{n-j}$ .

For associativity, both $\bigl((\alpha\beta)\gamma\bigr)_n$ and $\bigl(\alpha(\beta\gamma)\bigr)_n$ equal the sum of $a_ib_jc_k$ over the finitely many triples $(i,j,k)$ in $\mathbb{N}_0$ with $i + j + k = n$ , by splitting each double sum and reindexing. Distributivity is coordinatewise, since the $n$ -th coefficient of $\alpha(\beta + \gamma)$ is $\sum_j a_j(b_{n-j} + c_{n-j})$ .

The element $(1, 0, 0, \ldots)$ is an identity, since its only non-zero coefficient is at $0$ and the product sum collapses to $a_n \cdot 1$ .

If $a_n = 0$ for $n > N$ and $b_n = 0$ for $n > M$ , then every term of $\sum_j a_jb_{n-j}$ vanishes once $n > N + M$ , so $R[x]$ is closed under products; it is clearly closed under differences. So $R[x]$ is a subring, and it contains $(1,0,\ldots)$ .

Finally $x^n$ has $1$ in position $n$ and $0$ elsewhere, by induction on $n$ from the product rule, so $a_jx^j$ has $a_j$ in position $j$ and $0$ elsewhere, and adding those for $j \leqslant n$ reproduces $\alpha$ .

Proposition 10.83 (Degree and zero divisors).

Let $R$ be an integral domain and let $\alpha, \beta \in R[x]$ be non-zero, of degrees $\deg\alpha$ and $\deg\beta$ . Then $\alpha\beta \neq 0$ and $\deg(\alpha\beta) = \deg\alpha + \deg\beta$ . Hence $R[x]$ is an integral domain, and it is never a field.

Discussion.

We look at the top coefficient. Writing $N$ and $M$ for the two degrees, the coefficient of $\alpha\beta$ at $N + M$ is a sum in which every term but one has a factor above the cut-off of $\alpha$ or of $\beta$ , hence vanishes; the remaining term is the product of the two leading coefficients, which is non-zero because $R$ has no zero divisors. So the product is non-zero and its degree is exactly $N + M$ . The last claim follows: $x$ has degree $1$ , so any $\gamma$ with $x\gamma = 1$ would have $1 + \deg\gamma = \deg 1 = 0$ , which the order on $\mathbb{N}_0$ forbids.

Proof.

Put $N \defeq \deg\alpha$ and $M \defeq \deg\beta$ . The coefficient of $\alpha\beta$ at $N + M$ is $\sum_{j} a_j b_{N+M-j}$ ; a term with $j > N$ has $a_j = 0$ , and a term with $j < N$ has $N + M - j > M$ and so $b_{N+M-j} = 0$ . Only $j = N$ survives, giving $a_N b_M$ , which is non-zero because $R$ has no zero divisors. Coefficients above $N + M$ vanish by the same count. So $\alpha\beta \neq 0$ with degree $N + M$ , and $R[x]$ is an integral domain.

If $x\gamma = 1$ for some $\gamma \neq 0$ , then $1 + \deg\gamma = 0$ , impossible in $\mathbb{N}_0$ . So $x$ is not invertible and $R[x]$ is not a field.

Problem 10.36.

Show that $\alpha \in R[[x]]$ is a unit if and only if $a_0$ is a unit in $R$ , by solving for the coefficients of the inverse one at a time. Deduce that $1 - x$ is invertible in $R[[x]]$ and identify its inverse.

Problem 10.37.

Show that evaluation at $r \in R$ , sending $\sum_j a_j x^j$ to $\sum_j a_j r^j$ , is a ring homomorphism $R[x] \to R$ . Show that its kernel is an ideal containing $x - r$ .

Problem 10.38.

Show that $\mathbb{F}_2[x]$ is infinite while the set of functions $\mathbb{F}_2 \to \mathbb{F}_2$ is finite, and conclude that distinct polynomials may define the same function. Give two such polynomials.

Modules and Vector Spaces

The last definition of the chapter describes a ring acting on an abelian group, where the group need not be a ring itself.

Definition 10.84 (Module and vector space).

Let $R$ be a commutative ring with identity. An $R$ -module is an abelian group $(M, +)$ together with a map $R \times M \to M$ , written $(c, v) \mapsto cv$ , such that for all $c, c' \in R$ and $v, v' \in M$

(c + c')v = cv + c'v, \qquad c(v + v') = cv + cv', \qquad (cc')v = c(c'v), \qquad 1v = v .

When $R$ is a field $\mathbb{K}$ , an $R$ -module is called a $\mathbb{K}$ -vector space, and its elements vectors.

Example 10.85 (Modules already met).

Every ring $R$ is a module over itself, with $cv$ the ring product; every ideal of $R$ is a submodule. Every abelian group is a $\mathbb{Z}$ -module, with $nv$ the $n$ -fold sum, which is exactly the map $\iota$ construction read with $v$ in place of $1$ . For a field $\mathbb{K}$ and a set $X$ , the functions $X \to \mathbb{K}$ form a $\mathbb{K}$ -vector space under pointwise operations, and taking $X = \langle n \rangle$ gives the space $\mathbb{K}^n$ of $n$ -tuples with coordinatewise addition and scaling. And $R[x]$ is an $R$ -module, the scalars acting on the coefficients.

Remark (Where this goes).

A ring that is also a $\mathbb{K}$ -vector space, with the two structures compatible, is a $\mathbb{K}$ -algebra; the linear maps of a vector space to itself form one under composition, and choosing a basis identifies it with a ring of matrices. That is linear algebra, and we stop at the definition. Every structure in these chapters (semigroup, group, ring, field, module) was built from the same two things, a set and a function.

Problem 10.39.

Show that $0v = 0$ and $(-c)v = -(cv)$ in any $R$ -module, quoting the corresponding argument for rings.

Problem 10.40.

Show that an abelian group admits exactly one structure as a $\mathbb{Z}$ -module. Where does the uniqueness come from?

Exercises

Exercises in Lean

The proofs below are checked in your browser. Nothing is sent anywhere, and nothing is stored but your own work. Type \to for →, \and for ∧, \< and \> for ⟨ ⟩.

An integer of this chapter is a class of pairs of natural numbers, and the sum criterion that decides when two pairs name the same integer mentions nothing but addition on $\mathbb{N}_0$ . So the whole construction can be checked in the carrier the pairs are drawn from, and the sheet is written there: a pair is two naturals $m$ and $n$ , and $(m, n) \sim (m', n')$ is the equation

m + n' = n + m'

with no new notation needed for it. What the chapter proves about $\sim$ becomes an implication between such equations, which is what the exercises ask for.

Two laws of multiplication join the arithmetic already listed, both from the problems of the chapter on the natural numbers:

Nat.mul_assoc   (m * n) * p = m * (n * p)
Nat.add_mul     (m + n) * p = m * p + n * p

and divisibility arrives as the definition the order chapter gave it, written a ∣ b and typed \mid:

Nat.dvd_iff     a ∣ b ↔ ∃ c : ℕ, b = a * c

It is an equivalence rather than a definition the checker unfolds, so it is used through .mp and .mpr.

Rearranging a sum

Every calculation with pairs ends the same way: four naturals added in one order must be shown equal to the same four added in another order. We prove this once.

Example.

Associativity opens the bracketing, commutativity exchanges the middle pair, and associativity closes it again. It is listed below as Nat.add_shuffle.

Example.

Symmetry of $\sim$ , which is the sum criterion read backwards with each side commuted.

Exercise 10.12.

The exchange that Proposition 10.5 turns on.

Exercise 10.13.

A common shift satisfies the sum criterion, which is the third condition of Theorem 10.8 implying the second.

Exercise 10.14.

Transitivity, and with it Proposition 10.2 . Adding $n''$ to the first equation is what makes the second substitutable, and cancellation clears what is left.

Exercise 10.15.

Proposition 10.15 : pre-addition respects the equivalence. This is what Nat.add_shuffle is for.

Exercise 10.16.

Every number divides itself.

Exercise 10.17.

The first part of Proposition 10.54 , on the carrier.

Exercise 10.18.

And Corollary 10.55 , whose subtraction half has no reading in $\mathbb{N}_0$ .

Exercise 10.19.

A divisibility survives multiplying both sides by the same factor.

Exercise 10.20.

The third part of Proposition 10.54 , in the form $\mathbb{N}_0$ can state. (Harder.)

What the checker understands

Tactics

intro h	assume the hypothesis of an implication, naming it h
exact e	give the proof outright
apply f	reduce the goal to the hypotheses of f
assumption	close the goal with a hypothesis already present
trivial	close the goal True
exfalso	replace the goal with False
by_contra h	assume the negation of the goal
constructor	split ∧ into both halves, or ↔ into both directions
left / right	choose which half of a ∨ to prove
rcases h with a \| b	argue by cases on a disjunction
obtain ⟨a, b⟩ := h	take a conjunction or an existential apart
cases h	as above, keeping the name
refine e	give the proof with holes left in it
have h : p := …	record an intermediate result
show p	restate the goal in an equal form
use w	give a witness for ∃
specialize h a	instantiate a ∀ hypothesis
rw [h]	rewrite with an equation, ← to go backwards
rfl	both sides compute to the same thing
decide / norm_num	settle a closed computation
tauto	close a goal that is true by pure logic
induction n with k ih	the fifth Peano condition: prove the goal at 0, then at succ k from ih

Results you may cite

Classical.em	∀ (a : Prop), a ∨ ¬a — the law of excluded middle
Classical.byContradiction	∀ {a : Prop}, (¬a → False) → a — proof by contradiction; the tactic by_contra does this for you
Classical.byCases	∀ {a b : Prop}, (a → b) → (¬a → b) → b — split on whether a holds
not_not	∀ {a : Prop}, ¬¬a ↔ a — double negation
not_and_or	∀ {a b : Prop}, ¬(a ∧ b) ↔ ¬a ∨ ¬b — De Morgan
not_or	∀ {a b : Prop}, ¬(a ∨ b) ↔ ¬a ∧ ¬b — De Morgan
not_imp	∀ {a b : Prop}, ¬(a → b) ↔ a ∧ ¬b
and_comm	∀ {a b : Prop}, a ∧ b ↔ b ∧ a
or_comm	∀ {a b : Prop}, a ∨ b ↔ b ∨ a
Set.ext	∀ {A B : Obj}, (∀ x : Obj, x ∈ A ↔ x ∈ B) → A = B — extensionality: sets with the same elements are equal
Set.ext_iff	∀ {A B : Obj}, A = B ↔ (∀ x : Obj, x ∈ A ↔ x ∈ B) — extensionality and substitution, in one biconditional
Set.subset_antisymm	∀ {A B : Obj}, A ⊆ B → B ⊆ A → A = B — mutual inclusion is equality
Set.empty_subset	∀ {A : Obj}, ∅ ⊆ A — the empty set is a subset of every set
Set.pair_eq	∀ {a b c d : Obj}, ((a, b) = (c, d)) ↔ (a = c ∧ b = d) — two ordered pairs are equal exactly when their coordinates are
Nat.succ_inj	∀ {m n : ℕ}, succ m = succ n → m = n — the third Peano condition: the successor is injective
Nat.succ_ne_zero	∀ (n : ℕ), succ n ≠ 0 — the fourth: zero is nobody's successor
Nat.pred	∀ {n : ℕ}, n ≠ 0 → ∃ m : ℕ, n = succ m — predecessors: everything but zero is a successor
Nat.add_zero	∀ (m : ℕ), m + 0 = m — the first clause of addition
Nat.add_succ	∀ (m n : ℕ), m + succ n = succ (m + n) — the second clause of addition
Nat.zero_add	∀ (n : ℕ), 0 + n = n — addition from the left
Nat.succ_add	∀ (m n : ℕ), succ m + n = succ (m + n) — addition from the left, at a successor
Nat.add_assoc	∀ (m n p : ℕ), (m + n) + p = m + (n + p) — addition is associative
Nat.add_comm	∀ (m n : ℕ), m + n = n + m — addition is commutative
Nat.add_ne_zero	∀ {a : ℕ} (b : ℕ), a ≠ 0 → a + b ≠ 0 — positivity is absorbing
Nat.mul_zero	∀ (m : ℕ), m * 0 = 0 — the first clause of multiplication
Nat.mul_succ	∀ (m n : ℕ), m * succ n = m * n + m — the second clause of multiplication
Nat.zero_mul	∀ (m : ℕ), 0 * m = 0 — multiplication from the left
Nat.succ_mul	∀ (m n : ℕ), succ m * n = m * n + n — multiplication from the left, at a successor
Nat.add_right_cancel	∀ {m n k : ℕ}, m + k = n + k → m = n — cancellation, from the last sheet
Nat.add_eq_zero	∀ {m n : ℕ}, m + n = 0 → m = 0 ∧ n = 0 — a sum is zero only when both parts are, from the last sheet
Nat.mul_comm	∀ (m n : ℕ), m * n = n * m — multiplication is commutative, from the last sheet
Nat.mul_add	∀ (m n p : ℕ), m * (n + p) = m * n + m * p — multiplication distributes over addition, from the last sheet
Nat.mul_assoc	∀ (m n p : ℕ), (m * n) * p = m * (n * p) — multiplication associates, from the problems of the last chapter
Nat.add_mul	∀ (m n p : ℕ), (m + n) * p = m * p + n * p — distributivity on the other side
Nat.add_left_cancel	∀ {a m n : ℕ}, a + m = a + n → m = n — uniqueness of differences
Nat.lt_trichotomy	∀ (m n : ℕ), m < n ∨ m = n ∨ n < m — trichotomy, from the theorem that ℕ is strictly ordered
Nat.lt_irrefl	∀ (n : ℕ), ¬(n < n) — anti-reflexivity, from the last sheet
Nat.lt_trans	∀ {m n p : ℕ}, m < n → n < p → m < p — transitivity of the strict order, from the last sheet
Nat.lt_succ_self	∀ (n : ℕ), n < succ n — every number is below its successor, from the last sheet
Nat.not_lt_zero	∀ {n : ℕ}, ¬(n < 0) — nothing lies below zero
Nat.lt_succ_iff	∀ {m n : ℕ}, m < succ n ↔ m < n ∨ m = n — nothing lies strictly between n and succ n
Num.inj	∀ {m n : ℕ}, ↑m = ↑n → m = n — distinct numbers name distinct objects of ω
swap_apply_left	∀ (a b : Obj), swap a b a = b — the transposition sends a to b
swap_apply_right	∀ (a b : Obj), swap a b b = a — and b to a
swap_apply_of_ne_of_ne	∀ {a b x : Obj}, x ≠ a → x ≠ b → swap a b x = x — and fixes every other point
swap_swap	∀ (a b x : Obj), swap a b (swap a b x) = x — a transposition is its own inverse
Function.iterate_zero_apply	∀ (f : Obj → Obj) (x : Obj), f^[0] x = x — the first clause of the powers of a map
Function.iterate_succ_apply	∀ (f : Obj → Obj) (n : ℕ) (x : Obj), f^[succ n] x = f^[n] (f x) — the second clause: f^[succ n] is f^[n] ∘ f
Function.iterate_add_apply	∀ (f : Obj → Obj) (m n : ℕ) (x : Obj), f^[m + n] x = f^[m] (f^[n] x) — the first law of exponents
Nat.factorial_zero	0 ! = succ 0 — the first clause of the factorial
Nat.factorial_succ	∀ (n : ℕ), (succ n) ! = succ n * n ! — the second clause of the factorial
Nat.choose_zero_right	∀ (n : ℕ), choose n 0 = succ 0 — the empty set is the one 0-subset
Nat.choose_eq_zero_of_lt	∀ {n k : ℕ}, n < k → choose n k = 0 — no subset is larger than the whole
Nat.choose_succ_succ	∀ (n k : ℕ), choose (succ n) (succ k) = choose n k + choose n (succ k) — Pascal's identity
mul_assoc	∀ (a b c : Obj), (a ∗ b) ∗ c = a ∗ (b ∗ c) — the operation associates
e_mul	∀ (a : Obj), e ∗ a = a — the identity on the left
mul_e	∀ (a : Obj), a ∗ e = a — and on the right
inv_mul_cancel	∀ (a : Obj), a⁻¹ ∗ a = e — the inverse on the left
mul_inv_cancel	∀ (a : Obj), a ∗ a⁻¹ = e — and on the right
inv_inv	∀ (a : Obj), (a⁻¹)⁻¹ = a — worked above: inverting twice gives the element back
mul_left_cancel	∀ {a x y : Obj}, a ∗ x = a ∗ y → x = y — worked above: the left half of prop-9-5
Nat.add_shuffle	∀ (p q r s : ℕ), (p + q) + (r + s) = (p + r) + (q + s) — worked above: the rearrangement every pair calculation needs
Nat.dvd_iff	∀ {a b : ℕ}, a ∣ b ↔ ∃ c : ℕ, b = a * c — a divides b when b is a multiple of it

From the logical core

And.intro	∀ {a b : Prop}, a → b → a ∧ b
And.left	∀ {a b : Prop}, a ∧ b → a
And.right	∀ {a b : Prop}, a ∧ b → b
And.symm	∀ {a b : Prop}, a ∧ b → b ∧ a
Or.inl	∀ {a b : Prop}, a → a ∨ b
Or.inr	∀ {a b : Prop}, b → a ∨ b
Or.elim	∀ {a b c : Prop}, a ∨ b → (a → c) → (b → c) → c
Or.symm	∀ {a b : Prop}, a ∨ b → b ∨ a
Iff.intro	∀ {a b : Prop}, (a → b) → (b → a) → (a ↔ b)
Iff.mp	∀ {a b : Prop}, (a ↔ b) → a → b
Iff.mpr	∀ {a b : Prop}, (a ↔ b) → b → a
Iff.symm	∀ {a b : Prop}, (a ↔ b) → (b ↔ a)
Iff.rfl	∀ {a : Prop}, a ↔ a
Iff.trans	∀ {a b c : Prop}, (a ↔ b) → (b ↔ c) → (a ↔ c)
True.intro	True
False.elim	∀ {a : Prop}, False → a
absurd	∀ {a b : Prop}, a → ¬a → b
id	∀ {a : Prop}, a → a
mt	∀ {a b : Prop}, (a → b) → ¬b → ¬a
Eq.refl	∀ {α : Type} (a : α), a = a
Eq.symm	∀ {α : Type} {a b : α}, a = b → b = a
Eq.trans	∀ {α : Type} {a b c : α}, a = b → b = c → a = c
congrArg	∀ {α : Type} {β : Type} {a b : α} (f : α → β), a = b → f a = f b
Exists.intro	∀ {α : Type} {p : α → Prop} (w : α), p w → ∃ x : α, p x
Exists.elim	∀ {α : Type} {p : α → Prop} {b : Prop}, (∃ x : α, p x) → (∀ y : α, p y → b) → b

Symbols, Logic and Language

Syntax and Semantics

Propositions and Truth Values

Semantically Void and Self-Contradictory Sentences

Atomic Statements and the Structure of Mathematical Truth

Logical Connectives

Negation

Compound Sentences

”If-Then” Statements

Operator Precedence

The Structure of Proof Systems

Satisfiability

Logical Equivalence and Nonequivalence

Biconditional

Equivalence Proofs

Fundamental Theorems

Equivalences Involving Conditionals

Contrapositive, Converse, and Inverse

Negating Conditional Statements

Equivalence Proofs in Practice

Normal Forms

Disjunctive Normal Form

Constructing DNF from Truth Tables

Conjunctive Normal Form

Constructing CNF

The Cost of Canonicality

Application: Knights and Knaves

Exercises

Exercises in Lean

Statements and proofs

Implication

intro

exact

apply

True and False

trivial

exfalso

Negation

by_contra

Predicates

Variables and Predicates

Introduction to Sets

Quantifiers

The Universal Quantifier

Specifying the Quantification Set

The Existential Quantifier

Scope and Variable Binding

The Unique Existential Quantifier

Formalising Arguments

Quantifier Negation

Negation and Bounded Quantifiers

Distribution of Quantifiers over Connectives

Validity and Satisfiability in Predicate Logic

Nested Quantifiers

Order of Quantifiers

Quantification Fixes a Variable

Negating Nested Quantifiers

Proofs

Valid Arguments and Inference Rules

Arguments and Validity

Modus Ponens

Modus Tollens

Hypothetical Syllogism

Implication Elimination

Further Propositional Inference Rules

Building Complex Arguments

Inference Rules for Quantified Statements

Fallacies

Proof Techniques

Proving Conditional Claims

Indirect proofs

Proving Biconditional Claims

Proving Conjunctions

Proving Disjunctions

Proving Universal Claims

Proving Existential Claims

Disproving Claims

The Law of Excluded Middle

Ex Falso Quodlibet

Using Assumptions in Proofs

`intro`

`exact`

`apply`

`trivial`

`exfalso`

`by_contra`

`constructor`

`obtain`

`left` and `right`

`rcases`

`have`

`induction`