Axioms for Set Theory the Following Is a Subset of the Zermelo-Fraenkel

The following is a subset of the Zermelo-Fraenkel axioms for set theory. In this setting, all objects are sets which are denoted by letters, e.g. x, y, X, Y . Equality is logical identity: sets X,Y are equal (X = Y ) if and only if the letters X,Y denote the same set. There is a primitive notion of membership between sets (x ∈ X) which is read x is a member of X.

1. Equality Axiom: Sets X,Y are equal if they have the same members.

(∀X)(∀Y )(X = Y ↔ (∀x)(x ∈ X ↔ x ∈ Y ))

The set X is said to be a subset of Y every element of X is an element of Y . This statement is denoted by X ⊆ Y . 2. Empty Set Axiom: There is a set with no elements.

(∃Z)(∀x)x∈ / Z

This set Z is denoted by {} or ∅. 3. Doubleton Axiom: For any sets x, y there is a set whose elements are x, y.

(∀x)(∀y)(∃Z)(z ∈ Z ↔ (z = x ∧ z = y))

This set Z is denoted by {x, y}. The singleton set {x} is deﬁned to be {x, x}. 4. Set Union Axiom: For any set X there is a set whose elements are the elements of the sets in the set X. (∀X)(∃Z)((∀z)(z ∈ Z ↔ (∃Y )(Y ∈ X ∧ z ∈ Y ))) S S This set is denoted by X. If A, B are sets the union of A and B is deﬁned to be {A, B}. This set is denoted by A ∪ B. We have x ∈ A ∪ B ↔ (x ∈ A ∨ x ∈ B). 5. Power Set Axiom: For any set X there is a set whose elements are the subsets of X.

(∀X)(∃Z)(z ∈ X ↔ z ⊆ X)

This set Z is called the power set of X and is denoted by P (X). 6. Set Formation Axiom: For any set A and any statement P (x) involving a variable set x there is a subset of A consisting of those elements x in A for which P (x) is true.

(∃X)(∀x)(x ∈ X ↔ (x ∈ A ∧ P (x)))

This set is denoted by {x | x ∈ A ∧ P (x)} or {x ∈ A | P (x)}. This axiom is actually actually an axiom scheme; one axiom for each P (x). If C is a non-empty set, say C1 ∈ C then

(∀C)(C ∈ C → x ∈ C) ⇐⇒ x ∈ C1 ∧ (∀C)(C ∈ C =⇒ x ∈ C) T T so that the set C = {x | (∀C)(C ∈ C → x ∈ C) exists by Axiom 6. TheT set C consists of those elements which lie in every set in the collection C. If C = {A, B} then C = {x | x ∈ A ∧ x ∈ B} which is by deﬁnition A ∩ B, the intersection of the sets A and B.

1 7. Infinity Axiom: (∃Z)(∅ ∈ Z ∧ (z ∈ Z → z ∪ {z} ∈ Z)) A set Z is called an inductive set if ∅ ∈ Z and z ∈ Z → z ∪ {z} ∈ Z. The infinity axiom simply states that there is an inductive set. T Theorem 1. If C is a non-empty collection of inductive sets then C is inductive. T Proof. Let D = C. We have ∅ ∈ D since every set C ∈ C is inductive and hence ∅ ∈ C. Let x ∈ D. Then for every set C ∈ C we have x ∈ C which implies x ∪ {x} ∈ C since C is inductive. Thus x ∪ {x} ∈ D and D is inductive. QED Theorem 2. Let Z be an inductive set and let N be the intersection of the inductive subsets of Z. If D is any inductive set then N ⊆ D. Proof. Since D ∩ Z is an inductive subset of Z we have N ⊆ D ∩ Z and so N ⊆ D. QED Definition 1. The set N of natural numbers is the smallest inductive set (with respect to the inclusion relation). The existence and uniqueness of N follows immediately from Theorem 2. If we define x + 1 to be x ∪ {x} and

0 = ∅, 1 = 0 + 1 = {0}, 2 = 1 + 1 = {0, 1}, 3 = 2 + 1 = {0, 1, 2}, 4 = 3 + 1 = {0, 1, 2, 3}, ... then 0, 1, 2, 3, 4, .... ∈ N. We now establish the elementary properties of the natural numbers. The ﬁrst property which follows immediately from the fact that set N is the contained in every inductive set is the so-called Principle of Induction. If S is a subset of N which satisﬁes

(a) 0 ∈ S

(b) n ∈ S =⇒ n + 1 ∈ S then S = N. Proposition 1. (∀n ∈ N)(∀m ∈ N)(m ∈ n =⇒ m ⊆ n) Proof. Let S = {n ∈ N | (∀m ∈ N)(m ∈ n =⇒ m ⊆ n)}. By the Principle of Induction is suﬃces to prove that S is inductive. We have 0 ∈ S since

(∀m ∈ N)(m ∈ 0 =⇒ m ⊆ 0) is vacuously true as 0 = ∅ implies that m ∈ 0 is false for all m ∈ N. Now suppose that n ∈ S and that m ∈ n + 1. Then m ∈ n or m = n. If m ∈ n we have by our inductive hypothesis n ∈ S that m ⊆ n which is also true if m = n. Since n ⊆ n + 1 we obtain m ⊆ n + 1 and hence that n + 1 ∈ S. Hence S is inductive. QED Corollary 1. (∀m ∈ N)(∀n ∈ N)(m + 1 = n + 1 =⇒ m = n) Proof. If m + 1 = n + 1 then m ∈ n + 1 which implies m ∈ n or m = n and hence that m ⊆ n by Proposition 1. Similarly n ∈ m + 1 implies n ⊆ m. Hence m = n. QED Deﬁnition 2. The mapping σ : N → N deﬁned by σ(n) = n + 1 is called the successor mapping. We also denote σ(n) by n+. Theorem 3. The successor mapping is injective with image N − {0}.

2 Proof. The injectivity follows from Corollary 1. The mapping σ is not surjective since 0 6= n + 1 for any n ∈ N. Now let S = {0} ∪ σ(N). It suffices to show that S = N. Since 0 ∈ we only have to show that n ∈ S =⇒ n + 1 ∈ S in order to prove that S is inductive. But n + 1 = σ(n) ∈ S for any n ∈ N so n + 1 ∈ S for any n ∈ S. QED Corollary 2. If n ∈ N and n 6= 0 there is a unique m ∈ N with n = m + 1. This natural number m is denoted by n − 1 or n− and is called the immediate predecessor of n. Proposition 2. (∀n ∈ N)n∈ / n Proof. Let S = {n | n∈ / n}. Then 0 ∈ S since 0 ∈/ 0 = ∅. Suppose that n ∈ S. Then n∈ / n. We want to show that n + 1 ∈/ n + 1. Suppose to the contrary that n + 1 ∈ n + 1 = n ∪ {n}. Then n + 1 ∈ n or n + 1 = n which implies n + 1 ⊆ n by Proposition 1. But then n ∈ n + 1 implies that n ∈ n which is a contradiction. Hence n + 1 ∈ S and S is inductive. QED Corollary 3. (∀n ∈ N)(∀m ∈ N)(m ∈ n =⇒ m ⊂ n) Proposition 3. (∀n ∈ N)(∀m ∈ N)(m ∈ n ⇐⇒ m ⊂ n) Proof. By Corollary 3 we only have to prove (∀n ∈ N)(∀m ∈ N)(m ⊂ n =⇒ m ∈ n. Let S = {n ∈ N | (∀m ∈ N)(m ⊂ n =⇒ m ∈ n)}. Then 0 ∈ S since m ⊂ 0 = ∅ is false for all m. Let n ∈ S and suppose that m ⊂ n + 1 = n ∪ {n}. Then m ⊆ n or n ∈ m. If m ⊆ n then m = n or m ∈ n by the inductive hypothesis. In either case m ∈ n + 1. If n ∈ m then n ⊂ m ⊂ n + 1, which is not possible. Hence n + 1 ∈ S and S is inductive. QED Proposition 4. (∀n ∈ N)(∀m ∈ N)(m ⊆ n ∨ n ⊆ m) Proof. By induction on n. Let S = {n ∈ N | (∀m ∈ N)(m ⊆ n or n ⊆ m)}. Then 0 ∈ S since 0 = ∅ ⊆ m for any m. Let n ∈ S and let m ∈ N. Then m ⊆ n or n ⊂ m. But m ⊆ n implies that m ⊆ n + 1 and n ⊂ m implies n ∈ m and hence n + 1 = n ∪ {n} ⊆ m. Hence n + 1 ∈ S and S is inductive. QED Corollary 4. N is linearly ordered under inclusion. If m, n ∈ N we also denote m ⊆ n by m ≤ n. Definition 3. For n ∈ N we let [0, n) denote the set {m ∈ N | 0 ≤ m < n}. Note that by the definition of N we have [0, n) = n for any n ∈ N. Definition 4. Let X be a set. An infinite sequence of elements of X is a function with domain N and range a subset of X. If a is an infinite sequence it is denoted by

(an) = (a0, a1, . . . , an,...) where an = a(n) = (n + 1)−st term of a. A finite sequence of elements of X of length n is a function a with domain [0, n) and range a subset of X. If n = 0 the domain of the finite sequence is ∅ and the sequence is called the empty sequence. A finite sequence a of length n is usually denoted by

(a0, a1, . . . , an−1) with the convention that this denotes the empty sequence () when n = 0. An important method for constructing sequences is called recursion or deﬁnition by induction. The following theorem is an important special case called simple recursion. We will treat the general case later.

3 Theorem 3. Let X be a set, let x0 ∈ X and let φ be a mapping of X into itself. Then there is a unique infinite sequence (an) such that a0 = x0 and an+1 = φ(an) for n ≥ 0. Proof. We only give the main steps of the proof. The details of the proof are left to the reader. First one shows that for any n ≥ 1 there is a unique finite sequence a of length n such a0 = x0 and am+1 = φ(am) for 0 ≤ m < n. Then, if a and b are two such finite sequences with a of length n and b of length p > n, one shows that aS⊂ b, i.e. that am = bm for m < n. If F is the collection of all such functions a of length ≥ 1 then F is the required infinite sequence. QED We now introduce the operations of addition and multiplication of natural numbers. For n ∈ N we define σn : N → N inductively by

0 n+1 n σ = 1N (the identity mapping on N), σ = σ ◦ σ .

We can similarly deﬁne f n for any mapping f : X → X where X is any set. Proposition 4. (∀n ∈ N)σn(0) = n 0 n Proof. By induction on n. The proposition is true for n = 0 since σ (0) = 1N(0) = 0. If σ (0) = n for some n ∈ N then σn+1(0) = σ(σn(0)) = σ(n) = n + 1. QED Deﬁnition 3. Let m, n ∈ N. Then

m + n = σn(m) = σn ◦ σm(0), mn = (σm)n(0).

We have m + 1 = σ(m), m + 0 = 0 + m = m, m · 1 = 1 · m = m, m · 0 = 0 · m = 0 and

m + (n + 1) = (m + n) + 1, m(n + 1) = mn + m.

Theorem 4. Let m, n, p ∈ N. Then

(a) (m + n) + p = m + (n + p), (mn)p = m(np) (Associative Laws) (b) m + n = n + m, mn = nm (Commutative Laws) (c) m(n + p) = mn + mp , (m + n)p = mp + np (Distributive Laws)

(d) m ≥ n ⇐⇒ (∃p ∈ N)m = n + p (We denote p by m − n.)

Proof. (1) We ﬁrst prove (m + n) + p = m + (n + p), the associative law for addition, by induction on p. Since (m + n) + 0 = m + n = m + (n + 0), it is true for p = 0. If it is true for some p then

(m + n) + (p + 1) = ((m + n) + p) + 1 = (m + (n + p)) + 1 = m + ((n + p) + 1) = m + (n + (p + 1).

Hence by induction it is true for all p.

(2) We now prove m + n = n + m, the commutative law for addition, by induction. For n = 0 we have m + 0 = m = 0 + m. The proof of our inductive step will require the following auxiliary result.

Lemma 1. (∀n ∈ N)1 + n = n + 1

4 Proof of Lemma. We proceed by induction on n. For n = 0 we have 1 + 0 = 0 = 1 + 0 and if 1 + n = n + 1 for some n we have 1 + (n + 1) = (1 + n) + 1 = (n + 1) + 1. QED

Resuming our inductive proof of the commutative law for addition, assume that m + n = n + m for some n. Then

m + (n + 1) = (m + n) + 1 = (n + m) + 1 = 1 + (n + m) = (1 + n) + m = (n + 1) + m

which completes the proof of the inductive step.

(3) We now prove the distributive laws by induction on p. For p = 0 we have

m(n + 0) = mn = mn + 0 = mn + m · 0, (m + n) · 0 = 0 = 0 + 0 = m · 0 + n · 0.

If we assume that m(n + p) = mn + mp and (m + n)p = mp + np for some p then

m(n+(p+1)) = m((n+p)+1) = m(n+p)+1 = (mn+mp)+1 = mn+(mp+m) = mn+m(p+1)

and

(m + n)(p + 1) = (m + n)p + (m + n) = ((mp + np) + m) + n = (m + (mp + np)) + n = ((m + mp) + np) + n = (mp + m) + (np + n) = m(p + 1) + n(p + 1).

(4) We leave the proofs of the associative and commutative laws for multiplication as well as (d) to the reader.

Definition 5. A set X is said to be finite if there is a bijection f : [0, n) → X for some n ∈ N and infinite otherwise. If f : [0, n) → X and g : [0, n) → X are bijections then h = g−1f : [0, n) → [0, m) is a bijection. The following theorem shows that m = n so that the natural number n in Definition 4 is uniquely determined by the finite set X. This number is called the cardinality of X and is denoted by |X|. Theorem 5. If h : [0, n) → [0, m) is a bijection then n = m. Proof. By induction on n. If n = 0 then [0, n) = ∅ so [0, m) = ∅ and m = 0. Assume that for some natural number n h : [0, n) → [0, m) bijective =⇒ n = m and let g : [0, n + 1) → [0, m) be a bijection. Then m > 0 and we set p = m−. Let q = g(n) and let f : [0, m) → [0, m) be the bijection defined by

f(p) = q, f(q) = p, and f(x) = x otherwise.

Then h = f ◦ g : [0, n + 1) → [0, m) is bijective and h(n) = p. It follows that the restriction of h to [0, n) is a bijection of [0, n) with [0, p). By our inductive hypothesis this gives n = p so that n + 1 = p + 1 = m. QED Theorem 6. If X is a finite set and Y ⊆ X then Y is finite and |Y | ≤ |X|. If |X| = 0 then X = ∅ and so Y = ∅, a finite set with |Y | = 0 = |X|. Suppose for some n ∈ N the statement is true for any finite set X’ with |X0| = n. Let X be a finite set with X = n + 1, let

5 f : [0, n + 1) → X be a bijection, let c = f(n) and let X0 = X − {c}. Then g : [0, n) → X0 defined by g(i) = f(i) is a bijection which shows that X0 is finite with |X0| = n. Now Y 0 = Y −{c} ⊂ X0 implies by the inductive hypothesis that Y 0 finite with |Y 0| ≤ |X0|. Since |Y | ≤ |Y 0| + 1 ≤ |X0| + 1 = |X| we are done. QED Theorem 7. If X,Y are disjoint finite sets then X ∪ Y is finite and

|X ∪ Y | = |X| + |Y |.

Proof. Let f : [0, m) → and g : [0, n) → [0, n) be bijections. Then the mapping h : [0, m + n) → X ∪ Y defined by ( f(i) if 0 ≤ i < m, h(i) = g(i − m) if m ≤ i < n is bijective since X ∩ Y = ∅. QED Corollary 5. If X,Y are finite and f : X → Y is injective then |X| ≤ |Y | with equality if and only if f is surjective. Proof. We have |Y | = |f(X)| + |Y − f(X)| with |f(X)| = |X|. QED Corollary 6. N is an infinite set. Proof. σ : N → N is injective but not surjective. QED

Corollary 7. If X1,X2,...,Xn are pairwise disjoint ﬁnite sets then X1 ∪ X1 ∪ · · · ∪ Xn is ﬁnite and

|X1 ∪ X1 ∪ · · · ∪ Xn| = |X1| + |X2| + ··· + |Xn|.

Pn Proof. The proof is by induction on n. The details are left to the reader. Note that i=1 ai = a1 + ··· + an is deﬁned inductively by

X0 nX+1 Xn ai = 0 (the empty sum is zero), ai = ( ai) + an+1. i=1 i=1 i=1 Theorem 8. If X, Y are ﬁnite then X ∪ Y , X ∩ Y are ﬁnite and

|X ∪ Y | + |X ∩ Y | = |X| + |Y |.

Proof. The set X ∪ Y is the union of the pairwise disjoint sets X − X ∩ Y , Y − X ∩ Y , X ∩ Y so that |X ∪ Y | = |X − X ∩ Y | + |Y − X ∩ Y | + |X ∩ Y |. Hence |X ∪ Y | + |X ∩ Y | = |X − X ∩ Y | + |X ∩ Y | + |Y − X ∩ | = |X| + |Y |. QED Theorem 9. If X, Y are finite then X × Y is finite and |X × Y | = |X||Y |. Proof. The proof is by induction on |Y . The inductive step follows from the fact that if c∈ / Y then X × (Y ∪ {c}) is the union of the disjoint sets X × Y and X × {c} and the fact that |X × {c}| = |X|. The details are left to the reader. Theorem 10. A set X is infinite if and only if there is an injective mapping f : N → X. Proof. The proof requires the following additional axiom

6 Axiom of Choice. If C is a collection of non-empty subsets of a set X then there is a function φ with domain C such that φ(C) ∈ C for any C ∈ C. We apply this axiom in the case C is the set of non-empty subsets of the inﬁnite set X. We then deﬁne an injective function f : N → X inductively by

f(n) = φ(X − f([0, n)).

This is a stronger form of recursion in which the value of the function at n depends on the values f(m) for m < n. That such a function exists can be proven by a variant of the proof given for simple recursion. The details are left to the reader. Definition 6. An infinite set X is said to be countable if there is a bijection f : N → X and uncountable otherwise. Theorem 11. P (N) is uncountable. Proof. P (N) is infinite since n 7→ {n} defines an injective mapping of N into P (N). If f is a mapping of N into P (N) let A be the set of those n ∈ N with n∈ / f(n). If A = f(m) for some m ∈ N then m ∈ A implies m∈ / f(m) by definition of A which implies m∈ / A, a contradiction. On the other hand m∈ / A implies m∈ / f(m) which implies m ∈ A, again a contradiction. We are thus led to conclude that the hypothesis A = f(m) for some m is false which implies that f is not surjective. Hence there is no surjective mapping and hence no bijective mapping from N to X. QED Definition 7. If m, n ∈ N then mn ∈ N is defined inductively by

m0 = 1, mn+1 = mnm.

Theorem 12. If X,Y are finite sets then the set XY of all mappings of Y into X is finite of cardinality |X||Y |. Proof. The proof is by induction on Y . The inductive step uses the fact that if c∈ / Y the mapping Y ∪{c} Y of X into X × X, defined by f 7→ (f|Y , f(c)), is bijective. Here f|Y denotes the restriction of f to Y ; for y ∈ Y , we have f|Y (y) = f(y). The details are left to the reader.

Corollary 8. The number of finite sequences (a0, a1, . . . , an−1) of length n where the terms ai lie in a finite set of cardinality m is mn. Corollary 9. |P ([0, n))| = 2n Proof. This follows from the fact that there is a bijection between P (X) and 2X where 2 = {0, 1}. Theorem 13. Let X, Y be finite sets with |X| = m ≤ |Y | = n. Then the number of injections of X into Y is n(n − 1) ... (n − m + 1). Proof. The proof is by induction on m. The inductive step uses the fact that if a∈ / X, the set of injective mappings of X ∪ {a} into Y is in one-to-one correspondence with the set S of pairs (f, b) where f : X → Y is injective and b ∈ Y −f(X). One proves that if T is the set of injective mappings of X into Y then |S| = |T |(n − m). To do this we partition S as follows. For each f ∈ T let Sf be the set of those pairs (f, b) with b∈ / f(X). Then |Sf | = |Y − f(X)| = (n − m) which implies the result. Corollary 10. If X is a finite set then the number of bijections of X with itself is n(n−1) ··· 1 = n!.

7 Corollary 11. Let X be a set with |X| = n. If Cn,m denotes the number of Y subsets of X with |Y | = m then m!Cm,n = n(n − 1) ··· (n − m + 1). Proof. For any subset Y of X with |Y | = m the number of injective mappings f : [0, m) → X with f([0, m)) = Y is m!. QED If m, n ∈ N with n = mk for some k ∈ N we say that m divides n in N and denote it by m|n. If n m 6= 0 then k is unique and is denoted by m or n/m. Thus the number of m element subsets of an n element set is n(n − 1) ··· (n − m + 1) n! C = = . n,m m! m!(n − m)!

Theorem 14. Every non-empty set of natural numbers has a smallest element. i.e. the natural numbers are well-ordered. Proof.. Let S be a nonempty set of natural numbers and let m ∈ S. If m is the smallest element of S we are done otherwise there is an element p of S with p < m. Thus we are reduced to showing that the ﬁnite set S ∩ [0, m) has a smallest element. We leave to the reader of proving by induction that any ﬁnite non-empty set of natural numbers has a smallest element. Strong Induction.(∀n ∈ N)((∀m ∈ N)(m < n =⇒ P (m)) =⇒ P (n)) =⇒ (∀n ∈ N)P (n) Proof. If (∀n ∈ N)P (n) is false let n be the smallest natural number for which P (n) is false. Then

(∀m ∈ N)(m < n =⇒ P (m)) is true. But this implies P (n) true by hypothesis which is a contradiction. Hence P (n) is true for all n. It would seem that in the strong form of induction you don’t have to prove P (0) is true. But you do since (∀m ∈ N)(m < 0 =⇒ P (m)) =⇒ P (0) is true if and only iﬀ P (0) is true. More generally, we have P (n) true for all n ≥ k if P (k) is true and for all n > k the truth of P (n) follows from the truth of P (m) for k ≤ m < n. The proof of this fact is left to the reader.