Axioms for Set Theory

Transcription

1 Axioms for Set Theory 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. This set Z is denoted by {} or. ( Z)( x)x / Z 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 ))) 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 C 1 C then ( C)(C C x C) x C 1 ( C)(C C = x C) so that the set C = {x ( C)(C C x C) exists by Axiom 6. The 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

2 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. Theorem 1. If C is a non-empty collection of inductive sets then C is inductive. 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. 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. 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}, 2 = = {0, 1}, 3 = = {0, 1, 2}, 4 = = {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. 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. 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

3 Proof. The injectivity follows from Corollary 1. The mapping σ is not surjective since 0 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. Corollary 2. If n N and n 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. 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. 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. 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 (a n ) = (a 0, a 1,..., a n,...) where a n = 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 (a 0, a 1,..., a n 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

4 Theorem 3. Let X be a set, let x 0 X and let φ be a mapping of X into itself. Then there is a unique infinite sequence (a n ) such that a 0 = x 0 and a n+1 = φ(a n ) 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 a 0 = x 0 and a m+1 = φ(a m ) 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 a b, i.e. that a m = b m for m < n. If F is the collection of all such functions a of length 1 then F is the required infinite sequence. We now introduce the operations of addition and multiplication of natural numbers. For n N we define σ n : N N inductively by σ 0 = 1 N (the identity mapping on N), σ n+1 = σ σ n. We can similarly define f n for any mapping f : X X where X is any set. Proposition 4. ( n N)σ n (0) = n Proof. By induction on n. The proposition is true for n = 0 since σ 0 (0) = 1 N (0) = 0. If σ n (0) = n for some n N then σ n+1 (0) = σ(σ n (0)) = σ(n) = n + 1. Definition 3. Let m, n N. Then m + n = σ n (m) = σ n σ m (0), mn = (σ m ) n (0). We have m + 1 = σ(m), m + 0 = 0 + m = m, m 1 = 1 m = m, m 0 = 0 m = 0 and Theorem 4. Let m, n, p N. Then m + (n + 1) = (m + n) + 1, m(n + 1) = mn + m. (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

5 Proof of Lemma. We proceed by induction on n. For n = 0 we have = 0 = and if 1 + n = n + 1 for some n we have 1 + (n + 1) = (1 + n) + 1 = (n + 1) + 1. 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 = = 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 1 f : [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. 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 X = n. Let X be a finite set with X = n + 1, let 5

6 f : [0, n + 1) X be a bijection, let c = f(n) and let X = X {c}. Then g : [0, n) X defined by g(i) = f(i) is a bijection which shows that X is finite with X = n. Now Y = Y {c} X implies by the inductive hypothesis that Y finite with Y X. Since Y Y + 1 X + 1 = X we are done. 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 =. 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. Corollary 6. N is an infinite set. Proof. σ : N N is injective but not surjective. Corollary 7. If X 1, X 2,..., X n are pairwise disjoint finite sets then X 1 X 1 X n is finite and X 1 X 1 X n = X 1 + X X n. Proof. The proof is by induction on n. The details are left to the reader. Note that n i=1 a i = a a n is defined inductively by 0 a i = 0 (the empty sum is zero), i=1 n+1 n a i = ( a i ) + a n+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. 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

7 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 {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. Definition 7. If m, n N then m n N is defined inductively by m 0 = 1, m n+1 = m n m. Theorem 12. If X, Y are finite sets then the set X Y 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 of X Y {c} into X Y X, defined by f (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 (a 0, a 1,..., a n 1 ) of length n where the terms a i lie in a finite set of cardinality m is m n. Corollary 9. P ([0, n)) = 2 n Proof. This follows from the fact that there is a bijection between P (X) and 2 X 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 S f be the set of those pairs (f, b) with b / f(x). Then S f = 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

8 Corollary 11. Let X be a set with X = n. If C n,m denotes the number of Y subsets of X with Y = m then m!c m,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!. 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 m 0 then k is unique and is denoted by n 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. 8