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

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Axioms for Set Theory the Following Is a Subset of the Zermelo-Fraenkel 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 2 X) which is read x is a member of X. 1. Equality Axiom: Sets X; Y are equal if they have the same members. (8X)(8Y )(X = Y $ (8x)(x 2 X $ x 2 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. (9Z)(8x)x2 = Z This set Z is denoted by fg or ;. 3. Doubleton Axiom: For any sets x; y there is a set whose elements are x; y. (8x)(8y)(9Z)(z 2 Z $ (z = x ^ z = y)) This set Z is denoted by fx; yg. The singleton set fxg is defined to be fx; xg. 4. Set Union Axiom: For any set X there is a set whose elements are the elements of the sets in the set X. (8X)(9Z)((8z)(z 2 Z $ (9Y )(Y 2 X ^ z 2 Y ))) S S This set is denoted by X. If A; B are sets the union of A and B is defined to be fA; Bg. This set is denoted by A [ B. We have x 2 A [ B $ (x 2 A _ x 2 B). 5. Power Set Axiom: For any set X there is a set whose elements are the subsets of X. (8X)(9Z)(z 2 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. (9X)(8x)(x 2 X $ (x 2 A ^ P (x))) This set is denoted by fx j x 2 A ^ P (x)g or fx 2 A j P (x)g. This axiom is actually actually an axiom scheme; one axiom for each P (x). If C is a non-empty set, say C1 2 C then (8C)(C 2 C ! x 2 C) () x 2 C1 ^ (8C)(C 2 C =) x 2 C) T T so that the set C = fx j (8C)(C 2 C ! x 2 C) exists by Axiom 6. TheT set C consists of those elements which lie in every set in the collection C. If C = fA; Bg then C = fx j x 2 A ^ x 2 Bg which is by definition A \ B, the intersection of the sets A and B. 1 7. Infinity Axiom: (9Z)(; 2 Z ^ (z 2 Z ! z [ fzg 2 Z)) A set Z is called an inductive set if ; 2 Z and z 2 Z ! z [ fzg 2 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 ; 2 D since every set C 2 C is inductive and hence ; 2 C. Let x 2 D. Then for every set C 2 C we have x 2 C which implies x [ fxg 2 C since C is inductive. Thus x [ fxg 2 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 [ fxg and 0 = ;; 1 = 0 + 1 = f0g; 2 = 1 + 1 = f0; 1g; 3 = 2 + 1 = f0; 1; 2g; 4 = 3 + 1 = f0; 1; 2; 3g; ::: then 0; 1; 2; 3; 4; :::: 2 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 2 S (b) n 2 S =) n + 1 2 S then S = N. Proposition 1. (8n 2 N)(8m 2 N)(m 2 n =) m ⊆ n) Proof. Let S = fn 2 N j (8m 2 N)(m 2 n =) m ⊆ n)g. By the Principle of Induction is suffices to prove that S is inductive. We have 0 2 S since (8m 2 N)(m 2 0 =) m ⊆ 0) is vacuously true as 0 = ; implies that m 2 0 is false for all m 2 N. Now suppose that n 2 S and that m 2 n + 1. Then m 2 n or m = n. If m 2 n we have by our inductive hypothesis n 2 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 2 S. Hence S is inductive. QED Corollary 1. (8m 2 N)(8n 2 N)(m + 1 = n + 1 =) m = n) Proof. If m + 1 = n + 1 then m 2 n + 1 which implies m 2 n or m = n and hence that m ⊆ n by Proposition 1. Similarly n 2 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 ¡ f0g. 2 Proof. The injectivity follows from Corollary 1. The mapping σ is not surjective since 0 6= n + 1 for any n 2 N. Now let S = f0g [ σ(N). It suffices to show that S = N. Since 0 2 we only have to show that n 2 S =) n + 1 2 S in order to prove that S is inductive. But n + 1 = σ(n) 2 S for any n 2 N so n + 1 2 S for any n 2 S. QED Corollary 2. If n 2 N and n 6= 0 there is a unique m 2 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. (8n 2 N)n2 = n Proof. Let S = fn j n2 = ng. Then 0 2 S since 0 2= 0 = ;. Suppose that n 2 S. Then n2 = n. We want to show that n + 1 2= n + 1. Suppose to the contrary that n + 1 2 n + 1 = n [ fng. Then n + 1 2 n or n + 1 = n which implies n + 1 ⊆ n by Proposition 1. But then n 2 n + 1 implies that n 2 n which is a contradiction. Hence n + 1 2 S and S is inductive. QED Corollary 3. (8n 2 N)(8m 2 N)(m 2 n =) m ½ n) Proposition 3. (8n 2 N)(8m 2 N)(m 2 n () m ½ n) Proof. By Corollary 3 we only have to prove (8n 2 N)(8m 2 N)(m ½ n =) m 2 n. Let S = fn 2 N j (8m 2 N)(m ½ n =) m 2 n)g. Then 0 2 S since m ½ 0 = ; is false for all m. Let n 2 S and suppose that m ½ n + 1 = n [ fng. Then m ⊆ n or n 2 m. If m ⊆ n then m = n or m 2 n by the inductive hypothesis. In either case m 2 n + 1. If n 2 m then n ½ m ½ n + 1, which is not possible. Hence n + 1 2 S and S is inductive. QED Proposition 4. (8n 2 N)(8m 2 N)(m ⊆ n _ n ⊆ m) Proof. By induction on n. Let S = fn 2 N j (8m 2 N)(m ⊆ n or n ⊆ m)g. Then 0 2 S since 0 = ; ⊆ m for any m. Let n 2 S and let m 2 N. Then m ⊆ n or n ½ m. But m ⊆ n implies that m ⊆ n + 1 and n ½ m implies n 2 m and hence n + 1 = n [ fng ⊆ m. Hence n + 1 2 S and S is inductive. QED Corollary 4. N is linearly ordered under inclusion. If m; n 2 N we also denote m ⊆ n by m · n. Definition 3. For n 2 N we let [0; n) denote the set fm 2 N j 0 · m < ng. Note that by the definition of N we have [0; n) = n for any n 2 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.
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