Elements of Set Theory Chapter 6: Cardinal Numbers and The Axiom of Choice Equinumerosity March 18 & 20, 2014 Lecturer: Fan Yang 1/28 2/28 What is the size of a set? X = f , , , g Y = f , , , , g ! = f0; 1; 2; 3; 4;::: g Do A and B have the same size? Does A have more elements than B? Are there exactly as many houses as people? Y has more elements than X. Yes, since there are 5 houses and 5 people. The infinite set ! has more elements than the finite sets X and Y . Yes, since there is a one-to-one correspondence between the two Do ! and Z have the same size? What about Q and Z? sets. 0 1 2 3 ! Definition 6.1 -3 -2 -1 0 1 2 3 Z A set A is said to be equinumerous or equipotent to a set B (written 3 1 11 − 2 2 4 A ≈ B) iff there is a bijection from A onto B. -3 -2 -1 0 1 2 3 Q A bijection from A onto B is also called a one-to-one correspondence Given two infinite sets A and B, how to compare their sizes? between sets A and B. 3/28 4/28 Example 6.1: Let A = fa; b; c; dg and B = fa; b; cg. Then A 6≈ B, since there is no bijection from A onto B. In general, for any two finite sets X and Y , if Y ⊂ X, then X 6≈ Y . Consider the following infinite sets. Clearly, Even ⊂ !. ! = f0; 1; 2; 3; 4; 5; 6;::: g, Even = f0; 2; 4; 6; 8; 10; 12;::: g = f2n j n 2 !g. Example 6.2: ! ≈ Even. Are there exactly as many knives as forks? Proof. The function f : ! ! Even defined by taking Yes, as there is a one-to-one correspondence between knives and forks. f (n) = 2n is a bijection. Indeed, f is injective, since n 6= m =) 2n 6= 2m. f is surjective, since for all m 2 Even, m = 2n for some n 2 !, and f (n) = 2n = m. 5/28 6/28 Example 6.3: [Galileo] Similarly, ! ≈ Sq, where ! = f0; 1; 2; 3; 4; 5;::: g, Example 6.5: ! × ! ≈ !. Sq = f0; 1; 4; 9; 16; 25;::: g = fn2 j n 2 !g. Proof. The function f : ! × ! ! ! defined as 1 f (m; n) = [(m + n)2 + 3m + n] since there is a bijection f : ! ! Sq defined as f (n) = n2. 2 is a bijection. Example 6.4: ! ≈ ! n f0g, since the function f : ! ! ! n f0g defined as ! f (n) = n+ is a bijection. 10 0 1 2 3 4 ! 11 6 12 1 2 3 4 ! n f0g 3 7 Remark: For infinite sets A; B, 1 4 8 13 A ⊂ B 6=) A 6≈ B ! 0 2 5 9 14 ! 7/28 8/28 Example 6.6: Q ≈ !. Example 6.7: (0; 1) ≈ R, where (0; 1) = fx 2 R j 0 < x < 1g. Proof. In the following picture, we specify a bijection f : ! ! Q. 3 3 3 3 3 1 2 3 4 5 Proof. By picture. Here (0; 1) has been bent into a semicircle with center P. Each point in (0; 1) is paired with its projection (from P) on [10] 2 2 2 [9] 2 2 the real line. 1 2 3 4 5 [1] 1 1 [2] 1 [8] 1 1 1 2 3 4 5 0 1 [0] 0 0 0 0 0 1 2 3 4 5 :::::: x [4] 1 1 [3] 1 [7] 1 1 − 1 − 2 − 3 − 4 − 5 [5] 2 2 2 [6] 2 2 − 1 − 2 − 3 − 4 − 5 r R 3 3 3 3 3 − 1 − 2 − 3 − 4 − 5 9/28 10/28 Recall : for any sets X and Y , An alternative proof: Claim: the mapping f :(0; 1) ! defined by R X putting Y = ff : X ! Y j f is a functiong: 1 1 f (x) = − x 1 − x Example 6.8: For any set A, we have }A ≈ A2. is a bijection. 1 1 1 1 For any a; b 2 (0; 1), if f (a) = f (b), namely, if a − 1−a = b − 1−b , then Proof. Define a function H : }A ! A2 by taking (b − a) 1 + 1 = 0 thereby a = b. Hence f is injective. ab (1−a)(1−b) H(B) = fB; For any r 2 , there is x = p 2 2 (0; 1) such that where f : A ! f0; 1g is the characteristic function of B defined as R r 2 B r+ 4+r +2 ( 1 if x 2 B; 1 1 1 1 f (x) = f (x ) = − = − 2 B r x 1 − x x 1 − x 0 if x 2 A n B: r p r r p r r + 4 + r 2 − 2 r + 4 + r 2 + 2 Claim: H is a bijection. = · p 2 r + 4 + r 2 Indeed, H is injective, since for any B; C ⊆ A, p 2 2 (r + 4 + r 2) − 2 H(B) = H(C) =) fB = fC =) 8x 2 A(fB(x) = fC(x)) = p = r: 2(r + 4 + r 2) =) 8x 2 A(x 2 B $ x 2 C) =) B = C; A Hence f is surjective. H is surjective, since for any function g 2 2, there is B = fx 2 A j g(x) = 1g ⊆ A such that H(B) = fB = g. 11/28 12/28 Theorem 6B (Cantor 1873) (a) The set ! is not equinumerous to the set R of real numbers. (b) No set is equinumerous to its power set. Theorem 6A Proof. (a) [Diagonal argument] Given any map f : ! ! R. We show For any sets A; B and C: that there exists z 2 R such that z 2= ran f . (a) A ≈ A. f (0) = 236: 0012 0 1 2 ::: (b) If A ≈ B, then B ≈ A. f (1) = −7: 73 374 7 4 ::: (c) If A ≈ B and B ≈ C, then A ≈ C. f (2) = 3: 1 41 15 5 ::: Proof. Exercise. f (3) = 0: 5 2 46 6 ::: . That is, ≈ is an “equivalence relation” on the class of all sets. The integer part of z is 0, and the (n + 1)st decimal place of z is 3 unless the (n + 1)st decimal place of f (n) is 3, in which case the (n + 1)st decimal place of z is 4. For example, in the case shown, z = 0:3433 ::: Clearly, z 6= f (n) for all n, as it differs from f (n) in the (n + 1)st decimal place. Hence z 2= ran f . 13/28 14/28 (b) Suppose g : A ! }A is surjective. Let B = fx 2 A j x 2= g(x)g: Since B 2 }A and g is surjective, there exists x0 2 A such that g(x0) = B: But then, by the definition of B, Finite Sets x0 2 B () x0 2= g(x0) = B; which is a contradiction. Hence there is no surjection from A onto }A, thus A 6≈ }A. Remark: We will soon be able to prove R ≈ }!. R has smaller size than }R, which has smaller size than }}R, etc. 15/28 16/28 Definition 6.2 3 4 5 2 A set is finite iff it is equinumerous to some natural number. Otherwise it is infinite. 1 That is, for any set A: 0 A is finite iff 9n 2 !(A ≈ n) A is infinite iff 8n 2 !(A 6≈ n) Example 6.9: The set A = fa0; a1; a2; a3g is finite, since A ≈ 4 via the bijection f : 4 ! A defined as: f (n) = an. 4 = f0; 1; 2; 3g A = fa0; a1; a2; a3g Example 6.10: Any finite set A is not equinumerous to an infinite set B. There are finitely many forks in the picture. Why? Proof. Suppose A ≈ B. As A is finite, A ≈ n for some n 2 !. From the Because there are exactly6 forks. This, in turn, is because there is a transitivity of ≈, it follows that B ≈ n, contradicting B being infinite. one-to-one correspondence between the natural number Next, we want to check that each finite set A is equinumerous to a 6 = f0; 1; 2; 3; 4; 5g and the set F of forks, or in other words, 6 ≈ F. unique natural number n. This requires the Pigeonhole Principle. 17/28 18/28 Proof. (of Pigeonhole Principle) We shall show that for any n 2 !, if f : n ! n Pigeonhole Principle: is a one-to-one function, then ran f = n (not a proper subset of n), namely, that the set T is inductive: If n items are put into m pigeonholes with n > m, T = fn 2 ! j ran f = n for any one-to-one function f : n ! ng: then at least one pigeon- Base case: 0 2 T , since the only function f : 0 ! 0 is the empty function hole must contain more f = ;, and ran ; = ;. than one item. Inductive step: Assume k 2 T and f : k + ! k + is a one-to-one function. We + + Here n = 10 and m = 9. show that ran f = k . Note that f k is a one-to-one function from k into k . Case 1: ran (f k) ⊆ k. Then f k : k ! k is a one-to-one function, as + + f : k ! k is 1-1. Thus, the assumption k 2 T implies ran f k = k. Pigeonhole Princinple Moreover, since f is 1-1, we must have f (k) = k.