Elements of Set Theory

Chapter 6: Cardinal Numbers and The Axiom of Choice Equinumerosity March 18 & 20, 2014

Lecturer: Fan Yang

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What is the size of a set?

X = { , , , }

Y = { , , , , } ω = {0, 1, 2, 3, 4,... } 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 ω has more elements than the finite sets X and Y . Yes, since there is a one-to-one correspondence between the two Do ω and Z have the same size? What about Q and Z? sets. 0 1 2 3 ω 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 = {a, b, c, d} and B = {a, b, c}. 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 ⊂ ω.

ω = {0, 1, 2, 3, 4, 5, 6,... },

Even = {0, 2, 4, 6, 8, 10, 12,... } = {2n | n ∈ ω}.

Example 6.2: ω ≈ Even. Are there exactly as many knives as forks? Proof. The function f : ω → 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 ∈ Even, m = 2n for some n ∈ ω, and f (n) = 2n = m.

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Example 6.3: [Galileo] Similarly, ω ≈ Sq, where ω = {0, 1, 2, 3, 4, 5,... }, Example 6.5: ω × ω ≈ ω.

Sq = {0, 1, 4, 9, 16, 25,... } = {n2 | n ∈ ω}. Proof. The function f : ω × ω → ω defined as 1 f (m, n) = [(m + n)2 + 3m + n] since there is a bijection f : ω → Sq defined as f (n) = n2. 2 is a bijection. Example 6.4: ω ≈ ω \{0}, since the function f : ω → ω \{0} defined as ω f (n) = n+ is a bijection. 10 0 1 2 3 4 ω 11 6

12 1 2 3 4 ω \{0} 3 7

Remark: For infinite sets A, B, 1 4 8 13

A ⊂ B 6=⇒ A 6≈ B ! 0 2 5 9 14 ω

7/28 8/28 Example 6.6: Q ≈ ω. Example 6.7: (0, 1) ≈ R, where (0, 1) = {x ∈ R | 0 < x < 1}. Proof. In the following picture, we specify a bijection f : ω → 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

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Recall : for any sets X and Y , An alternative proof: Claim: the mapping f :(0, 1) → defined by R X putting Y = {f : X → Y | f is a function}. 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 ∈ (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 ∈ , there is x = √ 2 ∈ (0, 1) such that where f : A → {0, 1} is the characteristic function of B defined as R r 2 B r+ 4+r +2 ( 1 if x ∈ B, 1 1  1  1 f (x) = f (x ) = − = − 2 B r x 1 − x x 1 − x 0 if x ∈ A \ B. r √ r r √ r r + 4 + r 2 − 2 r + 4 + r 2 + 2 Claim: H is a bijection. = · √ 2 r + 4 + r 2 Indeed, H is injective, since for any B, C ⊆ A, √ 2 2 (r + 4 + r 2) − 2 H(B) = H(C) =⇒ fB = fC =⇒ ∀x ∈ A(fB(x) = fC(x)) = √ = r. 2(r + 4 + r 2) =⇒ ∀x ∈ A(x ∈ B ↔ x ∈ C) =⇒ B = C; A Hence f is surjective. H is surjective, since for any function g ∈ 2, there is B = {x ∈ A | g(x) = 1} ⊆ A such that H(B) = fB = g. 11/28 12/28 Theorem 6B (Cantor 1873) (a) The set ω 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 : ω → R. We show For any sets A, B and C: that there exists z ∈ R such that z ∈/ 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 ∈/ ran f . 13/28 14/28

(b) Suppose g : A → ℘A is surjective. Let

B = {x ∈ A | x ∈/ g(x)}.

Since B ∈ ℘A and g is surjective, there exists x0 ∈ A such that

g(x0) = B. But then, by the definition of B, Finite Sets

x0 ∈ B ⇐⇒ x0 ∈/ 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 ≈ ℘ω. 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 ∃n ∈ ω(A ≈ n) A is infinite iff ∀n ∈ ω(A 6≈ n)

Example 6.9: The set A = {a0, a1, a2, a3} is finite, since A ≈ 4 via the bijection f : 4 → A defined as: f (n) = an. 4 = {0, 1, 2, 3}

A = {a0, a1, a2, a3} 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 ∈ ω. 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 = {0, 1, 2, 3, 4, 5} 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 ∈ ω, 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 = {n ∈ ω | ran f = n for any one-to-one function f : n → n}. then at least one pigeon- Base case: 0 ∈ 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 ∈ 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 ∈ T implies ran f  k = k. Pigeonhole Princinple Moreover, since f is 1-1, we must have f (k) = k. Hence ran f = k ∪ {k} = k +. ˆ + + No natural number is equinumerous to a proper subset of itself. Case 2: ran (f  k) * k. Then f (p) = k for some p ∈ k. Define f : k → k as ˆf (p) = f (k), 4 = {0, 1, 2, 3} “Pigeons” ˆf (k) = f (p) = k, ˆf (x) = f (x) for other x ∈ k +. x = {3, 1, 0} “Pigeonholes” Then ˆf : k + → k + is one-to-one, and ˆf (k) = k. By Case 1, + ˆ In particular, n 6≈ m for any m ∈ n. k = ranf = ranf . 19/28 20/28 Corollary 6C Corollary 6E No finite set is equinumerous to a proper subset of itself. Any finite set A is equinumerous to a unique natural number n, called Proof. Let A be an arbitrary finite set. Suppose f : A → B is a bijection the cardinal number or the cardinality of A, denoted by card A or |A|. for some B ⊂ A. Let g : A → n be a bijection for some n ∈ ω. Proof. Assume that A ≈ m and A ≈ n for some natural numbers m, n. The function g ◦ f ◦ g−1 is a bijection from n onto a proper subset g[B] It follows that m ≈ n. Thus m 6⊂ n and n 6⊂ m, which by trichotomy and of n (see picture on blackboard), contradicting the Pigeonhole Corollary 4M implies that m = n is the case. Principle. It follows from the corollary that A ≈ n ⇐⇒ |A| = n. Corollary 6D For example, since n ≈ n, we have |n| = n. If a, b, c, d are distinct (a) Any set equinumerous to a proper subset of itself is infinite. objects, then |{a, b, c, d}| = 4, as {a, b, c, d} ≈ {0, 1, 2, 3}. (b) The set ω is infinite. Clearly, for finite sets A, B:

Proof. (a) follows immediately from Corollary 6C. |A| = |B| ⇐⇒ (|A| = n ∧ |B| = n) ⇐⇒ (A ≈ n ∧ B ≈ n) ⇐⇒ A ≈ B. (b) follows from (a), since e.g. ω ≈ ω \{0}, via the bijection f : ω → ω \{0} defined as f (n) = n+.

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Definition 6.3 A cardinal number κ is the cardinality of some set, i.e., κ is a cardinal We also want to have cardinal numbers for infinite sets. In fact, what number iff κ = |A| for some set A. sets these “numbers” are is not too crucial, but the essential demand is that we will define the cardinality |A| for arbitrary set A in such a way For example, that Every natural number n is a cardinal number, as n = |n|. |A| = |B| ⇐⇒ A ≈ B |ω| is a cardinal number, and name (due to Cantor) |ω| = ℵ0. is the case. We postpone until Chapter 7 the actual definition of the set There is a unique set whose cardinality is 0, namely the empty set ∅. |A|. The information we need for the present chapter is embodied in the following promise: Given a cardinal number κ > 0, there are many sets A of cardinality κ. Promise: For any set A, we will define a set |A| in such a way that: If one set A of cardinality κ is a finite set, then all sets of cardinality κ are finite sets. In this case, κ is called a finite cardinal. Otherwise, κ is 1 |A| = |B| ⇐⇒ A ≈ B, called an infinite cardinal. 2 for a finite set A, the cardinal number |A| is the natural number n for which A ≈ n. Natural numbers are finite cardinals, and they are the only finite cardinals.

ℵ0, |R|, |℘ω|, |℘℘ω| are infinite cardinals. Note ℵ0 6= |R| and ℵ0 6= |℘ω|= 6 |℘℘ω| as we have shown ω 6≈ R, and ω 6≈ ℘ω 6≈ ℘℘ω.

23/28 24/28 Elementary school math: 2 + 3 = 5

Cardinal Arithmetic + =

Definition 6.4 (Addition) Let κ and λ be two cardinal numbers. Define κ + λ := |K ∪ L|, where K and L are any disjoint sets with |K | = κ and |L| = λ. 25/28 26/28

Note: It is always possible to take disjoint sets K , L in the above definition, as K × {0} and L × {1} are always disjoint. Theorem 6H

Assume that K1 ≈ K2 and L1 ≈ L2. If K1 ∩ L1 = K2 ∩ L2 = ∅, then Example 6.11: Prove: 2 + 2 = 4. K1 ∪ L1 ≈ K2 ∪ L2.

Proof. Since K1 ≈ K2 and L1 ≈ L2, there are bijections f : K1 → K2 and g : L → L . Define a function h : K ∪ L → K ∪ L by taking Proof. Let K = 2 × {0} and L = 2 × {1}. Clearly, K ∩ L = ∅ and 1 2 1 1 2 2 ( |K | = 2 = |L|. We need to show that K ∪ L ≈ 4. f (x), if x ∈ K , h(x) = 1 We have that g(x), if x ∈ L1.

Since K1 ∩ L1 = ∅, h is indeed a function. Claim: h is a bijection. K ∪ L = (2 × {0}) ∪ (2 × {1}) = {(0, 0), (1, 0), (0, 1), (1, 1)}. For any x1, x2 ∈ K1 ∪ L1 such that h(x1) = h(x2) = y, since K2 ∩ L2 = ∅, y is in exactly one set of K2 and L2. W.l.o.g., assume that y ∈ K2. Then x1, x2 ∈ K1. Clearly, the function f : 4 → K ∪ L defined by taking Since f is injective, x1 = x2. Hence h is injective. f (0) = (0, 0), f (1) = (1, 0), f (2) = (0, 1), f (3) = (1, 1) We have that h[K1 ∪ L1] = h[K1] ∪ h[L1] = K2 ∪ L2, since h[K1] = f [K1] = K2 and h[L1] = g[L1] = L2. Hence h is surjective. is a bijection.

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