Elements of Set Theory

Chapter 6: Cardinal Numbers and The Axiom ofChoice

March 18 & 20, 2014

Lecturer: Fan Yang

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Equinumerosity

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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?

Y has more elements than X .The infinite set ω has more elements than the finite sets X and Y .Do ω and Z have the same size? What about Q and Z?

Z-3 -2 -1 0 1 2 3

ω0 1 2 3

Q-3 -2 -1 0 1 2 3

12

114− 3

2

Given two infinite sets A and B, how to compare their sizes?3/28

Are there exactly as many houses as people?Yes, since there are 5 houses and 5 people.Yes, since there is a one-to-one correspondence between the twosets.

Definition 6.1A set A is said to be equinumerous or equipotent to a set B (writtenA ≈ B) iff there is a bijection from A onto B.

A bijection from A onto B is also called a one-to-one correspondencebetween sets A and B.

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Are there exactly as many knives as forks?

Yes, as there is a one-to-one correspondence between knives andforks.

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Example 6.1: Let A = {a,b, c,d} and B = {a,b, c}. Then A 6≈ B, sincethere is no bijection from A onto B. In general, for any two finite sets Xand 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.

Proof. The function f : ω →Even defined by taking

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, andf (n) = 2n = m.

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Example 6.3: [Galileo] Similarly, ω ≈Sq, where

ω = {0,1,2,3,4,5, . . . },

Sq = {0,1,4,9,16,25, . . . } = {n2 | n ∈ ω}.

since there is a bijection f : ω →Sq defined as f (n) = n2.

Example 6.4: ω ≈ ω \ {0}, since the function f : ω → ω \ {0} defined asf (n) = n+ is a bijection.

ω \ {0}1 2 3 4

ω0 1 2 3 4

Remark: For infinite sets A,B,

A ⊂ B 6=⇒A 6≈ B !

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Example 6.5: ω × ω ≈ ω.

Proof. The function f : ω × ω → ω defined as

f (m,n) =12[(m + n)2 + 3m + n]

is a bijection.

ω

ω

0

1

2

4

5

7

8

9

11

12

13

14

3

6

10

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Example 6.6: Q ≈ ω.

Proof. In the following picture, we specify a bijection f : ω → Q.

− 31 − 3

2 − 33 − 3

4 − 35

− 21 − 2

2 − 23 − 2

4 − 25

− 11 − 1

2 − 13 − 1

4 − 15

01

02

03

04

05

11

12

13

14

15

21

22

23

24

25

31

32

33

34

35

. . . . . .[0] 0

1

[1] 11 [2]1

2

[3]− 12[4]− 1

1

[5]− 21 [6]− 2

3

[7]− 13

[8]13

[9]23[10] 2

1

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Example 6.7: (0,1) ≈ R, where (0,1) = {x ∈ R | 0 < x < 1}.

Proof. By picture. Here (0,1) has been bent into a semicircle withcenter P. Each point in (0,1) is paired with its projection (from P) onthe real line.

Rr

x

10

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An alternative proof: Claim: the mapping f : (0,1)→ R defined byputting

f (x) =1x− 1

1− xis a bijection.For any a,b ∈ (0,1), if f (a) = f (b), namely, if 1

a −1

1−a = 1b −

11−b , then

(b − a)(

1ab + 1

(1−a)(1−b)

)= 0 thereby a = b. Hence f is injective.

For any r ∈ R, there is xr =2

r+√

4+r2+2∈ (0,1) such that

f (xr ) =1xr− 1

1− xr=

(1xr− 2)

11− xr

=r +√

4 + r2 − 22

· r +√

4 + r2 + 2r +√

4 + r2

=(r +

√4 + r2)2 − 22

2(r +√

4 + r2)= r .

Hence f is surjective.

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Recall : for any sets X and Y ,X Y = {f : X → Y | f is a function}.

Example 6.8: For any set A, we have ℘A ≈ A2.

Proof. Define a function H : ℘A→ A2 by takingH(B) = fB,

where fB : A→ {0,1} is the characteristic function of B defined as

fB(x) =

{1 if x ∈ B,0 if x ∈ A \ B.

Claim: H is a bijection.

Indeed, H is injective, since for any B,C ⊆ A,

H(B) = H(C) =⇒ fB = fC =⇒ ∀x ∈ A(fB(x) = fC(x))=⇒ ∀x ∈ A(x ∈ B ↔ x ∈ C) =⇒ B = C;

H is surjective, since for any function g ∈ A2, there isB = {x ∈ A | g(x) = 1} ⊆ A such that H(B) = fB = g.

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Theorem 6AFor any sets A,B and C:

(a) A ≈ A.(b) If A ≈ B, then B ≈ A.(c) If A ≈ B and B ≈ C, then A ≈ C.

Proof. Exercise.

That is, ≈ is an “equivalence relation” on the class of all sets.

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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.

Proof. (a) [Diagonal argument] Given any map f : ω → R. We showthat there exists z ∈ R such that z /∈ ran f .

f (0) =

f (1) =

f (2) =

f (3) =...

236.

−7.

3.

0.

0 0 1 2 . . .

7 3 7 4 . . .

1 4 1 5 . . .

5 2 4 6 . . .

0 0 1 2 . . .

7 3 7 4 . . .

1 4 1 5 . . .

5 2 4 6 . . .

The integer part of z is 0, and the (n + 1)st decimal place of z is 3unless 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 decimalplace. Hence z /∈ ran f .

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(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,

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.

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Finite Sets

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5432

1

0

There are finitely many forks in the picture. Why?

Because there are exactly 6 forks. This, in turn, is because there is aone-to-one correspondence between the natural number6 = {0,1,2,3,4,5} and the set F of forks, or in other words, 6 ≈ F .

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Definition 6.2A set is finite iff it is equinumerous to some natural number. Otherwiseit is infinite.

That is, for any set A: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 thebijection 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.Proof. Suppose A ≈ B. As A is finite, A ≈ n for some n ∈ ω. From thetransitivity of ≈, it follows that B ≈ n, contradicting B being infinite.

Next, we want to check that each finite set A is equinumerous to aunique natural number n. This requires the Pigeonhole Principle.

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Pigeonhole Principle:

If n items are put into mpigeonholes with n > m,then at least one pigeon-hole must contain morethan one item.

Here n = 10 and m = 9.

Pigeonhole PrincinpleNo natural number is equinumerous to a proper subset of itself.

“Pigeons”

“Pigeonholes”

4 = {0,1,2,3}

x = {3,1,0}

In particular, n 6≈ m for any m ∈ n.19/28

Proof. (of Pigeonhole Principle) We shall show that for any n ∈ ω, if f : n→ nis a one-to-one function, then ran f = n (not a proper subset of n), namely,that the set T is inductive:

T = {n ∈ ω | ran f = n for any one-to-one function f : n→ n}.

Base case: 0 ∈ T , since the only function f : 0→ 0 is the empty functionf = ∅, and ran ∅ = ∅.Inductive step: Assume k ∈ T and f : k+ → k+ is a one-to-one function. Weshow 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, asf : k+ → k+ is 1-1. Thus, the assumption k ∈ T implies ran f � k = k .Moreover, since f is 1-1, we must have f (k) = k . Hence ran f = k ∪ {k} = k+.Case 2: ran (f � k) * k . Then f (p) = k for some p ∈ k . Define f̂ : k+ → k+ as

f̂ (p) = f (k),f̂ (k) = f (p) = k ,f̂ (x) = f (x) for other x ∈ k+.

Then f̂ : k+ → k+ is one-to-one, and f̂ (k) = k . By Case 1,

k+ = ranf̂ = ranf .20/28

Corollary 6CNo finite set is equinumerous to a proper subset of itself.

Proof. Let A be an arbitrary finite set. Suppose f : A→ B is a bijectionfor some B ⊂ A. Let g : A→ n be a bijection for some n ∈ ω.

The function g ◦ f ◦ g−1 is a bijection from n onto a proper subset g[B]of n (see picture on blackboard), contradicting the PigeonholePrinciple.

Corollary 6D(a) Any set equinumerous to a proper subset of itself is infinite.(b) The set ω is infinite.

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

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Corollary 6EAny finite set A is equinumerous to a unique natural number n, calledthe cardinal number or the cardinality of A, denoted by card A or |A|.

Proof. Assume that A ≈ m and A ≈ n for some natural numbers m,n.It follows that m ≈ n. Thus m 6⊂ n and n 6⊂ m, which by trichotomy andCorollary 4M implies that m = n is the case.

It follows from the corollary that A ≈ n ⇐⇒ |A| = n.

For example, since n ≈ n, we have |n| = n. If a,b, c,d are distinctobjects, then |{a,b, c,d}| = 4, as {a,b, c,d} ≈ {0,1,2,3}.Clearly, for finite sets A,B:

|A| = |B| ⇐⇒ (|A| = n ∧ |B| = n) ⇐⇒ (A ≈ n ∧ B ≈ n) ⇐⇒ A ≈ B.

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We also want to have cardinal numbers for infinite sets. In fact, whatsets these “numbers” are is not too crucial, but the essential demand isthat we will define the cardinality |A| for arbitrary set A in such a waythat

|A| = |B| ⇐⇒ A ≈ B

is the case. We postpone until Chapter 7 the actual definition of the set|A|. The information we need for the present chapter is embodied inthe following promise:

Promise: For any set A, we will define a set |A| in such a way that:1 |A| = |B| ⇐⇒ A ≈ B,2 for a finite set A, the cardinal number |A| is the natural number n

for which A ≈ n.

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Definition 6.3A cardinal number κ is the cardinality of some set, i.e., κ is a cardinalnumber iff κ = |A| for some set A.

For example,Every natural number n is a cardinal number, as n = |n|.|ω| is a cardinal number, and name (due to Cantor) |ω| = ℵ0.

There is a unique set whose cardinality is 0, namely the empty set ∅.

Given a cardinal number κ > 0, there are many sets A of cardinality κ.If one set A of cardinality κ is a finite set, then all sets of cardinality κare finite sets. In this case, κ is called a finite cardinal. Otherwise, κ iscalled an infinite cardinal.

Natural numbers are finite cardinals, and they are the only finitecardinals.ℵ0, |R|, |℘ω|, |℘℘ω| are infinite cardinals. Note ℵ0 6= |R| andℵ0 6= |℘ω| 6= |℘℘ω| as we have shown ω 6≈ R, and ω 6≈ ℘ω 6≈ ℘℘ω.

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Cardinal Arithmetic

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Elementary school math:

+ =

2 + 3 = 5

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.

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

Example 6.11: Prove: 2 + 2 = 4.

Proof. Let K = 2× {0} and L = 2× {1}. Clearly, K ∩ L = ∅ and|K | = 2 = |L|. We need to show that K ∪ L ≈ 4.

We have that

K ∪ L = (2× {0}) ∪ (2× {1}) = {(0,0), (1,0), (0,1), (1,1)}.

Clearly, the function f : 4→ K ∪ L defined by taking

f (0) = (0,0), f (1) = (1,0), f (2) = (0,1), f (3) = (1,1)

is a bijection.

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Theorem 6HAssume that K1 ≈ K2 and L1 ≈ L2. If K1 ∩ L1 = K2 ∩ L2 = ∅, thenK1 ∪ L1 ≈ K2 ∪ L2.

Proof. Since K1 ≈ K2 and L1 ≈ L2, there are bijections f : K1 → K2 andg : L1 → L2. Define a function h : K1 ∪ L1 → K2 ∪ L2 by taking

h(x) =

{f (x), if x ∈ K1,

g(x), if x ∈ L1.

Since K1 ∩ L1 = ∅, h is indeed a function. Claim: h is a bijection.

For any x1, x2 ∈ K1 ∪ L1 such that h(x1) = h(x2) = y , since K2 ∩ L2 = ∅, y is inexactly one set of K2 and L2. W.l.o.g., assume that y ∈ K2. Then x1, x2 ∈ K1.Since f is injective, x1 = x2. Hence h is injective.

We have that h[K1 ∪ L1] = h[K1] ∪ h[L1] = K2 ∪ L2, since h[K1] = f [K1] = K2and h[L1] = g[L1] = L2. Hence h is surjective.

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