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Chapter 7 Cardinality

Counting a given collection of finitely many objects amounts to establishing, for

some natural number n, a one-to-one correspondence from the set

{1, 2, 3, 4, . . . , n} to the given set of objects. We then say that the set has n

objects, where the number n refers to our intuitive conception of the set

{1, 2, 3, . . . , n}. In the same way, we have our understanding of the size of the set

N = {1, 2, 3, . . . , }. Carrying out the same analogy, if we can exhibit a one-to-one

correspondence from the set N to a given set S, then we say S has N elements.

But we easily see that the set N and the set of all the perfect squares are in

one-to-one correspondence, and hence, by the above argument, the latter set has

N elements. But, the set of perfect squares is a proper subset of N!! A proper

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subset of a set cannot have the same number of elements as the set itself. This

seemingly ridiculous conclusion forced Galileo to abandon this idea of counting

infinite sets.

In fact the above idea of counting infinite sets was not at fault. The paradox of

Galileo was not a problem. We just have to understand from a different point of

view infinite sets. The paradox was just a reflection of the characteristic property

of infinite sets. Infinite sets are very different from finite sets!

Definition (7.1.1) Two given sets A and B are said to have the same

cardinality if there is a one-to-one correspondence between. In such case, we write

card A = card B.

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Note At this point, there is no meaning attached to the symbol card A. We

only write card A = card B to express what the above definition says.

Examples

1. Let O be the set of all odd natural numbers and Z be the set of all the integers.

Then cardN = card O = cardZ.

2. For real numbers a < b and c < d, the function f : [a, b] → [c, d] given by

f(x) =d− c

b− a(x− a) + c

is a one-to-one correspondence. (Verify this !) Hence card [a, b] = card [c, d].

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Try to establish a one-to-one correspondence between (0, 1) and (0, 1]. Hence

show that card (0, 1) = card [0, 1].

Theorem (7.1.3) Let A,B and C be sets. Then

1. A has the same cardinality as itself.

2. If A has the same cardinality as B, then B has the same cardinality as A

3. If A has the same cardinality as B and B has the same cardinality as C, then

A has the same cardinality as C.

This theorem follows from the fact that 1. the identity function IA is a one-to-one

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correspondence, 2. the inverse of a one-to-one correspondence is also a one-to-one

correspondence, 3. the composition of two one-to-one correspondences is again a

one-to-one correspondence.

So if U is any collection of sets, then the relation “has the same cardinality” is an

equivalence relation on U .

Definition (7.1.4) If there is a one-to-one function from A into B, then we say

that the cardinality of A is less than or equal to the cardinality of B, and we

denote this by card A ≤ card B.

Theorem (7.1.5) Let A,B and C be sets.

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1. If A ⊆ B, then card A ≤ card B.

2. If card A ≤ card B and card B ≤ card C, then card A ≤ card C.

3. If C ⊆ A and B has the same cardinality as C, then cardB ≤ card A.

This is proved by using similar technique as Theorem 7.1.3. For instance, in 2 the

hypothesis says that there exist one-to-one functions f : A → B and g : B → C.

Then g ◦ f : A → C is also a one-to-one function. Hence card A ≤ card C.

Note Be careful, if f : A → B is a one-to-one function which is not surjective,

we cannot conclude that card B > card A. This can be illustrated by the earlier

example that A is the set of the perfect squares and B is the set of the natural

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numbers.

Definition (7.1.7) Let A and B be sets. Then we say that card A < card B if

card A ≤ card B and it is not true that card A = card B. (We may write this as

card A 6= card B.) So card A < card B means there exists a one-to-one function

from A into B but there does not exist a bijective function between A and B.

(Distinguish this from the statement that “ there exists a function from A to B

which is one-to-one but not onto.”)

In case the sets are finite, card A is a measure of the size of A, which is an

ordering for sets. Can we extend this to an ordering of sets, finite or infinite?

Let X be any set. Theorem 7.1.3 already shows that, on P(X), the relation

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card A = card B is an equivalence relation. Hence P(X) is partitioned into

equivalence classes.

7.2 Infinite Sets

Definition (7.2.1) A set A is said to be finite if A = φ or if there exists some

n ∈ N so that A can be put into one-to-one correspondence with {1, 2, 3, . . . , n}.

Definition (7.2.2) A set A is infinite if it is not finite.

Theorem (7.2.3) N is an infinite set.

Proof by contradiction. Assume that f : N→ {1, 2, 3, . . . , k} is a one-to-one

correspondence. Consider f−1({1}), f−1({2}), . . . , f−1({k}). Since f is

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one-to-one, each of these sets contains a single natural number. Let n be the sum

of these k (distinct) numbers. Then n does not belong to Dom(f), since n is

distinct from each of these k numbers (this applies only to the case k > 1). This

shows that f is not a function, a clear contradiction. Hence N is not a finite set.

Theorem (7.2.5) Let A be a set. The following statements about A are

equivalent.

(i) A is infinite.

(ii) A contains a sequence of distinct terms.

(iii) A can be put into one-to-one correspondence with a proper subset of itself.

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Proof ((i) ⇒ (ii)). Since A is infinite, A is nonempty. We can choose

(arbitrarily) an element from A, called a1. The set A\{a1} is not empty, otherwise

A = {a1} which is a finite set (with only one element). Hence we can choose

again an element, a2, say, from this. Clearly a2 6= a1. Next consider A\{a1, a2}.Again we see that it is nonempty and we can again choose from it an a3.

Inductively, after we have chosen k distinct elements from A, we find that the set

A\{a1, a2, a3, . . . , ak} is still nonempty and another element ak+1 can be chosen.

So, A contains the sequence a1, a2, a3, . . . of distinct terms.

((ii) ⇒(i)). We prove by contradiction. Let a1, a2, a3, . . . be the sequence with

distinct terms, where each term ai ∈ A. If A were finite then, for some natural

number k, there is a one-to-one correspondence f from A to {1, 2, 3, . . . , k}. For

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i = 1, 2, 3, . . . , let f−1(i) = ani. Let m > max{n1, n2, n3, . . . , nk}. Then am is

not equal to f−1(`) for any `, a clear contradiction. This proves the desired

implication.

((ii) ⇒ (iii)). Let (a1, a2, a3, . . .) be the sequence of distinct terms and let

B = A\{a1, a2, a3, . . .}. Note that B is a subset of A but B may be empty. Note

also that A = B ∪ {a1, a2, a3, . . .}. Clearly there is a one-to-one correspondence

between {a1, a2, a3, . . .} and {a2, a3, a4, . . .}. Together with the identity function

IB, we get a one-to-one correspondence from A to A\{a1}, as desired.

((iii) ⇒ (ii)). Let C be a proper subset of A and f : A → C is a one-to-one

correspondence. Choose (arbitrarily) a1 from A\C. Denote a2 = f(a1). Clearly a2

is distinct from a1 since a2 ∈ C while a1 does not. Denote a3 = f(a2). Again we

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see that a3 is distinct from a2, otherwise, by their definition and since f is

one-to-one, a2 = a1, a contradiction. Also, a3 6= a1. Denote a4 = f(a3). Again a4

is distinct from the previously found a1, a2, a3, otherwise, since f is one-to-one, a3

would be equal to a previous a1 or a2. Continue this process, we find, step by step,

ak+1 = f(ak) for k = 1, 2, 3, . . ., and these are all distinct terms from A. This

exhibits a sequence of distinct terms from A.

Applying the equivalence of (i) and (ii), we obtain the following.

Corollary (7.2.6) A set is infinite if and only if it has an infinite subset.

Also we can see that N is the smallest infinite set, since any infinite set, according

to Theorem 7.2.5 (ii), contains a sequence of distinct terms, and such sequence is

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in one-to-one correspondence with N. So we have the following.

Corollary (7.2.7) For any infinite set C, we have cardN ≤ card C.

7.3 Countable Sets

Definition (7.3.1) Let A be a set. A is said to be denumerable or countably

infinite if A and N have the same cardinality. A is said to be countable if it it

finite or denumerable.

Examples

1. N is denumerable.

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2. The set of the perfect squares, the set of all the odd integers, the set of all those

integers that are divisible by 3 are all denumerable.

If A is a countably infinite set, then we can list the elements of A as a sequence of

distinct terms

A = {a1, a2, a3, . . .}.

Theorem (7.3.3) Let C be a countable set and let B be any set. If

f : C → B is a one-to-one correspondence, then B is countable.

Proof If C is finite and non-empty, then there is a positive integer n and a

one-to-one correspondence g : {1, 2, 3, . . . , n} → C. The composition

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f ◦ g : {1, 2, 3, . . . , n} → B is also a one-to-one correspondence. This shows that

B is also finite. If C is countably infinite, we just change {1, 2, 3, . . . , n} to N.

This shows that B is also countably infinite.

Theorem (7.3.4) Every subset of a countable set is countable.

Proof Let A be a countable set and C be a subset of A. If A is finite, then by

Corollary 7.2.6, A has only finite subsets. If A is countably infinite, then A is a

sequence of distinct terms and C is either a finite set, or an infinite subsequence

of A and hence also countably infinite.

Theorem (7.3.5) Let f : N→ X be a onto function. Then X is countable.

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Proof For each x ∈ X, the set f−1({x}) is a nonempty subset of N. Let nx be

the least element of f−1({x}). Note that if x 6= y, then nx 6= ny. The subset

S = {nx : x ∈ X} of N is either finite or countably infinite, and the function

g : X → S given by g(x) = nx is clearly a one-to-one correspondence. Hence X is

also either finite or countably infinite.

Theorem (7.3.6) The set of rational numbers is countably infinite.

Exercise (7.3.9) The union of two countable sets is countable.

The idea is as follows. Let A and B be two countable sets. Suppose first that A

and B are disjoint (that is A ∩B = φ), then, by writing

A = {a1, a2, a3, . . .}, B = {b1, b2, b3, . . .}, we have

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A ∪B = {a1, b1, a2, b2, a3, b3, a4, . . .}, a sequence of distinct terms. If A and B

are not disjoint, we can let B1 = B\A, then B1 is a subset of B (and hence is

also countable) and A ∪B = A ∪B1 and A and B1 are disjoint.

More generally, we have the following.

Theorem (7.3.10) Countable union of countable sets is countable.

Proof First, Exercise 7.3.9 shows that finite union of countable sets is

countable.

Next, denumerable union of finite sets is denumerable. If A1, A2, A3, . . . , are finite

sets, then we can arrange the elements of their union in a sequence by first listing

all the terms of A1, then followed by the terms of A2 which has not appeared

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already, and then followed by the terms of A3 which has not yet appeared, and so

on.

Next, if we are given an infinite sequence of countable sets, the union of all those

sets which are finite gives rise to a finite or denumerable set (as shown in the

previous step). Therefore, it suffices to show that the union of the remaining

denumerable sets, if there exist some, is also denumerable.

Assume A1, A2, A3, . . . are denumerable sets, and assume for the time being that

they are mutually disjoint (that is Ai ∩Aj = φ for i 6= j). Write the set Ai as the

sequence

Ai = {ai1, ai2, ai3, . . .}.

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Then ∞⋃

i=1

Ai = {a11, a12, a21, a13, a22, a31, a14, a23, a32, a41, a15, . . .}.

This shows that the union of the A′is is also denumerable.

In general, if A1, A2, A3, . . . , are not mutually disjoint, we can consider the sets

A1, A2\A1, A3\(A1 ∪A2), A4\(A1 ∪A2 ∪A3), . . . ,

Ak\(A1 ∪A2 ∪A3 ∪ · · · ∪Ak−1), . . . .

These sets are mutually disjoint and their union is just the union of the original

A′is.

This proves Theorem 7.3.10.

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Exercise (7.3.11) If A and B are countable sets, then A×B is also a countable

set. For instance, the set of all finite sequences in the set S = {a, b} is a countable

set. (A finite sequence in S is a string (or word) of a′s and b′s.)

Definition (7.4.2) A set that is not countable is said to be uncountable.

Theorem (7.4.3) Cantor’s diagonalization argument The set of all the

real numbers between 0 and 1 is uncountable.

Proof We prove that there exists no onto function from N to (0, 1).

Let f : N→ (0, 1) be any function. Denote each f(n) by its decimal expansion

f(n) = 0.a1na2na3n · · · ,

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where each decimal digit ain ∈ {0, 1, 2, . . . , 9}. To avoid ambiguity, where there is

a choice, we take the terminating rather than the repeating expansion. For

instance we take 0.2000 · · · rather than 0.19999 · · · , which represents the same

real number. We see that the real number b = 0.b1b2b3 · · · where bi = 2 if aii 6= 2

and bi = 3 if aii = 2, is not in the range of f . Note that b ∈ (0, 1) and b 6= f(n)

for any n. This is because the n-th decimal digit of b is distinct from that of f(n).

Thus f is not onto.

Corollary (7.4.4) The set R of real numbers is uncountable.

Corollary (7.4.5) For any real numbers a < b, (a, b), [a, b), (a, b] and [a, b] are

all uncountable.

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Corollary (7.4.6) The set of irrational numbers is uncountable.

Theorem (7.4.8) Generalized Cantor diagonalization argument Let A be

a set. Then card A < cardP(A).

Proof Clearly, card A ≤ card P(A) since the function f : A → P(A) given by

f(a) = {a} is obviously injective. We show that no function g : A → P(A) can be

onto. Indeed, for any such function g, consider the subset J = {x ∈ A : x 6∈ g(x)}of A. We cannot have J = g(y) for any y ∈ A. (Is y in J?)

Hence, for any set A, we have

card A < cardP(A) < card P(P(A)) < · · · .

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So there are larger and larger infinite sets.

Exercise (7.4.9) P(N) has the same cardinality as the set of all sequences of 0’s

and 1’s.

7.5 Comparing Cardinalities

Consider a given set X and let A be a subset of X.

Definition (7.5.1) The cardinality of A ∈ P(X) is the equivalence class of all

subsets of X with the same cardinality, that is

card A = {B ⊆ X : there is a one-to-one correspondence between A and B}.

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So card A = card B now means the equivalence class containing A is the same as

the equivalence class containing B, that is A and B have the same cardinality,

there is a one-to-one correspondence between A and B.

Definition (7.5.3) We say that card A ≤ card B if there exists a one-to-one

function from A to B.

Now since card A and card B are equivalence classes, this definition requires

justification that it is well-defined, that is, if C ∈ card A and D ∈ card B, then

there also exists a one-to-one function from C to D. Indeed, there exist bijective

functions h : A → C, k : B → D and also a one-to-one function f : A → B.

Then k ◦ f ◦ h−1 is a one-to-one function from C to D. (Not difficult to verify.)

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The following theorem is now obvious.

Theorem (7.5.5) The relation ≤ on the set of cardinalities of subsets of X

is reflexive and transitive.

Theorem (Schroeder-Bernstein) If cardA ≤ card B and card B ≤ card A,

then card A = card B.

In other words, given two subsets A and B of X, if there exist one-to-one functions

f : A → B and g : B → A, then there exists a bijective function from A to B.

Sketch of proof

Step I Define A0 = A and B0 = B; A1 = f(A0) and B1 = g(B0). For each n ∈ N,

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define

A2n = g(A2n−1), A2n+1 = f(A2n),

B2n = f(B2n−1), B2n+1 = g(B2n).

Step II Then,

A0 ⊇ B1 ⊇ A2 ⊇ B3 ⊇ A4 ⊇ B5 ⊇ A6 ⊇ · · ·

and

B0 ⊇ A1 ⊇ B2 ⊇ A3 ⊇ B4 ⊇ A5 ⊇ B6 ⊇ · · · .

This can be proved by mathematical induction on the two chains simultaneously.

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Step III Using Step II, it is not difficult to show that

∞⋂

i=0

A2i =∞⋂

i=0

B2i+1 = (A∞, say), and∞⋂

i=0

B2i =∞⋂

i=0

A2i+1 = (B∞, say).

Step IV

{A∞, (A0\B1), (B1\A2) (A2\B3), . . .}

is a partition of the set A. Similarly,

{B∞, (B0\A1), (A1\B2), (B2\A3), . . .}

is a partition of the set B.

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Step V We now use the above partition of A to define the following function h : A → B.

h(x) =

f(x) if x ∈ A2i\B2i+1 for some i ∈ N ∪ {0},

g−1(x) if x ∈ B2i+1\A2i+2 for some i ∈ N ∪ {0},

f(x) if x ∈ A∞.

Step VI Clearly h is a function from A into B.

Step VII The function h is one-to-one and onto. This step requires more justification.

Schroeder-Bernstein theorem is thus proved and ≤ is a partial order on the set of

cardinalities of subsets of X.

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Example Let A = N = B. Let f : A → B and g : B → A be given by

f(n) = 3n, g(m) = 4m for n,m ∈ N. Then

A0 = N, A1 = 3N, A2 = 3 · 4N = 12N, A3 = 32 · 4N = 36N,

A4 = 32 · 42N = 144N, A5 = 33 · 42N = 432N, . . . ,

and

B0 = N, B1 = 4N, B2 = 3 · 4N = 12N, B3 = 42 · 3N = 48N,

B4 = 42 · 32N = 144N, B5 = 43 · 32N = 576N, . . . .

So,

A∞ = B∞ = φ.

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Since 69 ∈ A0\B1, by definition, h(69) = f(69) = 207. 8 ∈ B1\A2, hence

h(8) = g−1(8) = 2.

Exercise (7.5.7) Use the Schroeder-Bernstein theorem to prove that if

A ⊆ B ⊆ C and card A = card C, then A, B and C all have the same cardinality.

Theorem (7.5.8) Let A and B be subsets of X. Exactly one of the following

is true: (i) card A < card B, (ii) card A = card B, (iii) card A > card B. So

≤ is a total ordering in the set of cardinalities of subsets of X.

7.6 The Continuum Hypothesis

We know that cardN < cardR. Problem 8 of Assignment 4 shows that P(N) and

R have the same cardinality.

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Question Does there exist a subset S of R such that

cardN < card S < cardR?

The Continuum Hypothesis There does not exist a set S such that

cardN < card S < cardR.

The Generalized Continuum Hypothesis Given any infinite set A, there does

not exist a set B such that card A < card B < cardP(A).

In 1938, the German logician Kurt Godel proved that the Continuum Hypothesis is

consistent with the standard axioms of set theory. Hence it cannot be proved to be

false.

In 1964, the American mathematician Paul J. Cohen proved that the negation of

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the Continuum Hypothesis is also consistent with the standard axioms of set

theory. Hence it cannot be proved to be true either. In other words, the

Continuum Hypothesis is independent of the other axioms of set theory.