Dr. Neal, WKU MATH 337 Cardinality We now shall prove that the rational numbers are a countable set while ℜ is uncountable. This result shows that there are two different magnitudes of infinity. But we will show that there are, in fact, an infinite number of infinities. Definition 5.1. The cardinality of set A is the number of elements in A and is denoted by either n(A) or A .

Finite Sets Definition 5.2. (a) For all n ≥ 1, the first n natural numbers is the set Nn = {1, . . . , n } and Nn = n . (b) A non-empty set A is called finite and A = n if and only if there exists a bijection f : Nn→ A . The empty set is also finite and ∅ = 0. Note: If A and B are finite sets, then A × B is finite and A× B = A × B . Theorem 5.1. Two non-empty finite sets have the same cardinality if and only if they are equivalent. Proof. Suppose A = n = B . Then there exist bijections f : Nn→ A and g: Nn → B . Then f −1: A→ Nn is a bijection; hence, the composition g f

−1: A→ B is also a bijection. Therefore, A ~ B . Now suppose A ~ B where A and B are finite sets with A = n and B = m . Then there exist bijections f : A→ B, g: Nn → A, and h: Nm → B. Then the function

h−1 f g: Nn → Nm is also a bijection. Therefore, Nm = n ; i.e., m = n . Hence, A = B . QED Finite sets of different sizes cannot be equivalent. In fact, suppose A and B are finite sets and f : A→ B is a function (with the domain being all of A ). (i) If A < B , then f cannot be onto. (ii) If A > B , then f cannot be one-to-one. (iii) f is a bijection if and only if A = B . Note: Whenever A ~ Nn , then there is a bijection f : Nn→ A . We then can let f ( i) = ai ∈ A for 1 ≤ i ≤ n . Then A can be written as A = {a1,. . . ,an}. Theorem 5.2. A finite union of disjoint, non-empty finite sets is finite.

Dr. Neal, WKU Proof. Let A1 , A2 , . . . , Am be disjoint, finite sets denoted by

A1 = {a1,1, . . . , a1,n1} A2 = {a2,1, . . . , a2,n2} . . . Am = {am,1, . . . , am,nm}

Then

Aii=1

m has

€

M = nii =1

m∑ elements. We define a bijection from

€

f : Aii=1

m →{1, . . .,M} by

f (ai, j ) = j if i = 1n1 + . . . +ni−1 + j if i > 1

Thus,

Aii=1

m ~ NM . QED

Infinite Sets Definition 5.3. Set A is called infinite if it is not finite. Some infinite sets are: the natural numbers ℵ, the integers Z , the Rationals Q , the Reals ℜ , the even integers E , the odd integers O , the prime numbers P , and the irrationals I (because p is irrational for every prime p ).

Denumerable and Countable Sets Definition 5.4. (a) Set A is called denumerable (or countably infinite) if there exists a bijection between the natural numbers ℵ and set A . In other words, a denumerable set is equivalent to the natural numbers ℵ. (b) If set A is denumerable (A ~ ℵ), then A is said to have cardinality “Aleph Naught” denoted by A = ℵ0 . (c) Set A is called countable if it is either finite or denumerable. Example 5.1. The set of all odd numbers O is denumerable.

Proof. Define f :ℵ→ O by f (n) = n if n is odd

−n +1 if n is even

. Then f is a bijection; so the odd

numbers are countably infinite and O = ℵ0 .

ℵ 1 2 3 4 5 6 . . ↓ ↓ ↓ ↓ ↓ ↓ . . O 1 –1 3 –3 5 –5 . .

Dr. Neal, WKU Theorem 5.3. All denumerable sets are equivalent and have cardinality ℵ0 . Proof. Any two denumerable sets are equivalent to the natural numbers. Because set equivalence is an equivalence relation, any two denumerable sets are equivalent to each other. Moreover, by being denumerable, each has cardinality ℵ0 . QED Note: Whenever A is denumerable, then there is a bijection f :ℵ→ A . We then can let f ( i) = ai ∈ A for i ≥ 1. Then A can be written as A = {a1, a2 ,. . .}. Lemma 5.1. (a) If A is non-empty and finite, and B is denumerable with A∩ B = ∅ , then A∪ B is denumerable. (b) If A and B are denumerable with A∩ B = ∅ , then A∪ B is denumerable. (c) Any infinite subset S of a denumerable set A is denumerable. Proof. (a) Let A = {a1,. . . ,an} and B = {b1, b2 ,. . .} . Define a bijection f :A∪ B→ℵ by f (ai ) = i and f (bj ) = n + j . Note that f −1:ℵ→ A∪ B is also a bijection given by

f −1( i) =ai if 1 ≤ i ≤ nbi−n if i > n

.

(b) Let A = {a1, a2 ,. . .} and B = {b1, b2 ,. . .} . Define a bijection f :A∪ B→ℵ by f (ai ) = 2i and f (bj ) = 2 j − 1 . Note that f −1:ℵ→ A∪ B is also a bijection given by

f −1( i) =ai / 2 if i is evenb(i+1)/2 if i is odd

.

(c) Let A = {a1, a2 ,. . .}. Subset S is actually a subsequence S = {an1 , an2 ,. . .} where n1 is the first index such that an1 ∈S , n2 is the next index such that an2 ∈S, etc. So we can define a bijection f : ℵ→ S by f (k ) = ank . QED Corollary 5.1. The set integers Z is denumerable. Proof. First, the whole numbers W = {0}∪ℵ are denumerable by Lemma 1(a). The negative integers −ℵ are clearly in one-to-one correspondence with ℵ, so they are denumerable. Thus, Z = −ℵ∪W is denumerable Lemma 5.1 (b). Exercise: Find a direct one-to-one correspondence between ℵ and Z .

Dr. Neal, WKU Corollary 5.2. (a) The prime numbers are denumerable. (b) The odd numbers are denumerable. (c) The set of all multiples of a fixed natural number is denumerable. Proof. All are infinite subsets of the denumerable set Z . Lemma 5.2. Let A be an infinite set and let f :A→ℵ be one-to-one. Then A is denumerable. Proof. Redefine the function as f :A→ Range f . Then f is both one-to-one and onto. Thus, A is equivalent to Range f which must be infinite because A is infinite. But then Range f is an infinite subset of N ; hence, Range f is denumerable. Therefore A is denumerable because it is equivalent to the denumerable set Range f . QED Theorem 5.4. A countable union of disjoint denumerable sets is denumerable. Proof. Let A1 , A2 , . . . , An , . . . be disjoint denumerable sets and let A be their union. We can enumerate set each as follows:

A1 = {a11 , a12 , a13 , . . . } A2 = {a21 , a22 , a23 , . . . }

. . . An = {an1 , an2 , an3 , . . . }

etc.

Because all the An are disjoint, each ai j is distinct. Now define f :A→ℵ by

f (ai j ) = 2i 3 j .

By the uniqueness of prime factorizations, each ai j gives a different natural number. Thus, f is one-to-one. By Lemma 2, the infinite set A is denumerable. QED Recall: Countable means finite or denumerable. So the countable collection of sets may be finite A1 , A2 , . . . , An or denumerable A1 , A2 , . . . , An , . . . which does not affect the preceding proof. Corollary 5.3. A countable union of disjoint countable sets is countable. To prove this corollary, simply apply the preceding proof where each set An could be either finite or denumerable. Theorem 5.5. The set of positive rational numbers Q+ and the set of negative rational numbers Q− are denumerable.

Dr. Neal, WKU Proof. We can write Q+ as the set {a / b a ,b ∈ℵ, gcd(a ,b) = 1} . Define f :Q+ →ℵ by

f (a / b) = 2a 3b .

By the uniqueness of prime factorizations, each positive rational a / b gives a different natural number. Thus, f is one-to-one. By Lemma 2, the infinite set Q+ is denumerable. Because Q− ~ Q+ , we have that Q− is denumerable also. QED Corollary 5.4. The rational numbers Q are denumerable. Proof. Q = Q− ∪{0}∪ Q+ is denumerable by Lemma 5.1. QED

Uncountable Sets Definition 5.5. Set A is called uncountable if it is infinite but not denumerable. All finite sets can be counted. Denumerable sets are infinite, but are equivalent to the natural numbers 1, 2, 3, . . . , so they too can be counted. We aim to prove that the irrational numbers and the real numbers are both uncountable sets. For now, we can express the relationship between all types of infinite and countable sets with the following Venn diagram:

InfiniteCountable

Finite DenumerableUncountable

∅{1, . . . , n}

ℵℜ

Theorem 5.6. The interval (0, 1) is uncountable. Proof. Because (0, 1) contains all fractions of the form 1 / n for n ≥ 2, the interval is at least infinite. Suppose though that (0, 1) is denumerable (i.e., countably infinite). We shall obtain a contradiction.

Dr. Neal, WKU If (0, 1) is denumerable, then there exists a bijection f :ℵ→ (0, 1) . Thus, for every natural number n , 0 < f (n) < 1. So we consider the list f (1) , f (2) , f (3) , . . . Each such value in (0, 1) has a unique decimal expansion of the form 0.d1 d2 d3 . . . = d110

+d2100

+d31000

+ . . . , where

€

di ∈ {0, 1, 2, . . ., 9} and there is no ending string of all 9. (For example, we write 1/2 as 0.5000000 . . . rather than 0.4999999 . . . ) So the list of values in (0, 1) becomes

f (1) = 0.d11 d12 d13 d14 d15 . . . (e.g., 0.48732 . . . )

f (2) = 0.d21 d22 d23 d24 d25 . . . (e.g., 0.11111 . . . )

f (3) = 0.d31 d32 d33 d34 d35 .. . (e.g., 0.50000 . . . )

f (4) = 0.d41 d42 d43 d44 d45 . . . (e.g., 0.678678678 . . . )

etc.

Next, consider the terms d11 in f (1) , d22 in f (2) , d33 in f (3) , d44 in f (4) , etc. We then create a new element x = 0.e1 e2 e3 . . . in (0, 1) by letting ei = 5 if di i = 6 and letting ei = 6 if di i ≠ 6. (The choice of 5 and 6 is arbitrary.) For the example values above, the first four digits in x are 0.6665 . . . Because the i th decimal digit in x is different than the i th decimal digit in f (i) , x cannot equal f (i) for any i . So x is not in the range of f . Thus, f is not onto, which is a contradiction to it being a bijection. Thus, (0, 1)

cannot be denumerable. QED Corollary 5.5. The real numbers are uncountable. We can see this result in a couple of ways: (i) The interval (0, 1) is a proper subset of ℜ ; hence, ℜ must have at least as large a cardinality as (0, 1). (ii) We have seen previously that the interval (0, 1) is set equivalent to the real line. So if the real line were denumerable (equivalent to the naturals), then (0, 1) would be denumerable by transitivity. Corollary 5.6. The irrational numbers I are uncountable. Proof. We know that the Rationals Q are denumerable and the Reals ℜ are uncountable. If I were denumerable, then the disjoint union ℜ = Q∪ I would be denumerable by Lemma 5.1, which is a contradiction. QED Definition 5.6. We denote the cardinality of the real numbers by ℜ = c . If set A is equivalent to the real line, then A = c also. Corollary 5.7. Any open interval (a , b ) on the real line has cardinality c .

Dr. Neal, WKU

Continuum Hypothesis So far, we have three types of cardinalities: finite, denumerable, uncountable. These cardinalities create different numbers: 0, 1, 2, 3, . . ., n , . . . , ℵ0 , c . The symbol ℵ0 represents the countably infinite cardinality of the natural numbers and is also called first infinite ordinal. The symbol c represents the uncountably infinite cardinality of the real numbers. So we have two different types of infinity here, and we designate 0 < n < ℵ0 < c . The continuum hypothesis states that there are no other cardinalities between ℵ0 and c . Georg Cantor (1845–1918) believed but could not prove this result. In 1940, Kurt Godel proved that the continuum hypothesis is consistent with the axioms of set theory. In other words, accepting the continuum hypothesis as true causes no contradictions and set theory cannot disprove the hypothesis. In 1963, Paul Cohen proved that the continuum hypothesis is independent of the axioms of set theory. In other words, set theory also cannot prove the hypothesis. Therefore, the continuum hypothesis can be accepted as true or it can be accepted as false without causing any other contradictions. Convention. As of now, we have three types of cardinalities: finite, denumerable, and uncountable (n , ℵ0 , and c ). For sets of other possible types of cardinalities, we say that the sets have the same cardinality if and only if they are equivalent (which is already the case for cardinalities n , ℵ0 , and c ).

The Power Set Definition 5.7. Let A be any set. The power set of A , denoted P(A ), is the collection of all subsets of A . If A is finite with n elements, then there are 2n subsets of A . For example if A = ∅ , then P(A ) = {∅}, which is the set containing the null set. So for A = 0, then P(A) = 20 = 1. Also, if A = {a , b , c }, then the power set of A is

P(A ) = {∅ , {a }, {b }, {c }, {a , b }, {b , c }, {a ,c }, {a , b , c }} So if A = 3, then P(A) = 23 = 8. Cantor showed how to use the power set of infinite sets to create more types of infinities beyond ℵ0 and c . Cantor’s Theorem. For every set A , the cardinality of A is strictly less than the cardinality of its power set: A < P(A) .

Dr. Neal, WKU Proof. First, if A = ∅ , then A = 0 and P(A) = {∅} = 1. So the theorem holds. Next, assume A is finite with A = n > 0. To create a subset of A , we have two choices for every element in A : “in the subset” or “not in the subset.” Thus, there are 2n possibilities, which makes P(A) = 2n . Because n < 2n for all n ≥ 0 (which can be proved by induction), we have A < P(A) for a finite set A . Now suppose that A is infinite. Let f : A → P(A ) be defined by f (x) = { x }. If f (x) = f (y) , then { x } = { y } which implies that x = y . Thus, f is one-to-one. Hence, A is equivalent to the subset of P(A ) consisting of all singleton sets { x }. Because there are more elements in P(A ), we have A ≤ P(A) . We now must show that equality cannot hold. So suppose that A = P(A) . Then because of equal cardinalities, there would be a bijection g : A → P(A ). We shall show that this assumption leads to a contradiction. Let B be the subset of A defined by

B = {

€

x ∈ A | x ∉ g(x) }. The set B can be understood as follows: For every element x in A , the function value g(x) is a subset of A . So g(x) contains some of the elements of A . Either g(x) contains x or it does not contain x . If g(x) does not contain x , then we set aside x to belong to subset B . Because B is a subset of A , it belongs to P(A ). Then because g is onto, there must exist some

€

x ∈ A such that g(x) = B .

Question: Is

€

x ∈ B? If

€

x ∈ B , then by definition of B , we have that x ∉ g(x) = B , a contradiction. If x ∉ B = g(x) , then we set aside x to belong to B , another contradiction. Therefore the onto function g cannot exist. So A ≤ P(A) but A ≠ P(A) . We conclude that A < P(A) for infinite sets A .

How Many Infinities Are There? We know that the natural numbers ℵ are infinite. So by Cantor’s Theorem, the power set of ℵ has a larger cardinality. Then the power set of the power set has a still larger cardinality. Then the power set of the power set of the power set has an even larger cardinality. Denoting the cardinality of the natural numbers by ℵ0 , we obtain

ℵ0 = ℵ < P(ℵ) < P P(ℵ)( ) < P P P(ℵ)( )( ) < . . .

Dr. Neal, WKU As written in the above inequality, these increasing cardinalities are in one-to-one correspondence with the denumerable set {0, 1, 2, 3, . . . }. Thus we have a countably infinite number of infinities! However the real numbers ℜ are also infinite with a larger cardinality than ℵ. Denoting the cardinality of the real numbers by c , we obtain

c = ℜ < P(ℜ ) < P P(ℜ)( ) < P P P(ℜ)( )( ) < . . .

So it appears that we have another countably infinite set of infinite cardinalities. However, it has been proven that P(ℵ) = c . Hence, we really have just one countable string of infinities listed:

ℵ0 = ℵ < P(ℵ) = c = ℜ < P(ℜ ) < P P(ℜ)( ) < P P P(ℜ)( )( ) < . . . To prove that P(ℵ) = c , we first state a result (without proof) that was first postulated by Cantor. He was unable to prove it, but it was proven independently by Ernst Schroeder in 1896 and Felix Bernstein in 1898: Cantor-Schroeder-Bernstein Theorem. Let A and B be sets. If A ≤ B and B ≤ A , then A = B . This proof of this result is trivial for finite sets; however it is non-trivial for infinite sets. We demonstrate its usefulness though with our last result. Theorem 5.9. P(ℵ) = c . Proof. We first note that every number

€

x ∈ (0, 1] can be in a unique infinite binary expansion that does not end in an infinite string of zeros. For example, 1/2 can be written as 0.10000 . or as 0.01111. In the first case, 0.1 = 1× 2−1 = 1/2. In the second

case, 0.01111 =

2−kk=2

∞

∑ = (1 / 2)2

1− 1 / 2 = 1/2. So we write 1/2 as 0.01111 in binary.

Now define f : (0, 1] → P(ℵ) as follows: Let x =

ak 2−kk=1

∞

∑ , where ak = 0 or 1 for

all k where the binary expansion does not end in an infinite string of zeros. Let f (x) be the set of indices k in the expansion of x for which ak are non-zero. Then f (x) is a subset of ℵ (i.e., an element of P(ℵ)). By uniqueness of the binary expansion, f is well-defined and one-to-one. For if

f (x1) = f (x2) , then x1 = 2−kk∈ f (x1)∑ = 2−k

k ∈ f (x2 )∑ = x2 . By Theorem 5.6, since f is a one-

to-one function (but not necessarily onto), then c = (0, 1] ≤ P(ℵ) . Next let Ω be the set of denumerable subsets of ℵ. We now define g : P(ℵ) → ℜ by

Dr. Neal, WKU

€

g(A) =

2−k

k∈ A∑ if A ∈ Ω

2 + 2−k

k∈ A∑ if A ∉ Ω

.

Then g(A) gives a real number such that 0 < g(A) ≤ 1 or 2 < g(A) ≤ 3. If 0 < g(A) = g(B) ≤ 1, then

2−k

k ∈A∑ =

2−k

k ∈B∑ where A and B are infinite sets. Thus the sums cannot end

in an infinite string of zeros since the index sets are infinite. Thus by uniqueness of binary expansion, the index sets A and B must be equal. If 2 < g(A) = g(B) ≤ 3, then by subtracting 2,

2−k

k ∈A∑ =

2−k

k ∈B∑ where A and B are

finite sets. The sums here do end in an infinite string of zeros since the index sets are finite. Thus g(A) and g(B) must be rational numbers, and again by uniqueness of binary expansion, the index sets A and B must be equal. Thus g is one-to-one (but not necessarily onto); hence, P(ℵ) ≤ ℜ = c . Since c ≤ P(ℵ) and P(ℵ) ≤ c , we may conclude that P(ℵ) = c . QED

Addendum

The Schroeder-Bernstein Theorem states that if A1 ⊆ A2 ⊆ A3 and A1 is equivalent to A3 , then A2 is equivalent to A3 . We can apply the result to show that the open interval (0, 1) is equivalent to the half-open interval [0,1) . We simply note that [1 / 2,1) ⊆ (0, 1) ⊆ [0, 1) and [1 / 2,1) ~ [0,1) because [a ,b) ~ [c, d) by a direct linear

function f :[a , b)→ [c, d) defined by f ( x) = d − cb − a

( x − a) + c . Thus, (0, 1) ~ [0, 1), and in general (a, b) ~ [a ,b) , although it is difficult to define a direct one-to-one correspondence between the sets.

Exercises

1. For each problem, prove the result by defining a bijection f from ℵ into the set and verifying that f is a bijection. Also give the function formula for f

−1 . (a) Prove that the negative odd numbers O– = {–1, –3, –5, –7, . . . } are denumerable. (b) Prove that the even numbers E = { . . ., –6, –4, –2, 0, 2, 4, 6, . . . } are denumerable. 2. Prove that the intervals [0, 1) and [0, 1] are uncountable.