Cardinality

Equivalent Sets Definition: We say that two sets are equivalent (sometimes called equipotent), denoted by A ~ B iff there exists a bijection f: A → B.

This is an equivalence relation on the class of all sets:

A ~ A for all sets A. (IA: A → A is a bijection for all sets A)

If A ~ B then B ~ A. (If f:A → B is a bijection, f -1:B → A is also) If A ~ B and B ~ C then A ~ C. (If f:A → B is a bijection and g:B → C is a bijection, then g⊙f: A → C is a bijection)

The equivalence classes under this relation are called cardinalities.

Finite SetsWe define some special sets of natural numbers: ℕ

1 = {1} ℕ

2 = {1,2} ℕ

3 = {1,2,3}, ..., ℕ

m = {1,2,...,m}

These sets are sometimes called initial segments.

Definition: A set A is finite iff A = ∅ or A ~ ℕm for some m ∈ℕ.

A set is infinite iff it is not finite.

We say that ∅ is of cardinality 0.

If A ~ ℕm

we say that A is of cardinality m. This makes sense since A and ℕ

m are in the same equivalence

class, i.e., "are of the same cardinality".

Finite CardinalitiesWhen two finite sets are of the same cardinality, say of cardinality k, then by definition, there is a bijection between them, and from each of them onto ℕ

k.

Since a bijection sets up a one-to-one pairing of the elements in the domain and codomain, it is easy to see that all the sets of cardinality k, must have the same number of elements, namely k.

Indeed, for any set that has k elements we can set up a bijection between that set and ℕ

k. So, for finite sets, all the sets in the

same cardinality have the same number of elements.

This is why we often refer to a cardinality as a cardinal number.

ℕSince the definition of an infinite set is a "negation", we should expect that most proofs about them will use contradiction methods.

Theorem: ℕ is an infinite set.

Pf: BWOC assume that ℕ is a finite set. Then there exists a bijection f: ℕ

k → ℕ for some k.

Let n = f(1) + f(2) + ... + f(k) + 1. Then n is a natural number (being the sum of natural numbers) and n > f(i) for any i. So n is in the codomain of f, but since it can not equal any f(i), it is not in the Rng(f). So, f is not onto →←

Denumerable SetsDefinition: A set is denumerable iff it is of the same cardinality as ℕ. (Also known as countably infinite.)

The cardinality of the denumerable sets is denoted ℵ0 which is read

as "aleph naught" or "aleph null". (ℵ is the first letter of the Hebrew alphabet.)

One may be tempted to say, in analogy with finite sets, that all denumerable sets have the same number of elements, or all denumerable sets have ℵ

0 elements. As our next example will

show, you probably should avoid this temptation.

ℕ~2ℕLet 2ℕ denote the set of all even natural numbers.

Theorem: 2ℕ is denumerable.

Pf: To prove this we must find a bijection f: ℕ → 2ℕ. Our candidate will be f(x) = 2x with domain ℕ. This f is one-to-one: Suppose f(x) = f(y). Then 2x = 2y, so x=y. This f is onto: Let y be an arbitrary element of the codomain 2ℕ Since y is an even natural number, y = 2k for some k in ℕ. Thus, f(k) = 2k = y, and f is onto. f is a bijection, and so, ℕ~2ℕ i.e., 2ℕ is denumerable. ÿ

So, the number of even natural numbers is the same as the number of all natural numbers ?????

So, what gives?On the one hand, our intuition tells us that this last statement can't be correct. Every even natural number is in ℕ, but ℕ also has odd natural numbers ... so ℕ must have more elements than 2ℕ !

But, the bijection between these sets, pairs up the elements exactly, so there are just as many elements in one set as in the other!

Something has to give ... these are contradictory results.

Realize that intuition is built up from experience. Our direct experience with sets is limited to finite sets ... so our intuition is usually ok for these. But we have no direct experience with infinite sets (for instance, you've never counted the elements in one), and our intuition leads us astray when it comes to infinite sets!

Denumerable + 1 = DenumerableTheorem: If A is denumerable and x ∉ A then A ∪ {x} is denumerable.

Pf: Since A is denumerable there exists a bijection g: ℕ → A. Now define f: ℕ → A ∪ {x} by: f(1) = x f(2) = g(1) f(3) = g(2) .... and in general f(n) = g(n-1) for n > 1. f is one-to-one since no two images of f can be the same (no two images of g are the same, and none of them is x) f is onto because g is onto A and x is certainly in Rng(f). So, f is a bijection. ÿ

Listable Sets?This is a Cherowitzo special definition – you will not find this anywhere in the literature.

A denumerable set is one whose elements can written in a list (an infinite list) where all the elements appear somewhere and no element appears twice. If you can create such a list of elements of the set, then you can define a function whose arguments are the elements of the set and whose values are the positions in the list where the elements appear. This function is a bijection between the set and ℕ ... thus proving that the set is denumerable. Thus, you can prove that a set is denumerable by creating this list.

So, maybe, denumerable sets should be called listable sets.

Union of Denumerable SetsTheorem: If A and B are disjoint denumerable sets then A ∪ B is denumerable.

Pf: Since A is denumerable we can list its elements as: a

1, a

2, a

3, ...

Since B is denumerable we can also list its elements as: b

1, b

2, b

3, ....

We can now form a list of the elements of A ∪ B this way: a

1, b

1, a

2, b

2, a

3, b

3, .....

Clearly, every element of A ∪ B appears exactly once on this list, so A ∪ B is denumerable. More formally, we can define the bijection f from A ∪ B onto ℕ by:

f x = {2k−1 when x=a k

2k when x=bk } .

ℤ is denumerable

We can apply the theorems we have just proved to obtain this result.

First, notice that -ℕ is denumerable. (Consider f(x) = -x ).

Since 0 ∉ ℕ we have that A = ℕ ∪ {0} is denumerable.

Then ℤ = ℕ ∪ {0} ∪ -ℕ = A ∪ -ℕ is denumerable.

ℚ is denumerableFrom the last chapter we know that ℕ×ℕ is denumerable sincethe function f: ℕ×ℕℕ given by f(n,m) = 2n-1(2m – 1) is a bijection.

Theorem: If A and B are denumerable sets, then A×B is denumerable.Pf: Since A and B are denumerable, there exist bijections f:Aℕ and g:Bℕ. Now consider the function h:A×Bℕ×ℕ given byh(a,b) = (f(a), g(b)). We show that h is a bijection. Suppose that h(a,b) = h(c,d) ⇒ (f(a),g(b)) = (f(c),g(d)) ⇒ f(a) = f(c) and g(b) = g(d). Since f is an injection a = c, and since g is an injection b = d.Now, let (c,d) be an arbitrary element of ℕ×ℕ. Since f is onto, there is an a ∈ ℕ with f(a) = c, and since g is onto there is a b ∈ ℕ with g(b) = d. So, h(a,b) = (f(a),g(b)) = (c,d), so h is a bijection. The composition of h with f above shows that A×B is denumerable.

ℚ is denumerableSince ℤ is denumerable, the last theorem shows that ℤ×ℤ is denumerable.

Theorem: Any infinite subset of a denumerable set is denumerable.

Pf: Since the original set is denumerable, its elements can be put into a list. By passing through this list and removing any element which is not in the subset, we obtain a list of the elements of the subset. ÿ

By identifying each fraction p/q with the ordered pair (p,q) in ℤ×ℤ we see that the set of fractions is denumerable. By identifying each rational number with the fraction in reduced form that represents it, we see that ℚ is denumerable.

Countable SetsDefinition: A countable set is a set which is either finite or denumerable.

In most theorems involving denumerable sets the term denumerable can be replaced by countable. Proofs involve extending the proofs for denumerable sets by checking the cases when one or more of the sets involved are finite. Thus: Theorem: If A is countable and x ∉ A then A ∪ {x} is countable. Theorem: If A and B are disjoint countable sets then A ∪ B is countable. Theorem: If A and B are countable sets, then A×B is countable. Theorem: Any subset of a countable set is countable.

Countable UnionTheorem: Let A be a countable family of countable sets. Then ∪A is countable.

This theorem while plausible can not be proved from the generally accepted axioms of set theory (Zermelo-Fraenkel axioms). To prove it we need an additional axiom known as the Axiom of Choice. We will postpone this proof until we have talked about this axiom.

The plausibility of the result comes from the fact that we can prove all the various cases except that of a denumerable family of disjoint denumerable sets without appealing to the Axiom of Choice.

Uncountable SetsA set which is not countable is called uncountable.

Theorem: The set of real numbers (0,1) is an uncountable set.

Before proving this result we need to say a few things about the decimal representation of real numbers. First of all, contrary to what you learned in elementary school, there is no such thing as a "terminating" decimal number. ¼ = 0.25 is not a valid representation of this real number, rather = 0.250000000000 ... a repeating decimal with 0 as the repeating portion.Secondly, decimal representations are not unique! Some numbers have more than one decimal representation. ¼ = 0.2499999999999999 ... . However, the only numbers with more than one representation are those having a repeating 0 or a repeating 9.

Uncountable Sets

Theorem: The set of real numbers (0,1) is an uncountable set.

Pf: BWOC suppose that this set of reals is countable. We may then list all the elements of the set, one above the other as below. We can now find a real number in the set which is not on the list (by construction) →← The construction is: the ith digit is chosen to be anything other than 0, 9 or the digit in ith place of the ith number of the list. ÿ

Example of the Construction0.1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 9 ...0.2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 9 5 ...0.3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 9 8 5 ...0.1 2 3 4 5 6 7 8 9 5 4 3 2 1 6 7 8 9 1 2 3 4 5 6 7 9 ...0.4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 9 7 7 7 ...0.9 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 ...0.5 6 5 6 5 6 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 ... ⋮

We construct a number not on the list as follows:

0.3 2 7 211 .......

⋱

1

cThe cardinality of the set (0,1) is denoted by c which stands for "the continuum".

There are many sets equivalent to (0,1). While we can prove the following equivalences algebraically (by writing out a specific bijection), we will give some geometric arguments that show that the bijections exist without explicitly writing them out.

Any open interval of real numbers, (a,b) is equivalent to (0,1).

0 1

a b

c

ℝ is equivalent to any open interval.

( ) → →

Order of Cardinal NumbersWe define an ordering on the cardinal numbers in the following way:

For any set A we denote the cardinal number (cardinality) of A by |A|. |A| = |B| iff A ~ B, otherwise |A| ≠ |Β|. |Α| ≤ |B| iff there exists an injection f: A → B. |A| < |B| iff |Α| ≤ |B| and |A| ≠ |Β|.

With these definitions, it is easy to see that "=" for cardinal numbers is an equivalence relation.Also, the relation "≤" is reflexive and transitive. We will see later that it is also antisymmetric.

Cardinal NumbersThe finite cardinal numbers are 1, 2, 3, ..., etc.

The inclusion map, f: ℕk → ℕ

m given by f(x) = x is an injection

iff k ≤ m. This means that | ℕk | ≤ |ℕm | iff k ≤ m. By the pigeon-hole principle, |ℕk| < |ℕm| iff k < m.

The same map f: ℕk → ℕ shows that | ℕk | ≤ ℵ

0 for all k. Since

ℕ is an infinite set we have | ℕk | < ℵ0

Similarly f: ℕ → ℝ shows that ℵ0 ≤ c. Since ℝ is uncountable

we have ℵ0 < c.

Cantor's TheoremTheorem: For any set A, |A| < |P (A) |.

Pf: For x ∈ A, the function f: A → P (A) given by f(x) = {x} is an injection, so |A| ≤ |P (A)|. Now assume that |A| = |P (A)|, that is, there exists a bijection g:A → P (A). Note that g(a) is a subset of A. Define S = {a ∈ A: a ∉ g(a)}. S is a subset of A and since g is onto there must exist some c in A so that g(c) = S. If c ∈ S, then by the definition of S, c ∉ g(c) = S. →← If c ∉ S, then by the definition of S, c ∈ g(c), so c ∈ S. →←Our assumption has led to a contradiction, so it must be false.Thus, |A| ≠ |P (A)| so |A| < |P (A)|.

ConsequencesCantor's theorem immediately implies that there are infinitely many infinite cardinal numbers because we can repeatedly use it to show that:

ℵ0 < |P (ℕ)| < |P ( P ( ℕ) ) | < |P ( P ( P ( ℕ))) | < ...

We shall see a little later where c fits in this string of inequalities.

It is also immediate that there can be no largest cardinal number.

Schröder-Bernstein TheoremTheorem: If |A| ≤ |B| and |B| ≤ |A| then |A| = |B|.

The proof of this result is fairly long and complicated. I will not present it, but I do encourage you to look at it in the text.

Example: The closed interval [0,1] has cardinality c.

Let the cardinality of [0,1] be |A|. The inclusion map f: (0,1) → [0,1] shows that c ≤ |A|. The inclusion map g: [0,1] → (-1,2) shows that |A| ≤ c. Thus, by the Schröder-Bernstein Theorem, |A| = c.

It follows that any closed or half-open interval has cardinality c.

cWe can use the Schröder-Bernstein Theorem to prove that c = | P (ℕ) |

Pf: We first define a function from f: P (ℕ) → (0,1) as follows: Let A be a subset of ℕ. f(A) = 0.a

1a

2a

3 ... where

a i={3 when i∈A5 when i∉A}.

It is easy to see that f is an injection.Now define g: (0,1) → P (ℕ) as follows: Each real number in (0,1) can be represented by a binary decimal. As with the base 10 decimals, the representation is not unique. (0.00111111...) = (0.010000 ...) We chose the form that has the repeated 0 ending. Let g(0.b

1b

2b

3....) = {i: b

i = 1}.

g is also an injection, so by Schröder-Bernstein (0,1) ~ P (ℕ).

The Axiom of ChoiceThe following axiom can not be proved or disproved (i.e., it is independent ) of the axioms of the Zermelo-Fraenkel set theory.

Axiom of Choice: Given any collection of non-empty sets A, there is a function F (called a choice function) which selects an element from each set in A.

For any finite collection A, this axiom is provable, but for infinite sets this can't be done.

A well known example which shows the subtlety of this axiom is:If a shoe store had an infinite number of pairs of shoes and socks, then choosing one shoe from each pair does not require the axiom but choosing one sock from each pair does.

The Comparability Theorem

There are several statements that are equivalent to the axiom of choice: (1) For every collection A of nonempty, disjoint sets, there is a set consisting of exactly one member of each set in A. (2) Every set can be well ordered. (3) For any two cardinal numbers, |A| and |B|, one and only one of the following must hold: |A| < |B| |A| = |B| |B| < |A|.(This is known as the comparability theorem)

Banach-Tarski ParadoxMost mathematicians are comfortable with the axiom of choice, however, by using it one can prove:

Banach-Tarski Theorem: Any solid sphere can be decomposed into five pieces that can be reassembled into two solid spheres with the same radius as the original.

Repeated use of this construction can lead to a pea being decomposed into a finite number of pieces which can be reassembled to form enough peas to fill a volume the size of the sun!!!!!

Results like this lead many to be wary of the axiom of choice, so proofs that use it often make this use explicit.

Theorem 5.36Theorem: Every infinite set has a denumerable subset.

Pf: Suppose A is an infinite set. Since A is not finite, A ≠∅. Choose an element a

1 in A. Now, A – {a

1}, is also infinite, so we may

choose an element a2 from it (a

1 and a

2 are of course different

elements of A). We continue in this way to chose elements. By the Axiom of Choice, we have constructed a set B = {a

1,a

2,...,a

n,...} of

elements of A which are all distinct. The function f: ℕ → B given by f(n) = a

n is a bijection, showing that B is a denumerable subset of

A. ÿ

The Continuum HypothesisWe have already seen that ℵ

0 < c.

We can now ask the question, is there a cardinal number strictly between these two?

Cantor conjectured that the answer is no, and this is known as the continuum hypothesis.

The statement that there is no cardinal number strictly between |A| and |P (A)| for any infinite set A is known as the generalized continuum hypothesis. (The continuum hypothesis is the case where A = ℕ.)