Cardinality

Equivalent Sets

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

When two sets are equivalent, we say that they have the same cardinality.

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 ∅ has cardinality 0.

If A ~ ℕm

we say that A has cardinality m.

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.

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.

So, what gives?

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=ak

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, then AxB 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 Construction

We construct a number not on the list as follows:

0. .......

⋱

8 5 2 1 1 2 8

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

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.

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