Analysis with an introduction to proof 5th by steven lay ch02b

These intuitive ideas are fine for small finite sets, but how can we compare the size of infinite sets like or?. If a cardinal number is not finite, it is called transfinite.. The relati

Chapter 2

Sets and Functions

Section 2.4 Cardinality

How can we compare the sizes of two sets?

If S = {x  : x2 = 9}, then S = {–3, 3} and we say that S has two elements.

If T = {1,7,11}, then T has three elements and we think of T as being “larger” than S.

These intuitive ideas are fine for small (finite) sets, but how can we compare the size of (infinite) sets like or ?

Certainly, it is reasonable to say that two sets S and T are the same size if there is a

bijective function f : S  T, for this function will set up a one-to-one correspondence

between the elements of each set

Definition 2.4.1

Two sets S and T are called equinumerous, and we write S ~ T, if there exists a bijective function from S onto T.

It is easy to see that “~” is an equivalence relation on any family of sets, and it partitions the family into disjoint equivalence classes

With each equivalence class we associate

a cardinal number that we think of as giving the size of the set

In dealing with finite sets, it will be convenient to abbreviate the set {1, 2,…,n} by I n

Definition 2.4.3

A set S is said to be finite if S =  or if there exists and a bijection f: I n  S

If a set is not finite, it is said to be infinite

Definition 2.4.4

The cardinal number of I n is n, and if S ~ I n , we say that S has n elements

The cardinal number of  is taken to be 0

If a cardinal number is not finite, it is called transfinite

Definition 2.4.6

A set S is said to be denumerable if there exists a bijection f :  S

If a set is finite or denumerable, it is called countable

If a set is not countable, it is uncountable

The cardinal number of a denumerable set is denoted by 0

“Aleph” is the first letter of the Hebrew alphabet.

The relationships between the various “sizes” of sets is shown below.

countable sets infinite sets

denumerable sets

finite sets

uncountable sets

There are two kinds of countable sets: finite sets and denumerable sets

The denumerable sets are infinite sets

The infinite sets that are not denumerable sets are uncountable sets

We have not yet shown that there exist any infinite sets that are not denumerable

Before doing so, we look at some properties of countable sets

Countable Sets

It would seem at first glance that the set of natural numbers should be “bigger”

than the set E of even natural numbers Indeed, E is a proper subset of and, in fact,

it contains only “half” of But what is “half” of 0?

Example 2.4.7

Our experience with finite sets is a poor guide here, for and E are actually equinumerous!

E : 2 4 6 8 10 12 14

: 1 2 3 4 5 6 7

The function f (n) = 2n is a bijection from onto E, so E also has cardinality 0

f (n) = 2n

If a nonempty set S is finite, then there exists n  and a bijection f : I n  S.

Using the function f , we can count off the members of S as follows:

f (1), f (2), f (3), …, f (n).

Letting f (k) = s k for 1  k  n, we obtain the more familiar notation S = {s1,s2,…,s n}

The same kind of counting process is possible for a denumerable set, and this is why

both kinds of sets are called countable

For example, if T is denumerable, then there exists a bijection g :  T,

and we may write T = {g (1), g (2), g (3),…} or T = {t1, t2, t3,…}, where g (n) = t n

This ability to list the members of a set as a first, second, third and so on, characterizes countable sets If the list terminates, then the set is finite

If the list does not terminate, then the set is denumerable

Theorem 2.4.9

Let S be a countable set and let T  S Then T is countable.

The proof of this “obvious” result is not difficult, but it uses the Well-Ordering Property

of that we won’t encounter until Chapter 3.

Theorem 2.4.10

Let S be a nonempty set The following three conditions are equivalent

(a) S is countable.

(b) There exists an injection f : S 

(c) There exists a surjection g :  S.

The following theorem follows readily from Theorem 2.4.9 The details of the proof

are in the text

Example 2.4.11(a)

Suppose S and T are countable sets Then S  T is countable.

Theorem 2.4.10 implies that there exist surjections f :  S and g :  T

Define h:  S  T by 1

, if is odd 2

( )

, if is even

2

n

h n

n

   





 



 

  



1 2 3 4

1 2 3 4 5 6

3 4









S









T

f

g

h

f(1) = h(1)

f(2) = h(3)

f(3) = h(5)

g(1) = h(2)

g(2) = h(4)

g(3) = h(6)

let h use the odd integers with f to count S and the even integers with g to count T.

The first row contains all the positive integers

We can adapt this process to show that the set of rational numbers is countable

We begin by constructing a rectangular array of fractions

The second row contains all the fractions with denominator equal to 2 The third row contains all the fractions with denominator equal to 3, etc

Moving along each diagonal of the array in the manner indicated,

we obtain a listing of all the elements in the positive rationals :+

This listing defines a surjection f :  ,

so is countable

+ +

But, =  {0}  , + – so is countable, too.

In the text this is generalized to show that any union

of a countable family of countable sets is countable

3

3

1, 2, , , , 3, 4, , ,

Theorem 2.4.12

The set of real numbers is uncountable

Proof: Since any subset of a countable set is countable (Theorem 2.4.9), it suffices to

show that the interval J = (0,1) is uncountable

If J were countable, we could list its

members and have

We shall show that this leads to a contradiction by constructing a real number that is

in J but is not included in the list of x n’s

Each element of J has an infinite decimal

expansion, so we can write

where each a i j  {0,1,…,9}

We now construct a real number y = 0 b1b2b3 by defining

Since each digit in the decimal expansion of y is either 2 or 3, y  J.

But y is not one of the numbers x n , since it differs from x n in the nth decimal place.

This contradicts our assumption that J is countable, so J must be uncountable 

1 11 12 13

2 21 22 23

3 31 32 33





3, if 2

nn n

nn

a b

a











Ordering of Cardinal Numbers

Our approach to comparing the size of two sets is built on the definition of equinumerous and our understanding of functions

Intuitively, if f : S  T is injective, then S can be

no larger than T.

Not only is this true for finite sets, but we have observed that it also

holds for countable sets (Theorem 2.4.10)

Since we think of cardinal numbers as

representing the size of a set, we shall use them when comparing sizes

Definition 2.4.14

We denote the cardinal number of a set S by | S|, so that we have |S| = |T | iff S and T

are equinumerous That is, |S| = |T | iff there exists a bijection f : S  T.

We define |S|  |T | to mean that there exists an injection f : S  T.

As usual, |S| < |T| means that |S|  |T| and |S|  |T|

The basic properties of our ordering of cardinals are included in Theorem 2.4.15

The proofs are all straightforward and are left to the exercises

Theorem 2.4.15

Let S, T, and U be sets.

(a) If S  T, then | S|  |T|

(b) |S|  |S|

(c) If |S|  |T| and |T|  |U|, then |S|  |U|

(d) If m, n  and m  n, then |{1,2,…, m}|  |{1,2,…,n}|

(e) If S is finite, then | S| < 0

Notes: (a) This corresponds to our intuitive feeling about the relative sizes of subsets

(b) This is the reflexive property

(c) This is the transitive property

(d) This means that the order of m and n as integers is the same as the order for the finite cardinals m and n.

(e) Recall that 0 is the transfinite cardinal number of

It is customary to denote the cardinal number of by c, for continuum.

Since , we have  0  c.

In fact, since is countable and is uncountable, we have 0 < c

Thus Theorem 2.4.15(e) implies that 0 and c are unequal transfinite cardinals.

Are there any others?

The answer is an emphatic yes, as we see in our next theorem

Definition 2.4.16

Given any set S, let P (S) denote the collection of all the subsets of S

The set P (S) is called the power set of S.

Theorem 2.4.18 For any set S, we have | S| < |P (S)|

Recall: |S| < |T | means there exists an injection from S to T,

Theorem 2.4.18 For any set S, we have | S| < |P (S)|.

Proof: The function g:S P (S) given by g (s) = {s}is clearly injective, so |S|  |P (S)|

To prove that |S|  |P (S) |, we show that no function from S to P (S) can be surjective Suppose that f : S  P (S) Then for each x  S, f (x) is a subset of S.

Now for some x in S it may be that x is in the subset f (x) and for others it may not be.

Let T = {x  S : x  f (x)}.

We have T  S, so T P (S) If f were surjective, then T = f(y) for some y  S.

Now either y  T or y  T, but both possibilities lead to contradictions:

If y  T, then y  f(y) by the definition of T.

But f(y) = T, so y  f(y) implies y  T.

On the other hand, if y  T, then since f(y) = T, we have y f(y).

But then y  T, by the definition of T.

Thusweconcludethatnofunctionfrom S to P (S) canbe surjective, so |S|  |P (S)| 

By applying Theorem 2.4.18 again and again, we obtain an infinite sequence of

transfinite cardinals each larger than the one preceding:

0 = | | < | P ( )| < | P ( P ( ))|< | P ( P ( P ( )))| <   

Does the cardinal c = | | fit into this sequence?

In Exercise 24 we sketch the proof that |P ( )| = c.

In Exercise 11 we show that every infinite set has a denumerable subset

Since (by Theorem 2.4.9) every infinite subset of a denumerable set is denumerable,

we see that 0 is the smallest transfinite cardinal

What is the first cardinal greater than 0?

We know that c > 0, but is there any cardinal number  such that

0 <  < c ?

That is, is there any subset of with size “in between” and ?

The conjecture that there is no such set was first made by Cantor and is known

as the continuum hypothesis

In 1900 it was included as the first of Hilbert’s