Introduction to Real Analysis

Dr. Weihu Hong

Clayton State University

8/27/2009

Countable and Uncountable sets

Equivalent sets: two sets A and B are said to be equivalent ， denoted A~B, if there exists a one-to-one function of A onto B.

The notion of equivalence of sets is an equivalence relation (reflexive, symmetric, and transitive).

Definition of finite, infinite, countable, uncountable, and at most countable: for each positive integer n, let If A is a set, we say: (a) A is finite if A ~ for some n, or A = Ø. (b) A is infinite if A is not finite. (c) A is countable if A ~ N. (d) A is uncountable if A is neither finite nor countable. (e) A is at most countable if A is finite or countable.

}.,,3,2,1{ nNn

nN

Examples of countable sets

N is countable S = the set of positive integers that are

perfect squares is countable Z is countable N × N is countable Q is countable E = the set of even integers is countable O = the set of odd integers is countable

Definition of Sequences

If A is a set, by a sequence in A we mean a function from N into A, that is, f: NA. For each nєN, . Then is called the nth term of the sequence f.

If A is countable set, then there exists a one-to-one function f from N onto A. Thus

A = Range f = The sequence f is called an enumeration of the set

A; i.e.,

)(nfxn nx

,...}.3,2,1:{ nxn

mnwheneverxxwithnxA mnn ,...}3,2,1:{

Theorem 1.7.6

Every infinite subset of a countable set is countable. Proof: Let A be a countable set and let

be an enumeration of A. Suppose E is an infinite subset of A. Then each xєE is of the form for some nєN. We can write E as a sequence in the natural order as in A, i.e.,

Define function f: N E by Then f is one-to-one and onto. Thus E is countable.

,...}.3,2,1:{ nxn

nx

}.,...,3,2,1:{ jiwhenevernnandkx jink

knxkf )(

Theorem 1.7.7

If f maps N onto A, then A is at most countable. Proof: If A is finite, the A is at most countable. Suppose A is

infinite. Since f maps N onto A, each aєA is an image of some nєN, that is, f(n) = a. Since the set is not empty, it follows from the well-ordering principle that it has a smallest integer, which denote by Consider the mapping a of A into N. If then since f is a function, . Since A is infinite, is an infinite subset of N. Thus the mapping a is a one-to-one mapping of A onto an infinite subset of N. It follows from Theorem 1.7.6 that A is countable.

})(:{})({1 anfNnaf

anan

,ba ba nn }:{ Aana

an

Indexed Families of Sets

Let A and X be nonempty sets. An indexed family of subsets of X with index set A is a function from A to into P(X), denote by

Examples: For each xє(0,1), let Then is an indexed family of subsets of Q. The sequence

).(,}{ afEwhereE aAaa

}0:{ xrQrEx )1,0(}{ xxE

1}{}{ nnNnn NN

Operations of indexed families of subsets of X

Suppose is an indexed family of subsets of X.

The union and intersection of the family are defined

respectively

AaaE }{

}:{

}:{

AaallforExXxE

AasomeforExXxE

aaAa

aaAa

Distributive and De Morgan’s laws

')'(

')'(

)()(

)()(

aAa

aAa

aAa

aAa

aAa

aAa

aAa

aAa

EE

EE

EEEE

EEEE

The Countability of Q

Theorem 1.7.15 If is a sequence of countable sets and then S is countable.

Proof: Since is countable for each nєN, we can write Since is an infinite subset of S, the set S itself is infinite. Consider the function h: N×N S by The function h (may not be one-to-one) is a mapping of N×N onto S. Thus since N×N ~ N, there exists a mapping of N onto S. It follows from Theorem 1.7.7 that the set S is countable.

NnnE }{

nnES

1

nE

},2,1:{ , kxE knn 1E

knxknh ,),(

Corollary 1.7.16 Q is countable.

Proof: Since each rational can be written as a fraction of two integers, for each mєN, let

Then is countable, and it follows from Theorem 1.7.15 that the set Q is countable.

Znm

nEm :

mE mm EQ 1

Theorem 1.7.17 The closed interval [0,1] is uncountable.

Proof: (Cantor) Since there are infinitely many rational numbers in [0,1],the set is not finite. To prove that it is uncountable, we only need to show that it is not countable. To this end, we will prove that every countable subset of [0,1] is a proper subset of [0,1]. Thus [0,1] cannot be countable.

Let be a countable subset of [0,1]. Then each has a decimal expansion

}9,,1,0{,.0 ,3,,2,1, knnnnn xwhere，xxxx

},2,1:{ nxE n

nx

Continuation of the proof:

Define a new number as follows

Therefore, E is a proper subset of [0,1].

Eyunique

isyoftionrepresentadecimalthe

andNnanyforxySince

ydefinexif

ydefinexif

where

yyyy

n

nnn

nnn

,

,

;3,6

;6,5

,

.0

,

,

3,21