(ii) L et O , . . . , O be a ¬nite collection of open sets. T hen 7L O is open.

G G

L

(iii) L et +F , be a collection of closed sets. T hen 7 F is closed.

? ?Z ?Z ?

(iv) L et F , . . . , F be a ¬nite collection of closed sets. T hen 8L F is closed.

G G

L

Note that an in¬nite collection of open sets need not have an inter-

section that is open. For example, in R, for each n : 1, 2, 3, . . . , let

O : (1 9 1/n, 1 ; 1/n) : +x + R : 1 9 1/n : x : 1 ; 1/n,.

L

In R each O is open, and 7 O : +1,, a point. By Exercise 4.5, the set

L L

L

consisting of the point x : 1 is closed. Similarly, the union of an in¬nite

number of closed sets need not be closed. To see this, for each n : 1, 2, 3, . . . let

F : [0, 191/(n ;1)] : +x + R : 0 - x - 191/(n ;1),. Each F is closed, and

L L

8 TOPICS IN REAL ANALYSIS

8 F : [0, 1) : +x + R : 0 - x : 1,, which is not closed since it does not

L L

contain the limit point x : 1.

We now prove the theorem. To establish (i), we take an arbitrary element

x in 8 O . Since x is in the union, it must belong to at least one of the sets

?Z ?

+O , , say O . Since O is open, there exists an r 9 0 such that B(x, r ) 3

? ?Z ? ?

O . Hence B(x,r ) 3 8 O , and thus the union is open. To establish (ii), let

? ?Z ?

x be an arbitrary point in 7L O . Then x + O for each i : 1, . . . , n. Since each

G G G

O is open, for each i : 1, . . . , n there exists an r 9 0 such that B(x, r ) 3 O .

G G G G

Let r : min+r :i : 1, . . . , n,. Then B(x, r) 3 B(x, r ) 3 O for each i : 1, . . . , n.

G G G

Hence B(x, r) : 7L O , and thus the intersection is open.

G G

The most ef¬cient way to establish (iii) is to write

7 F : 7 c(cF ) : c 8 (cF ) ,

? ? ?

?Z ?Z ?Z

where the second equality follows from Lemma 3.1. Since F is closed, cF is

? ?

open and so is 8 (cF ). Hence the complement of this set is closed, and

?Z ?

therefore so is 7 F . To establish (iv), write

?Z ?

L L L

8 F : 8 c(cF ) : c 7 (cF ) .

G G G

G G G

Again, cF is open for each i : 1, . . . , n. Hence the intersection 7L (cF ) is

G

G G

open and the complement of the intersection is closed.

The reader who wishes to test his or her mastery of the relevant de¬nitions

should prove statements (iii) and (iv) of the theorem directly.

5. LIMITS AND CONTINUITY

A sequence in RL is a function that assigns to each positive integer k a vector

or point x in RL. We usually write the sequence as +x , or +x ,, rather than

I I

I I

in the usual function notation. Examples of sequences in R are

(i) +x , : +(k, k),,

I

k k

(ii) +x , : cos , sin ,

I 2 2

(1)

(91)I 1

(iii) +x , : , ,

I 2I 2I

1 1

(iv) +x , : (91)I 9 , (91)I 9 .

I k k

The reader should plot the points in these sequences.

LIMITS AND CONTINUITY 9

A sequence +x , is said to have a limit y or to converge to y if for every 9 0

I

there is a positive integer K( ) such that whenever k 9 K( ), x + B(y, ). We

I

write

lim x : y or x ; y.

I I

I

The sequences (i), (ii), and (iv) in (1) do not converge, while the sequence (iii)

converges to 0 : (0, 0). Sequences that converge are said to be convergent;

sequences that do not converge are said to be divergent. A sequence +x , is said

I

to be bounded if there exists an M 9 0 such that #x # : M for all k. A sequence

I

that is not bounded is said to be unbounded. The sequence (i) is unbounded;

the sequences (ii), (iii), and (iv) are bounded.

The following theorem lists some basic properties of convergent sequences.

T 5.1. (i) T he limit of a convergent sequence is unique.

(ii) L et lim x : x , let lim y : y , and let lim : , where + ,

I I I I I I

I

is a sequence of scalars. T hen lim (x ; y ) exists and equals x ; y , and

I I I

x exists and equals x .

lim

I I I

(iii) A convergent sequence is bounded.

Exercise 5.1. Prove Theorem 5.1.

Let +x , be a sequence in RL and let r : r : r : % : r : % be a strictly

I K

increasing sequence of positive integers. Then the sequence +x , is called

PK K

a subsequence of +x ,. Thus, +x , : +(2k, 2k), is a subsequence of (i). The

I I

I

sequence +x , : +(cos k , sin k ), is a subsequence of (ii). Each of the

I I I

sequences +x , : +((91)I91/2k, (91)I91/2k), : +(191/2k, 191/2k), and

I

+x , : +(91 9 1/(2k ; 1), 91 9 1/(2k ; 1)), is a subsequence of (iv). The

I>

reader should plot the various subsequences.

We can now formulate a very useful criterion for a point to be a limit point

of a set. (The reader should not confuse the notions of limit and limit point.

They are different.)

L 5.1. A point x is a limit point of a set S if and only if there exists a

sequence +x , of points in S such that, for each k, x " x and x ; x.

I I I

Proof. Let x be a limit point of S. Then for every positive integer k there is

a point x in S such that x " x and x + B(x, 1/k). For every 9 0, there is a

I I I

positive integer K( ) satisfying 1/K( ) : . Hence for k 9 K( ), we have 1/k :

and B(x, 1/k) : B(x, ). Hence the sequence +x , converges to x. Conversely, let

I

there exist a sequence +x , of points in S with x " x and x ; x. Let 9 0 be

I I I

arbitrary. Since x ; x, there exists a positive integer K( ) such that, for

I

k 9 K( ), x + B(x, ). Since by hypothesis x " x for all k, it follows that we

I I

have satis¬ed the requirements of the de¬nition that x is a limit point of S.

10 TOPICS IN REAL ANALYSIS

Remark 5.1. Let +x , be a sequence of points belonging to a set S and

I

converging to a point x. Then x must belong to S. If x is in S, there is nothing

to prove. If x were not in S, then since x + S for each k, it follows that x " x.

I I

Lemma 5.1 then implies that x is a limit point of S. Thus x + S. If S is a closed

set, then since S : S, the point x belongs to S.

Remark 5.2. Let S be a set in RL. If x belongs to S, then there is a sequence of

points +x , in S such that x ; x. To see this, note that if x is in S, then x

I I

belongs either to S or to S. If x + S, the assertion follows from Lemma 5.1. If

x + S and x , S, we may take x : x for all positive integers k.

I

Let S be a subset (proper or improper) of RL. Let f be a function with

domain S and range contained in RK. We denote this by f : S ; RK. Thus,

f : ( f , . . . , f ), where each f is a real-valued function.

K G

Let s be a limit point of S. We say that ˜˜the limit of f(x) as x approaches s

exists and equals L™™ if for every 9 0 there exists a 9 0, which may depend

on , such that for all x + B(s, ) 5 S, x " s, we have f(x) + B(L, ). We write

lim f(x) : L.

x;s

It is not hard to show that, if L : (L , . . . , L ), limx;s f(x) : L if and only

K

if, for each i : 1, . . . , m, limx;s f (x) : L .

G G

It is also not hard to show that if a limit exists it is unique and that the

algebra of limits, as stated in the next theorem, holds.

T 5.2. L et limx;s f(x) : L and let limx;s g(x) : M. T hen

lim [f(x) ; g(x)]

x;s

exists and equals L ; M. Also limx;s [ f(x)] exists and equals L.

A useful sequential criterion for the existence of a limit is now given.

T 5.3. Limx;s f(x) : L if and only if for every sequence +x , of points in

I

S such that x " s for all k and x ; s it is true that lim f(x ) : L.

I I

I I

Proof. Let limx;s f(x) : L and let +x , be an arbitrary sequence of points in

I

S with x " s for all k and with x ; s. Let 9 0 be given. Since limx;s f(x):L,

I I

there exists a ( ) 9 0 such that, for all x " s and in B(s, ) 5 S, f(x) + B(L, ).

Since x ; s, there exists a positive integer K( ( )) such that whenever

I

k 9 K( ( )), x + B(s, ). Thus, for k 9 K( ( )), f(x ) + B(L, ), and so f(x ) ; L.

I I I

We now prove the implication in the other direction. Suppose that

limx;s f(x) does not equal L. Then there exists an 9 0 such that for every

9 0 there exists a point x in S satisying the conditions x " s, x + B(s, ),

B B B

BASIC PROPERTY OF REAL NUMBERS 11

and f(x ) , B(L, ). Take through the sequence of values : 1/k, where k

B

runs through the positive integers. We then get a sequence +x ,, with x + S,

I I

x " s, and x ; s, with the further property that f(x ) , B(L, ). Thus f(x )

I I I I

does not converge to L, and the theorem is proved.

Let x be a point of S that is also a limit point of S; that is, x + S 5 S. We

say that f is continuous at x if

lim f(x) : f(x ).

x;x

If x + S and x is not a limit point of S, we will take f to be continuous at x

by convention.

From the corresponding property of limits it follows that f is continuous at

x if and only if each of the real-valued functions f , . . . , f is continuous at x .

K

It follows from Theorem 5.2 that if f and g are continuous at s, then so are

f ; g and f.

The function f is said to be continuous on S if it is continuous at every point

of S.

Exercise 5.2. Show that if limx;s f(x) : L, then L is unique.

Exercise 5.3. Prove Theorem 5.2.

Exercise 5.4. Let f be a real-valued function de¬ned on a set S in RL. Show

that if f is continuous at a point x in S and if f (x ) : 0, then there exists a

9 0 such that f (x) : 0 for all x in B(x , ) 5 S.

Exercise 5.5. Show that the real-valued function f : RL ; R de¬ned by