This norm is called the euclidean norm. In R and R the euclidean norm

reduces to the familiar length. It is straightforward to show that the norm has

the following properties:

#x# . 0 for all x in RL, (1)

#x# : 0 if and only if x is the zero vector 0 in RL, (2)

# x# : " " #x# for all real numbers and vectors x in RL. (3)

The norm has two additional properties, which we will prove:

For all vectors x and y in RL

"1x, y2" - #x# #y# (4)

VECTORS IN RL 3

with equality holding if and only if one of the vectors is a scalar multiple of

the other.

For all vectors x and y in RL

#x ; y# - #x# ; #y#, (5)

with equality holding if and only if one of the vectors is a nonnegative scalar

multiple of the other.

The inequality (4) is known as the Cauchy”Schwarz inequality. The reader

should not be discouraged by the slickness of the proof that we will give or by

the slickness of other proofs. Usually, the ¬rst time that a mathematical result

is discovered, the proof is rather complicated. Often, many very smart people

then look at it and constantly improve the proof until a relatively simple and

clever argument is found. Therefore, do not be intimidated by clever argu-

ments. Now to the proof.

Let x and y be any pair of ¬xed vectors in RL. Let be any real scalar. Then

0 - #x ; y# : 1x ; y, x ; y2

: #x# ; 2 (x, y2 ; #y# for all . (6)

This says that the quadratic in on the right either has a double real root or

has no real roots. Therefore, by the quadratic formula, its discriminant must

satisfy

41x, y2 9 4#x##y# - 0.

Transposing and taking square roots give

"1x, y2" - #x# #y#.

Equality holds if and only if the quadratic has a double real root, say : .

But then, from (6), #x ; y# : 0. Hence by (2) x ; y : 0, and x : 9 y.

To prove (5), we write

#x ; y# : 1x ; y, x ; y2 : #x# ; 21x, y2 ; #y#

- #x# ; 2#x# #y# ; #y# : (#x# ; #y#), (7)

where the inequality follows from the Cauchy”Schwarz inequality. If we now

take square roots, we get the triangle inequality. We obtain equality in (7), and

hence in the triangle inequality, if and only if 1x, y2 is nonnegative and either

y is a scalar multiple of x or x is a scalar multiple of y. But under these

circumstances, for 1x, y2 to be nonnegative, the multiple must be nonnegative.

The inequality (5) is called the triangle inequality because when it is applied to

vectors in R or R it says that the length of a third side of a triangle is less

than or equal to the sum of the lengths of the other two sides.

4 TOPICS IN REAL ANALYSIS

Exercise 1.1. For any vectors x and y in RL, show that #x ; y# ; #x 9y # :

2#x# ; 2#y#. Interpret this relation as a statement about parallelograms in

R and R.

In elementary courses it is shown that in R and R the cosine of the angle

, 0 - - , between two nonzero vectors x and y is given by the formula

1x, y2

cos : .

#x# #y#

Thus, x and y are orthogonal if and only if 1x, y2 : 0. In RL, the Cauchy”

Schwarz inequality allows us to give meaning to the notion of angle between

nonzero vectors as follows. Since "1x, y2# - #x# #y#, the absolute value of the

quotient

1x, y2

(8)

#x# #y#

is less than or equal to 1, and thus (8) is the cosine of an angle between zero

and . We de¬ne this angle to be the angle between x and y. We say that x

and y are orthogonal if 1x, y2 : 0.

3. ALGEBRA OF SETS

Given two sets A and B, the set A is said to be a subset of B, or to be contained

in B, and written A 3 B, if every element of A is also an element of B. The set

A is said to be a proper subset of B if A is a subset of B and there exists at least

one element of B that is not in A. This is written as A : B. Two sets A and B

are said to be equal, written A : B, if A 3 B and B 3 A. The empty set, or null

set, is the set that has no members. The empty set will be denoted by `.

The notation x + S will mean ˜˜x belongs to the set S™™ or ˜˜x is an element of

the set S.™™

Let +S , be a collection of sets indexed by an index set A. By the union

? ?Z

of the sets +S , , denoted by 8 S , we mean the set consisting of all

? ?Z ?Z ?

elements that belong to at least one of the S . Note that if not all of the S are

? ?

empty, then 8 S is not empty. By the intersection of the sets +S , ,

?Z ? ? ?Z

denoted by 7 S , we mean the set of points s with the property that s

?Z ?

belongs to every set S . Note that for a collection +S , of nonempty sets, the

? ?Z

?

intersection can be empty.

Let X be a set, say RL for example, and let S be a subset of X. By the

complement of S (relative to X), written cS, we mean the set consisting of those

points in X that do not belong to S. Note that cX is the empty set. By

convention, we take the complement of the empty set to be X. We thus always

have c(cS) : S. Also, if A : B, then cA 9 cB. The reader should convince

himself or herself of the truth of the following lemma, known as de Morgan™s

METRIC TOPOLOGY OF RL 5

law, either by writing out a proof or drawing Venn diagrams for some simple

cases.

L 3.1. L et +S , be a collection of subsets of a set X. T hen

? ?Z

8 S : c 7 (cS ) ,

? ?

?Z ?Z

7 S : c 8 (cS ) .

? ?

?Z ?Z

4. METRIC TOPOLOGY OF RL

Let X be a set. A function d that assigns a real number to each pair of points

(x, y) with x + X any y + X is said to be a metric, or distance function, on X if it

satis¬es the following:

d(x, y) . 0, with equality holding if and only if x : y, (1)

d(x, y) : d(y, x), (2)

d(x, y) - d(x, z) ; d(z, y) for all x, y, z in X. (3)

The last inequality is called the triangle inequality.

In RL we de¬ne a function d on pairs of points x, y by the formula

L

d(x, y) : #x 9 y# : (x 9 y ) (4)

.

G G

G

Exercise 4.1. Use the properties of the norm to show that the function d

de¬ned by (4) is a metric, or distance function, on RL.

A set X with metric d is called a metric space. Although metric spaces occur

in many areas of mathematics and applications, we shall con¬ne ourselves to

RL. In R and R the function de¬ned in (4) is the ordinary euclidean distance.

We now present some important de¬nitions to the reader, who should

master them and try to improve his or her understanding by drawing pictures

in R.

The (open) ball centered at x with radius r 9 0, denoted by B(x, r), is de¬ned

to be the set of points y whose distance from x is less than r. We write this in

symbols as

B(x, r) : +y : #y 9 x# : r,.

The closed ball B(x, r) with center at x and radius r 9 0 is de¬ned by

B(x, r) : +y : #y 9 x# - r,.

6 TOPICS IN REAL ANALYSIS

Let S 3 RL. A point x is an interior point of S if there exists an r 9 0 such

that B(x, r) : S. A set S need not have any interior points. For example,

consider the set of points in the plane of the form (x , 0) with 0 : x : 1. This

is the interval 0 : x : 1 on the x -axis. It has no interior points, considered

as a set in R. On the other hand, if we consider this as a set in R, every point

is an interior point. This leads to the important observation that for a given

set, whether or not it has interior points may depend on in which euclidean

space RL we take the set to be lying, or embedded.

If the set of interior points of a set S is not empty, then we call the set of

interior points the interior of S and denote it by int(S).

A set S is said to be open if all points of S are interior points. Thus an

equivalent de¬nition of an open set is that it is equal to its interior. If we go

back to the de¬nition of interior point, we can restate the de¬nition of an open

set as follows. A set S is open if for every point x in S there exists a positive

number r 9 0, which may depend on x, such that the ball B(x, r) is contained

in S. The last de¬nition, while being wordy, is the one that the reader should

picture mentally.

A point x is said to be a limit point of a set S if for every 9 0 there

exists a point x " x such that x belongs to S and x + B(x, ). The point x will

C C C C

in general depend on . A set need not have any limit points, and a limit

point of a set need not belong to the set. An example of a set without a

limit point is the set N of positive integers, considered as a set in R. For an

example of a limit point that does not belong to a set, consider the set S Y

+x : x : 1/n, n : 1, 2, 3, . . ., in R. Zero is a limit point of the set, yet zero does

not belong to S. We shall denote the set of limit points of a set S by S.

Exercise 4.2. (a) Sketch the graph of y : sin(1/x), x 9 0.

(b) Consider the graph as a set in R and ¬nd the limit points of this set.

The closure of a set S, denoted by S, is de¬ned to be the set S 6 S; that is,

S : S 6 S. A set S is said to be closed if S contains all of its limit points; that

is, S : S. A set S is closed if and only if S : S. To see this, note that S : S

and the de¬nitions S : S 6 S imply that S : S 6 S. Hence S : S. On the

other hand, if S : S, then S 6 S : S, and so S : S 6 S : S.

A set can be neither open nor closed. In R consider B(0, 1), the ball with

center at the origin and radius 1. It should be intuitively clear that all points

x in R with #x# : 1 are limit points of B(0, 1). Now consider the set

S : B(0, 1) 6 +x : (x , x ) : #x# : 1, x . 0,.

(The reader should sketch this set.) Points x : (x , x ) with #x# : 1 and

x . 0 are not interior points of S since for such an x, no matter how small

we choose 9 0, the ball B(x, ) will not belong to S. Hence S is not open. Also,

S is not closed since points x : (x , x ) with #x# : 1 and x : 0 are limit

points of S, yet they do not belong to S.

METRIC TOPOLOGY OF RL 7

It follows from our de¬nitions that RL itself is both open and closed. By

convention, we will also take the empty set to be open and closed.

T 4.1. Complements of closed sets are open. Complements of open sets

are closed.

Proof. If S : `, where ` denotes the empty set, or if S : RL, then there is

nothing to prove. Let S " `, S " RL and let S be closed. Let x be any point

in cS. Then x , S and x is not a limit point of S. If we recall the de¬nition of

˜˜x is a limit point of S,™™ we see that the statement ˜˜x is not a limit point of S™™

means that there is an 9 0 such that all points y " x with y + B(x, ) are in

cS. Since x + cS, this means that B(x, ) : cS, and so cS is open. Now let

S : `, S " RL and let S be open. Let x be a limit point of cS. Then for every

9 0 there exists a point x " x in cS such that x + B(x, ). Since x + cS, the

C C C

ball B(x, ) does not belong to S. Thus x cannot belong to the open set S, for

then we would be able to ¬nd an such that B(x, ) did belong to S. Hence

x + cS, so cS is closed.

Exercise 4.3. Show that for x + RL and r 9 0 the set B(x, r) is open; that is, show

that an open ball is open.

Exercise 4.4. Show that for x + RL and r 9 0 the closed ball B(x, r) is closed.

Exercise 4.5. Show that any ¬nite set of points x , . . . , x in RL is closed.

I

Exercise 4.6. Show that in RL no point x with #x# : 1 is an interior point of

B(0, 1).

Exercise 4.7. Show that for any set S in RL the set S is closed.

Exercise 4.8. Show that for any set S the closure S is equal to the intersection

of all closed sets containing S.

T 4.2. (i) L et +O , be a collection of open sets. T hen 8 O is open.