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Proof a): Let S be any union of open sets Si, and let x E S. The point x is in S

Iby virtue of its being a member of some set Si. Since Sj is open, there is an open

ball B around x contained in Si. Then B is contained in the union S of the Si.

i b): Let S,, , S, be open se&. and let S = n!вЂ˜=,S. If x is in S, then x is in

вЂ™ each S:. Since each Si is open, for each i there is an ci such that B,(x) C S,.

min F˜. The ball B,(x) is contained in each B,(x), and therefore is

r-= in each S,. Thus. it is contained in the intersection S of the SiвЂ™s. H

contamed

of an infinire

Note that any union of open sets is open: however, the intersection

number of open sets need not be open. For example, consider the open intervals

S,, = (˜ 1 /n, +1/n) in RвЂ™. The set S,, is the interval of radius 1 /n around 0. It

267

I1 2.41 CLOSED SETS

is easy to see that fl,-,S,, is just the point {Cl]; if y is any point different from 0,

y @ S,, for all II > l/y, and so y hE S. But (0) is not an open set.

Interior of a Set

The fact that arbitrary unions of open sets are open allows us to construct a

set-theoretic operation on all sets, which produces open sets.

Definition Suppose that S is a subset of RвЂќ. Let intS denote the union of all

open sets contained in S. The open set intS is called the interior of S.

By its definition, the interior of a set can be considered as the largest open set

which is contained in the given set. For example, the interior of the closed &-hall

about x is the open E-ball about x. The interior of the half-open interval [a, b) is

the open interval (a, b). The interior of a lint in the plane is the empty set, since no

subset of the lint in R* is open in RвЂ™. As Exercise 12.16 below shows, the interior

of an open set S is the set S itself.

12.4 CLOSED SETS

Consequently, a closed set must contain all its вЂњboundary points,вЂќ just the opposite

of the situation with open sets. In fact, closedness and openness are complementary

properties in that the complement of an open set is closed and the complement of

a closed set is open. Recall that the complement of a set S is the set of all points

that are not in S.

Theorem 12.9 A set S in Rm is closed if and only if its complement, SвЂ™ =

RвЂ™вЂќ S, i s o p e n .

Proof (Only if) Let S be closed. We need to show that its complement, T is open.

IIn other words, we need to show that for any x E T, there is an E > 0 such that

B,(x) C T. Choose x E T and suppose that this is not the cast, that is. that no

B,(x) lies completely in T. Then for all E 3, 0, B,(x) n S # 0. In perticular,

for each positive integer n, there is an element x. of S in B,,,,(x). The sequence

{x,,};=˜ lies in S and converges to x since IIx,, - ˜(1 < 1 /n. Since S is closed, x

is in S--a contradiction to our choice of x in T.

(If) Let S be the complement of an open set T. Let {x,J;˜ be a convergent

sequence in S with limit x. To show that S is closed, we need to show that

x E S Suppose not. Then x E T. Since T is open, there exists an open

B,(x) of x contained in T. Since the sequence converges to x, for

neighborhood

1.

n large enough, x. E B,(x), so x. E T-again a contradiction. since the x,,вЂ˜s

1are in S, the complement of T. n

Using Theorem 12.9 together with Theorem 12.8, we obtain the next theorem

simply by set-theoretic complement&m, since

complement of U Si = njcomplemcnt of&}.

The proof is left as an exercise.

Theorem 12.10

(a) Any intersection of closed sets is closed.

(b) The finite union of closed sets is closed.

Just as arbitrary intersections of open sets need not be open, so too arbitrary

unions of closed sets need not be closed. For example, consider the closed sets

S,, = [-n/(n + I), n/(n + I)1 fern 2 1. The readershould check that LJ,,,lS,, =

(- 1, I)_ is an open interval.

Closure of a Set

Complementing the operation that produces the interior of a set is the closure-

operation.

Suppose that S is a subset of Rm. Let clS (or sometimes 5) denote

Definition

the intersection of all closed sets containing S. The closed set cl.5 is called the

closure of s.

I1 2.41 CLOSEвЂќ 269

SETS

By its definition, the closure of S is the smallest closed set which contains S.

The following theorem is a sequential characterization of the closure of a set.

Let S be a set in RвЂ™. Then x is in clS if and only if there is

Theorem 12.11

a sequence of points in S converging to x.

Proof (If) Let {x.}r=, be a convergent sequence of points in a set S, with limit

x. Then {x,,}z=, C T for any closed set T 3 S; so x E T. Since this is true for

any closed subset T containing S, x E c1.S.

(Only If) Supposex E clS.Thenweclaim that for all E > O,B,(x)flS # 0,

because if for some E > 0, B,(x) n S = 0, then the complement of B,(x) is

a closed set containing S and not containing x-a contradiction to x E cl S.

Now construct a sequence {x.}F=, by choosing x, E B,,,(x) n S. This is a

sequence in S with limit x. m

Boundary of a Set

At this point, we can make precise the notion of the boundary of a set, a concept

which guided our intuition in our discussion of open and closed sets. Roughly

speaking, a point x is on the boundary of a set S if there are points inside S

arbitrarily close to x and points outside S arbitrarily close to x.

Definition A point x is in the boundary of a set S if every open ball about x

contains both points in S and points in the complement of S.

The following theorem follows essentially for the definition of a boundary

point.

The set of boundary points of a set S equals clS n ˜1.5вЂќ.

Theorem 12.12

There are plenty of sets which are neither open nor closed; for example, the

half-open interval (a, b] in RвЂ™, the sequence {l/n : II = 1, 2,. .} without its limit

0 in RвЂ™. and a line minus a point in the plane. There are only two sets which are

both open and closed in Rm: RвЂќвЂ™ itself and, by default, the empty set. See Exercises

12.27 and 12.28.

EXERCISES

12.18 Show th;lt closed intrrvois in RI -sets of the form (x : a 5 x 5 b} for fixed

numhcrs a and b-are closed sets.

12.19 Show that closed bulk in RвЂќ -sets of the form {x : IIx zll 5 6) for fixed z and

E ˜ are closed sets.

12.20 Prove that any tinite set is a closed set. Prow that the set of integers is a closed set.

12.21 Forcach of the following suhsrts of the plane. draw the set. state whether it is open.

closed. or neither, and justify your answer in a word or two:

270 LIMITS AND OPEN SETS (121

b) {(x. y) : x and y are integers),

a) {(x, y) : -1 < x < + I, y = O),

y = I),

c) ((4 y) : x + 4 Kx, Y) : x + < l)a

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