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e) {(x, y) : x = 0 or y = 0).

12.22 Prove Theorem 12.1вЂќ using what you know about complements of sets. Also prow

the theorem directly from the definition of a closed set.

Suppose that S is a suhsct of RвЂќ with complement T. Show that cl S is the comple-

12.23

ment of int T.

12.24 We generalize the definition of an accumulation point of a sequence to arbitrary

sets by defining x to be an accumulation point of a set S if every hall about x

contains points ofS other than x. Note that x need not lie in S. If S is closed, show

that it contains all its accumulation points. For general S, prove that cl S is the union

of S and its accumulation points.

12.25 Show that the three examples in the last paragraph are neither open nor closed.

12.26 Prove Theorem 12.12.

12.27 Show that RвЂќвЂ™ and the empty set satisfy the definitions of an open set and of a

closed set in Rm.

12.28 Prove that no nonempty, proper subset of RвЂќвЂ™ can be both open and closed. [Hint:

Suppose that such a set S exists. First find a line segment P in Rm with one end at

a point a in S and the other end at a point b not in S. Find the point E on 1 with the

properties that all the points on P between a and E are in the open set S and E is the

furthest point one from a with this property. Since S is closed and open. c E S

and all points cIosc enough to con !? are in S - a contradiction to the choice of c.]

12.5 COMPACT SETS

The next most important sets after open sets and closed sets are the compact sets.

basically hccause they have many of the desirable properties that finite sets have.

Economic theory. especially microeconomics, is concerned with the consequences

of optimization-the search for a point in some set at which a given function

takes on the maximum or minimum value it can achieve on the set. Compact

sets play an important role in questions about the exi,strncr of optima. We saw

h) or on

in Chapter 3 that a continuous function defined on an open interval (a,

an infinite interval (px, +вЂ˜x) need not achieve a maximum value on that interval:

hounded clr˜.wd

take for example, the lincat function JвЂ™(x) = Lx. However. on any

h], a continuous function does achieve

interval [u, its maximum; possibly at one of

the endpoints. Compact sets are the appropriate generalizations of bounded closed

intervals to higher dimensions. We will prove in Chapter 30 that a continuous

function defined on a compact set always achieves its maximum value on that set.

Recall that a set S in RвЂќ is bounded if there exists a number B such that

llxll 5 R for all x E S, that is, if S is contained in some ball in RвЂќ. Examples of

bounded sets include any interval or finite union of intervals in RвЂ™ except for those

which have +r OT -cc as an endpoint. Any disk in the plane with finite radius is

hounded. Examples of sets which are not hounded include the integers in RвЂ™. any

hyperplanc, or any ray.

[I 2.51 COMPACT SETS 271

Definition A set S in RвЂќ is compact if and only if it is both closed and bounded.

Thus, any closed interval in RвЂ™ with finite endpoints is compact, but open

intervals are not compact. Any closed disk of finite radius in the plane is compact,

but other disks are not.

An important feature of compact sets is that uny sequence defined on a compact

set must contain a subsequence that actually converges, a result known as the

Bolzano-Weierstrass Theorem. This feature of compact sets will be used a dozen

times in this book. In this section we prove this result for the compact interval

[O, 11 on RвЂ™ and sketch the general proof. Chapter 29 presents a careful proof for

a general compact set in RвЂќ.

Theorem 12.13 Any sequence contained in the closed and bounded interval

[0, I] has a convergent subsequence.

Proof Let (x,,],,be an arbitrary sequence in [O, I]. One of the two subintervals

[0, I /2] or [l/Z, l] of [O, l] roust contain an (infinite) subsequence of {x,˜}˜=˜,

since the union of two finite sets is a finite set. Say, for simplicity of argument,

that [O, l/2] contains an infinite subsequence and let yl = xnl be an entry of that

subsequence. Now, divide 10, l/2] into two subintervals [0, l/4] and [I /4, l/2].

One of these two contains an infinite subsequence of {x,,}, I, say [l/4, l/2]. Let

yz = .x,,˜ be an clement of such a subsequence with nl > n,. Now, divide this

subinterval [I /4, l/2] intotwocompact subintervals [l/4, 3/8] and [3/X, I /2].

of {xn},,. Let

yj = x,˜: be an element of such a subsequence, with n3 > n2. Keep dividing

into smaller and smaller subintervals and choosing a subinterval that contains

an (inJinire) subsequence of the original sequence {x,,},=, and an element of

the sequence in that subinterval further down the sequence than previously

chosen elements. Finally. one proves that the sequence of yjвЂ™s so constructed is

of {I,,};; I. n

a convergcnr subsequence

This same argument works for any compact interval [a, h] in RвЂ™ and, in fact,

for any rectangle [a,, h,] X [a?, h2] in R2 and any generalized box

B = {(I,, ,A-,,) E RвЂќ : a, 5 xl 5 h,, a,8 s x. 5 h,,}

in RвЂќ. Find a collapsing sequence of boxes RI > & 3 BJ > вЂќ вЂ˜, each of which

contains an infinite suhsequencc of {x,}:-,. Choose a subsequence b,,},=, such

that yk = x,,˜ for \ome k and 11, < n2 < ni < . . For a general compact subset

C of RвЂќ. we can find a generalized box B that contains C, since C is hounded.

Then. perform the above hisection argument in C fвЂ™ R. The bottom line is the

Thrown. whose complete proof is presented in Chapter 29.

Bolzirr˜o-Wei˜,r.sr˜fl.˜.s

Let C be a compact subset in RвЂќ and let {I,˜};=, be any

sequence in C. Then, (x,,i,:-, has a conwr#vr subsequence whose limit lies

272 LIMITS AND OPEN SETS I1 21

EXERCISES

12.29 Prow that a closed subset of a compact set is compact.

12.30 Prove that every finite set is compact.

12.31 Which of the five sets in Exercise 12.21 are compact sets?

12.32 Prove that the intersection of compact sets is compact and that the finite union of

compact sets is compact. Show that the infinite union 01 compact sets need not he

compact.

12.6 EPILOGUE

This chapter has covered the basic notions of limits and of open and closed sets

needed for a careful exposition of multivariable calculus. The discussion in this

chapter is continued in Chapter 29, where more advanced topics are discussed.

Time permitting, Chapter 29 can be read at this point while these ideas are still

fresh. The first section of Chapter 29 continues our description of the important

propcrtics of sequences. So far, the only way we can tell whether or not a sequence:

converges is to actually identify a limit for it: but we sometimes want to prow

that abstract sequences with certain special properties converge to a limit without

Cauchy

specifying any concrete limit. Section 2Y.l introduces the concept of a

convergence, which allows us to capture the concept of convcrgcncc without

actually finding a limit. Section 2Y.2 continues our discussion of compact sets.

including a careful proof of the Bolzan˜,-Wcicr˜t˜Is!, Theorem. The rest of Chapter

2Y deals with connected sets and with diffcrcnt norms on RвЂќ.

A P P E N D I X A 6

Selected Answers

Chapter 2 Answers

2.5 a) .r # I. h) all .x, c) .r # -1, -2, d) all 1.

2.8 u) ?вЂ˜ = 2.1˜ + 3, h) y = -3x, c) y = 4x - 3, d) 4вЂ˜ = -21 + 2,

e) y = A + I, f) J = (-7/2)x + 3.

2.9 u) additional revenue from the (q + l)st output, that is, marginal revenue.

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