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Math 317 HW #7 Solutions

1. Exercise 3.3.5. Decide which of the following sets are compact. For those that are notcompact, show how Definition 3.3.1 breaks down. In other words, give an example of asequence contained in the given set that does not possess a subsequence converging to a limitin the set.

(a) Q

Answer. Not compact. The sequence (1, 2, 3, . . .) is contained in Q, but has no convergentsubsequences.

(b) Q [0, 1]Answer. Not compact. Define the sequence (an) recursively by setting a1 = 1 andan+1=

1

4anfor all n = 1, 2, 3, . . .. Eachan is in the set Q [0, 1] but, as shown in HW

#5 Problem 1, the sequence (an) converges to 2

3, which is irrational and hence notin the set Q [0, 1].

(c) R

Answer. Not compact. The sequence (1, 2, 3, . . .) has no convergent subsequences.(d) Z [0, 10]

Answer. Compact.

(e){1, 1/2, 1/3, 1/4, 1/5, . . .}Answer. Not compact. The sequence (1/2, 1/3, 1/4, . . .) consists of elements of the set,but the sequence (and, thus, any subsequence) converges to 0, which is not in the set.

(f){1, 1/2, 2/3, 3/4, 4/5, . . .}Answer. Compact.

2. Exercise 3.3.7. Decide whether the following propositions are true or false. If the claim isvalid, supply a short proof, and if the claim is false, provide a counterexample.

(a) An arbitrary intersection of compact sets is compact.

Answer. True. We know that arbitrary intersections of closed sets are closed. Moreover,the arbitrary intersection of bounded sets will be bounded (since the intersection iscontained by every set in the intersection). Therefore, arbitrary intersections of closedand bounded sets are closed and bounded, which implies the above statement.

(b) Let A R be arbitrary, and let K R be compact. Then, the intersection A K iscompact.

Answer. False. ConsiderA = (0, 1) andK= [0, 2]. ThenKis closed and bounded, andhence compact, but

A K= (0, 1),which is not closed and hence cannot be compact.

(c) If F1 F2 F3 F4 is a nested sequence of nonempty closed sets, then theintersection

n=1Fn= .

Answer. False. Consider Fn = [n, +) for each n= 1, 2, 3, . . .. ThenF1 F2 F3 , but

n=1Fn= since for any r R, there exists N Nsuch that r / [n, +) for

n N.

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(d) A finite set is always compact.

Answer. True. Finite sets are always bounded (just letM= max{|a1|, . . . , |an|} wherethe set is{a1, . . . , an}; then M is an upper bound for the set) and closed (since theyhave no limit points). Since being closed and bounded is the same as being compact,this implies that every finite set is compact.

(e) A countable set is always compact.Answer. False. As we saw in Problem 1(a) above, the setQ is not compact even thoughit is countable.

3. Exercise 3.3.8. Follow these steps to prove the final implication in Theorem 3.3.8.

Assume Ksatisfies (i) and (ii), and let {O: } be an open cover for K. For contradiction,lets assume that no finite subcover exists. Let I0be a closed interval containing K, and bisectI0 into two closed intervals A1 and B1.

(a) Why must either A1 Kor B1 K(or both) have no finite subcover consisting of setsfrom{O: }?

Answer. The negation of the above statement is that both A1 K and B1 K havefinite subcovers. If that were true, then

(A1 K) (B1 K)

would have a finite subcover (since the sum of two finite numbers is finite). However,this set is equal to

(A1 K) (B1 K) = (A1 B1) K=I0 K=K,

sinceK I0. Thus, both A1 KandB1 Khaving finite subcovers would imply thatKhas a finite subcover, which were assuming is not true.

(b) Show that there exists a nested sequence of closed intervalsI0 I1 I2 with theproperty that, for each n, In Kcannot be finitely covered and lim |In| = 0.

Proof. In (a) we argued that eitherA1 KorB1 Khas no finite subcover. IfA1 Khas no finite subcover, let I1 = A1; otherwise, let I1 = B1.

Inductively, assume weve defined Ik Ik1 such that Ik K has no finite subcover.Lets bisect Ik into two closed intervals Ak+1 and Bk+1. By the same proof as in (a),either Ak+1 K or Bk+1 K(or both) has no finite subcover; let Ik+1 = Ak+1 if theformer, and let Ik+1= Bk+1 if the latter.

This gives us the desired sequence I0 I1 I2 such that In Khas no finitesubcover for alln. Moreover, ifd = |I0|, then

lim |In| = lim

d

2n

= 0.

(c) Show that there exists anx K such that x In for all n.

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This process defines a sequence (xn) with xn C and xn= x. Moreover, if > 0, letN >1/. Then for any n N,

|x xn| 13n

< 1

n 1

N < .

Since our choice of >0 was arbitrary, this implies that lim xn= x, sox is a limit pointofC. Since the choice ofxwas arbitrary, this proves that every element ofC is a limitpoint, soChas no isolated points.

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