# Math 317 HW #7 Solutions Judith Stewart | Download | HTML Embed
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1 Math 317 HW #7 Solutions 1. Exercise 3.3.5. Decide which of the following sets are compact. For those that are not compact, show how Definition 3.3.1 breaks down. In other words, give an example of a sequence contained in the given set that does not possess a subsequence converging to a limit in the set. (a) Q Answer. Not compact. The sequence (1, 2, 3, . . .) is contained in Q, but has no convergent subsequences. (b) Q [0, 1] Answer. Not compact. Define the sequence (an ) recursively by setting a1 = 1 and 1 an+1 = 4a for all n = 1, 2, 3, . . .. Each an is in the set Q [0, 1] but, as shown in HW n #5 Problem 1, the sequence (an ) converges to 2 3, which is irrational and hence not in 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 is valid, 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 is contained by every set in the intersection). Therefore, arbitrary intersections of closed and 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 is compact. Answer. False. Consider A = (0, 1) and K = [0, 2]. Then K is closed and bounded, and hence compact, but A K = (0, 1), which is not closed and hence cannot be compact. (c) If F1 F2 T F3 F4 is a nested sequence of nonempty closed sets, then the intersection n=1 Fn 6= . Answer. T False. Consider Fn = [n, +) for each n = 1, 2, 3, . . .. Then F1 F2 F3 , but n=1 Fn = since for any r R, there exists N N such that r / [n, +) for n N. 1

2 (d) A finite set is always compact. Answer. True. Finite sets are always bounded (just let M = max{|a1 |, . . . , |an |} where the set is {a1 , . . . , an }; then M is an upper bound for the set) and closed (since they have 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 set Q is not compact even though it is countable. 3. Exercise 3.3.8. Follow these steps to prove the final implication in Theorem 3.3.8. Assume K satisfies (i) and (ii), and let {O : } be an open cover for K. For contradiction, lets assume that no finite subcover exists. Let I0 be a closed interval containing K, and bisect I0 into two closed intervals A1 and B1 . (a) Why must either A1 K or B1 K (or both) have no finite subcover consisting of sets from {O : }? Answer. The negation of the above statement is that both A1 K and B1 K have finite 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, since K I0 . Thus, both A1 K and B1 K having finite subcovers would imply that K has a finite subcover, which were assuming is not true. (b) Show that there exists a nested sequence of closed intervals I0 I1 I2 with the property that, for each n, In K cannot be finitely covered and lim |In | = 0. Proof. In (a) we argued that either A1 K or B1 K has no finite subcover. If A1 K has 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 the former, and let Ik+1 = Bk+1 if the latter. This gives us the desired sequence I0 I1 I2 such that In K has no finite subcover for all n. Moreover, if d = |I0 |, then d lim |In | = lim = 0. 2n (c) Show that there exists an x K such that x In for all n. 2

3 Proof. We know, by the Nested Interval Property, that there exists x T n=0 In . If x c / K, then x K , which is open since K is closed. Therefore, there exists > 0 such that V (x) K c , meaning that (x , x + ) K c . However, this will lead to a contradiction. For any n > d/, |In | = d/2n < d/n, so In [x d/n, x + d/n] (x , x + ) since d/n < . Thus, for n > d/, In V (x) K c , meaning that In K = . This is impossible since we constructed In so that In K cannot be finitely covered and the empty set can clearly be finitely covered. From this contradiction, then, we conclude that x K. (d) Because x K, there must exist an open set O0 from the original collection that contains x as an element. Argue that there must be an n0 large enough to guarantee that In0 O0 . Explain why this furnishes us with the desired contradiction. Proof. Since x K, there exists 0 such that x O0 . Therefore, since O0 is open, there exists 0 > 0 such that O0 V0 (x) = (x 0 , x + 0 ). Now, as we argued in part (c), if n0 > d/0 , then In0 (x 0 , x + 0 ) O0 . However, this must not be possible: on the one hand, this implies that {O0 } is a finite open subcover for In0 and, hence, for In0 K; on the other hand, we defined In0 such that In0 K has no finite subcover. From this contradiction, then, we can conclude that if K satisfies (i) and (ii), then every open cover of K has a finite subcover, completing the proof of Theorem 3.3.8. 4. Exercise 3.3.10. Lets call a set clompact if it has the property that every closed cover (i.e., a cover consisting of closed sets) admits a finite subcover. Describe all of the clompact subsets of R. Answer. A set is clompact if and only if it is finite. Proof. () Suppose, first, that K is a clompact set, meaning that every closed cover has a finite subcover. From Problem 2(d), we know that every finite set is compact (and hence closed), so the collection of singleton sets {x} with x K gives a closed cover of K. In other words, {{x} : x K} 3

4 is a closed cover for K. By hypothesis, there exists a finite subcover {{x1 }, {x2 }, . . . , {xn }}. By definition of a cover, this implies that K {x1 } {x2 } {xn }, so K is finite. () On the other hand, suppose K is finite. Let {V : } be a closed cover for K. Since K is finite, we can write K = {x1 , x2 , . . . , xn }. Now, since {V : } is a cover, for each i = 1, 2, . . . , n, there exists i such that xi Vi . Therefore, K = {x1 , . . . , xn } V1 Vn , so {V1 , . . . , Vn } is a finite subcover of K. Since the choice of closed cover was arbitrary, we can conclude that every closed cover of a finite set has a finite subcover, so finite sets are clompact. Having proved both directions, we can conclude that a set is clompact if and only if it is finite. 5. Exercise 3.4.3. Review the portion of the proof given for Theorem 3.4.2 and follow these steps to complete the argument. (a) Because x C1 , argue that there exists an x1 C C1 with x1 6= x satisfying |x x1 | 1/3. Proof. Since x C1 = [0, 1/3] [2/3, 1], clearly either x [0, 1/3] or x [2/3, 1]. If the former, let x1 = 0 if x 6= 0; otherwise let x1 = 1/3. If the latter, let x1 = 1 if x 6= 1; otherwise, let x1 = 2/3. Whatever the case, x1 C and both x and x1 lie in an interval of length 1/3, so |x x1 | 1/3. (b) Finish the proof by showing that for each n N, there exists xn C Cn different from x, satisfying |x xn | 1/3n . Proof. Because n 1 3[ 2k 2k + 1 x Cn ( , , 3n 3n k=0 such that x 3n , 2k+1 2k 2k of length 1/3n . If x 6= there exists k 3n , which is an interval 3n , 2k 2k+1 then let xn = 3n ; otherwise, let xn = 3n . By construction, xn C, xn 6= x, and |x xn | 1/3n . 4

5 This process defines a sequence (xn ) with xn C and xn 6= x. Moreover, if > 0, let N > 1/. Then for any n N , 1 1 1 |x xn | n < < . 3 n N Since our choice of > 0 was arbitrary, this implies that lim xn = x, so x is a limit point of C. Since the choice of x was arbitrary, this proves that every element of C is a limit point, so C has no isolated points. 5