Math 421, Homework #7 Solutions. We can then us the triangle inequality to find for k N that (x k + y k ) (L + M) = (x k L) + (y k M) x k L + y k M

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1 Math 421, Homework #7 Solutions (1) Let {x k } and {y k } be convergent sequences in R n, and assume that lim k x k = L and that lim k y k = M. Prove directly from definition 9.1 (i.e. don t use Theorem 9.2) that: (a) lim k x k + y k = L + M. Proof. Let > 0. The assumptions that lim k x k = L and lim k y k = M allow us to find an N N so that for k N, x k L < 2 and y k M < 2. We can then us the triangle inequality to find for k N that (x k + y k ) (L + M) = (x k L) + (y k M) x k L + y k M < = We therefore conclude that lim k x k + y k = L + M.

2 (b) lim k x k y k = L M. Proof. Let > 0. We first consider the case that M = 0. Then there exists an N 1 N so that y k = y k 0 < for k N L Similarly, there is an N 2 N so that and hence x k L < 1 for k N 2 x k = (x k L) + L x k L + L < 1 + L for k N 2. Using the above with Cauchy-Schwartz inequality, we find for k N := max {N 1, N 2 } x k y k 0 = x k y k x k y k < (1 + L ) 1 + L =. Therefore lim k x k y k = 0 = L 0 = L M. To deal with the case where M 0 we can argue as in the previous paragraph to find an N 1 N so that x k < 1 + L for k N 1. We then use the assumption lim k x k = L to find an N 2 N so that x k L < for k N 2, 2 M and similarly, we can use the assumption lim k y k = M to find an N 3 N so that y k M < 2 (1 + L ). Then we can use the Cauchy-Schwartz inequality to find for k N := max {N 1, N 2, N 3 } that x k y k L M = x k y k x k M + x k M L M = x k (y k M) + (x k L) M x k (y k M) + (x k L) M Therefore lim k x k y k = L M as claimed. x k y k M + x k L M < (1 + L ) 2 (1 + L ) + M =. 2 M 2

3 (2) Prove directly from definition 9.10 (i.e. don t use the Heine-Borel Theorem or the Borel Covering Lemma) that if K 1 and K 2 are compact sets, then the union K 1 K 2 is also compact. Proof. Let {U α } α A be an open cover of K 1 K 2, i.e. we assume that all U α are open sets, and that K 1 K 2 α A In order to show that K 1 K 2 we need to find a finite subset A of A so that {U α } α A is cover of K 1 K 2. Since K 1 K 1 K 2 U α, {U α } α A is an open cover of K 1. Since K 1 is assumed to be compact there is a finite subcover, i.e. there is a finite subset A 1 of A so that K 1 α A 1 Similarly, {U α } α A is an open cover of K 2, and since K 2 is compact, we can find a finite subset A 2 of A so that K 2 α A 2 Combining the above, we have that ( ) ( ) K 1 K 2 U α U α = α A 1 α A 2 α A 1 A 2 Since A 1 and A 2 are both finite, so is A 1 A 2. Thus {U α } α A1 A 2 is a subcover of {U α } α A with a finite number of sets in it. We conclude that K 1 K 2 is compact. α A 3

4 (3) (9.2.4) Suppose that K R n is compact and that for every x K there is an r(x) > 0 so that B r(x) (x) K = {x}. Prove that K is a finite set. Proof using the definition of compactness. For each x, we choose an r(x) > 0 so that B r(x) (x) K = {x}. Then, since K B r(x) (x) and each set B r(x) (x) is open, { B r(x) (x) } is an open cover for K. Since K is assumed to be compact, it follows that there exists a finite collection of point {x 1,..., x j } K so that K B r(xk )(x k ). (1) Then we have that ( ) K = B r(xk )(x k ) K by (1) = = ( Br(xk )(x k ) K ) {x k } by assumptions about r(x) = {x 1,..., x j }. We conclude that K is a finite set. Proof using sequences. Assume to the contrary that K is infinite. Then we can construct a sequence {x k } with x k K for all k N, and with x k x j for k j. 1 Since K is compact, the sequence {x k } has a convergent subsequence {x jk } with x := lim k x jk K. Since elements of the original sequence are pairwise distinct, and j k is strictly increasing, it follows that the elements of the subsequence are also pairwise distinct, i.e. x jk x jl if k l. Since x K, we can by assumption find an r(x) > 0 so that B r(x) (x) K = {x}. But, by the definition of lim k x jk, we can find an N N so that for k N, x jk B r(x) (x). Since x jk K for all k N this implies that x jk B r(x) (x) K = {x} for all k N. We can conclude that x jk = x for all k N. This contradicts the fact that the x jk are pairwise distinct. Therefore K must be a finite set. 1 Such a sequence can be constructed as follows. Choose any x1 K, and inductively choose x k+1 K \ {x 1,..., x k }. If there exists a k N for which K \ {x 1,..., x k } is empty, then K would not be an infinite set. 4

5 (4) Let K R n be a compact set, let F R n be a closed set, and assume that K F =. Prove that there exists an open set O and a closed set C satisfying K O C F c. Proof. Let x K. Since K F =, we can conclude that x F c. Since F is closed, F c is open. Therefore, there exists an (x) > 0 so that B (x) (x) F c. Choosing such an (x) > 0 for each x K, we have that { B (x)/2 (x) } is an open cover for K since K B (x)/2 (x). Since K is compact, we can find a finite subcover, i.e. there is a finite collection of points {x k } j K so that K B (xk )/2(x k ). Since for any x K we have that we can conclude that K B (x)/2 (x) B (x)/2 (x) B (x) (x) F c, B (xk )/2(x k ) B (xk )/2(x k ) F c. Since any union of open sets is open (Theorem 8.24i), we can choose O = j B (x k )/2(x k ), and since finite unions of closed sets are closed (Theorem 8.24iv), we can choose C = j B (xk )/2(x k ). 5