sol-08 - MATH 453 SOLUTIONS TO ASSIGNMENT 8 NOVEMBER 6,...

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M  453 S   A  8 N  6, 2004 Exercise 2 from Section 26, page 171 (a) Let X be a subspace of R in the finite complement topology, and let { U α } be an open cover for X . Pick a particular U α 0 = R - A α 0 , where A α 0 is finite. Then X - U α 0 A α 0 is finite, say X - U α 0 = { a 1 , a 2 , . . . , a n } . For each a i , we can find an U α i that contains it, so { U α 0 , U α i , . . . , U α n } is a finite subcover. (b) [0 , 1] is not compact in this “countable complement” topology. For each n Z + , let U n = [0 , 1] - { 1 i | i > n } . Then { U n } n Z + is an open cover for [0 , 1] that has no finite subcover. ± Exercise 3 from Section 26, page 171 Let A 1 , . . . , A n be compact subspaces of X ; let A = S n i = 1 A i . Suppose C is an open cover for A . Then C is also an open cover for each A i , so we can find finite subcovers C i . Their union is the required finite subcover for A . Hence A is compact. ± Exercise 5 from Section 26, page 171 By Lemma 26.4, for each a A , we can find disjoint open sets U a and V a such that a U a and B V a . The collection { U a | a A } is an open cover for A , so by compactness, we can find a finite subcover { U a 1 , U a 2 , . . . , U a n } . Let U = S n i = 1 U a i and V = T n i = 1 V a i . Then U and V are disjoint, and A U and B V . ± Exercise 11 from Section 26, page 171 Suppose that Y is not connected, so there is a separation C D . Then C and D are both closed in Y and hence in X since Y is closed in X . Therefore C and D are compact and by problem 26.5 we can find disjoint open sets U and V containing C and D respectively.
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sol-08 - MATH 453 SOLUTIONS TO ASSIGNMENT 8 NOVEMBER 6,...

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