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Unformatted text preview: Solutions homework 10. Page 64 Problem 3. Let a = sup A . Then for all n 1 the number a 1 n is not an upperbound of A , so there exists a n A such that a 1 n < a n a . Then  a a n  < 1 n , so a n a . Thus a A . The proof for inf A A is similar. Page 64 Problem 6. (i) First let x A B . Then there exists x n A B such that x n x . Either x n ultimately in A or x n frequently in B . In the first case x n A for all n N , so x A . In the second case there exists a subsequence x n k B . Then x B . Either way x A B . Another way to prove this inclusion is: note A A and B B . Hence A B A B . As the union of two closed sets is closed, the righthand side is closed. This implies that A B A B . For the opposite inclusion observe A A B implies A A B . Similarly B A B . Hence A B A B (ii) Let A = Q and B = I . Then A B = ., so A B = . On other hand....
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 Fall '10
 Girardi

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