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
 Empty set, Metric space, Open set, Topological space, Closed set, Xn

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