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Unformatted text preview: MATH 117 LECTURE NOTES FEBRUARY 17, 2009 DR. JULIE ROWLETT 1. Exam solutions (1) (a) A subset of a metric space is compact, if every open cover admits a finite subcover. (b) An accumulation point x of a set S in a metric space is a point such that, for every > , N * ( x ) ∩ S 6 = ∅ . (c) A sequence { s n } n ∈ N converges to s ∈ R , if for any > 0 there exists N ∈ N such that  s n s  < whenever n ≥ N. 1 2 DR. JULIE ROWLETT (2) This was also a homework problem. Let A and B be compact subsets of a metric space. Then let { O a } be an open cover of A ∪ B. This means A ∪ B ⊂ ∪ a O a . Hence, since A ⊂ A ∪ B, A ⊂ ∪ a O a . Therefore, by the compactness of A, there exists n ∈ N such that A ⊂ ∪ n i =1 O i . Similarly, B ⊂ A ∪ B, so B ⊂ ∪ a O a . Therefore, by the compactness of B there exists m ∈ N such that B ⊂ ∪ m j =1 O j . Hence, A ∪ B ⊂ ∪ n,m i,j =1 O i ∪ O j . This is a finite subcover. Since the open cover was arbitrary, this shows that any open cover admits a finite subcover, hence A ∪ B is compact. MATH 117 LECTURE NOTES FEBRUARY 17, 2009 3 (3) This was done in class at least twice. (0 , 1) is not compact, since (0 , 1) ⊂ ∪ ∞ n =1 1 n , 1 1 n , but there is no finite subcover. (Why?) 4 DR. JULIE ROWLETT (4) This is also a homework problem. If { s n } ∞ n =1 ⊂ F, and s n → s as n → ∞ , then there are two possibilities. Either s ∈ F, or s is not in F. If s ∈ F we are done (why?) Otherwise, letwe are done (why?...
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 Spring '08
 Akhmedov,A
 Math, Order theory, Sn, Dominated convergence theorem, Monotone convergence theorem, DR. JULIE ROWLETT

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