117-021709-1 - MATH 117 LECTURE NOTES DR JULIE ROWLETT 1 Exam solutions(1(a A subset of a metric space is compact if every open cover admits a nite

117-021709-1 - MATH 117 LECTURE NOTES DR JULIE ROWLETT 1...

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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 > 0 , N * ( x ) S = . (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
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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.
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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?)
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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, let > 0 . Then, since s n s, there exists N N
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  • Fall '08
  • Akhmedov,A
  • Math, Order theory, Sn, Dominated convergence theorem, Monotone convergence theorem, DR. JULIE ROWLETT

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