117-021709

117-021709 - MATH 117 LECTURE NOTES FEBRUARY 17, 2009 DR....

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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, let > . Then, since s n s, there exists N N such that for all n N, | s n- s | < . This is by definition equivalent to s n N ( s ) . Since by hypotheses s n F for all n, and moreover we have assumed that s is not in F, consequently s n N * ( s ) F. We have just shown that s satisfies the definition of being an accumulation point of F. (why?) Hence, s F, by Theorem 13.17. MATH 117 LECTURE NOTES FEBRUARY 17, 2009 5 (5) This is Theorem 16.13 in the text. Assume { s n } n =1 is a convergent sequence of real numbers and its limit is s. Then, by definition of convergence (geez, go on and memorize it already!) using = 1 , there exists N N such that for all n N, | s n- s | < 1 | s n | < | s | + 1 for all...
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117-021709 - MATH 117 LECTURE NOTES FEBRUARY 17, 2009 DR....

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