hw12solutions - R such that K 1 K 2 K 3 ... . Prove that...

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Math 521 - Advanced Calculus I Instructor: J. Metcalfe Due: February 22, 2010 Assignment 12 1. Let K 6 = be a compact set in R . Show that inf K and sup K exist and belong to K . Since K is compact, we know that K is bounded. As such, a = inf K and b = sup K exist. Morever, we have K [ a,b ]. Look now at the collection of open sets { ( a - 1 ,b - (1 /n )) } n N . If b 6∈ K , then this is an open cover of K . Moreover, since b is the supremum of K , for any ε > 0, we can always find an element k K so that k > b - ε . Thus, if any finite subcollection of these sets is chosen, let M be the largest index in that finite subcover. Then, we apply the above with ε = 1 /M , and can select a k K so that k > b - (1 /M ). As such, this finite subcollection does not cover K . Since there is no finite subcover, it follows that K is not compact. This proves that sup K must be in K for K compact. The proof for inf K is quite similar. 2. Let { K n } n N be a sequence of nonempty compact sets in
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Unformatted text preview: R such that K 1 K 2 K 3 ... . Prove that there exists at least one point x R such that x K n for all n N ; that is, the intersection T n =1 K n is not empty. Suppose that T K n is empty. Since K n is compact, it is closed. Thus, K c n is open. Since T K n is empty, we have that S K c n = R . In particular, { K c n } n =2 is an open cover for K 1 . Thus, there are a nite number of indices 1 ,..., N so that K 1 K c 1 K c N = ( K 1 K N ) c . This means that K 1 K 1 K N is empty. But by the nexting property, if 1 N , then K 1 K 1 K N = K N which was nonempty by assumption. 1...
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This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.

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