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Unformatted text preview: Econ 204 Summer 2009 Problem Set 3 Due in Lecture Friday August 7 1. Cauchy Sequence Suppose f x n g 2 R n is a Cauchy sequence. It has a subsequence f x n k g such that lim n k !1 x n k = x . Show that lim n !1 x n = x . Solution: Consider a Cauchy sequence f x n g 2 R n . For any " > ; there exists an N ( " ) such that for any m; n > , & & x m & x n & & < " 2 . If there exists a subsequence f x n k g of f x n g , such that lim n k !1 x n k = x , then there exists a K ( " ) such that k > K ( " ) ) & & x n k & x & & < " 2 . Choose M > max n K ( " ) ;N ( " ) . For every n > M , we can &nd a n k > M . By triangle inequality, & & x n & x & & & & x n & x n k & & + & & x n k & x & & < " 2 + " 2 = " . Hence lim n !1 x n = x . 2. Compactness Use the open cover de&nition of compactness to show that the subset n n n 2 +1 ; n = 0 ; 1 ; 2 ::: o of R is compact. Solution: Denote A = n n n 2 +1 ; n = 0 ; 1 ; 2 ::: o . Let the collection of sets U & & 2 & be an open cover. Since 2 A , there exists an open set U & 2 U & such that 2 U & . Hence we can &nd an " > such that B " (0) U & . Note that there are only &nitely many points of K not included in U & which are those with n n 2 +1 > " . Denote them as a 1 ;:::;a n and for each a j choose a U & j from U & & 2 & such that a j 2 U & j...
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 Summer '08
 ANDERSON

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