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Unformatted text preview: To show that ( ), let . For every there exists an ( ) such that ( ). Since is compact, there exists a subsequence . Since ( ) for every , . The sequence ( , ) ( , ). Since is closed (Exercise 2.107), ( ). Therefore ( ) which implies that ( ). 2.112 ( ) is compact for every by Tychonoffs theorem (Proposition 1.2). Let be a sequence in and let = ( 1 , 2 ,..., ) with ( ) be a corresponding sequence of points in . For each , = 1 , 2 ,..., , there exists a subsequence with ( ) (Exercise 2.104). Therefore = ( 1 , 2 ,... ,,....
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This note was uploaded on 01/16/2012 for the course ECO 2024 taught by Professor Dr.dumond during the Fall '10 term at FSU.
- Fall '10