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Unformatted text preview: 2.106 1. Assume is closed. For any , let ( ) be a sequence in ( ). Since is closed, ( ). Therefore ( ) is closed. 2. Assume is closed-valued and uhc. Choose any ( , ) / graph( ). Since ( ) is closed, there exist disjoint open sets 1 and 2 in such that 1 and ( ) 2 (Exercise 1.93). Since is uhc, + ( 2 ) is a neighborhood of . Therefore + ( 2 ) 1 is a neighborhood of ( , ) disjoint from graph( ). Therefore the complement of graph( ) is open, which implies that graph( ) is closed. 3. Since is closed and compact, is compact-valued. Let ( ) be a sequence in and ( ) a sequence in with ( ). Since is compact, there exists a subsequence . Since is closed,...
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- Fall '10