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 closedvalued 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 compactvalued. 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
 Dr.DuMond
 Macroeconomics, Topology, Open set, Closed set, General topology, ????0

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