204Section72009

# 204Section72009 - Section 7 Econ 204 GSI Hui Zheng Key...

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Section 7 Econ 204, GSI: Hui Zheng Key Words Compactness, Open Cover, Sequentially Compact, Totally Bounded Section 7.1 A collection of sets U = f U : 2 g in a metric space ( X;d ) is an open cover of A if U is open for all 2 and [ 2 U ± A . A set A in a metric space is compact if every open cover of A f U : 2 g is an open cover of A , there exist n 2 N and 1 n 2 such that A ² U 1 [ ::: [ U n . A set A in a metric space ( X;d ) is sequentially compact if every sequence of elements of A contains a convergent subsequence whose limit lies in A . Example 7.1.1 Exhibit an open cover of (0 ; 1) Solution: Consider the cover f (0 ; 1 ³ 1 =n ) g . This covers (0 ; 1) the form (0 ; 1 ³ 1 =N ) , for some N . This interval is clearly a proper subset of (0 ; 1) . Example 7.1.2 Show that Q \ [ 0 ; 2 ] is not compact. Solution: U n = ( ³ 1 ; p 2 ³ 1 n ) [ ( p 2 + 1 n ; 3) . The collection of sets given by [ n 2 N U n = ( ³ 1 ; p 2) [ ( p 2 ; 3) is an open cover of Q \ [ 0 ; 2 ] . Since Q is dense, N , there exists a rational number q 2 Q \ [ p 2 ³ 1 N ; p 2 + 1 N ] . So Q \ [ 0 ; 2 ] . Example 7.1.3 Solution: Let A 1 ;:::;A n be compact sets and consider any open cover of A 1 [´´´[ A n . This open cover must cover each A i individually, and because each A i is compact, there must be a &nite sub- cover of each A i . The union of these n A 1 [´´´[ A n . Therefore every open cover of A 1 [´´´[ A n A 1 [´´´[ A n is compact. Example 7.1.4 (Cantor±s Intersection Theorem) compactness to prove a decreasing sequence of nonempty compact subsets A 1 µ A 2 µ ´´´ of a metric space ( X;d ) has nonempty intersection. Solution: By contradiction. Suppose their intersection in empty: A 1 \ A 2 \ ´´´ = ± . Since A 1 µ A 2 µ ´´´ and they are nonempty sets, A 2 \ A 3 \ ´´´ = ± . Let U = X n ( A 2 \ A 3 \ ´´´ ) = X n A 2 [ X n A 3 [ ´´´ and it is open, so it constructs an open cover for A 1 . Because A 1 X n A 2 [ X n A 3 [ ´´´ [ X n A N µ A 1 . Then its complement X n ( X n A 2 [ X n A 3 [´´´[ X n A N ) = A 2 \ A 3 \´´´\ A N has no common element with A 1 , A 1 \ A 2 \ A 3 \ ´´´ \ A N = ± . But we know A 1 \ A 2 \ A 3 \ ´´´ \ A N = A N .

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204Section72009 - Section 7 Econ 204 GSI Hui Zheng Key...

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