hmwk8sol_a

hmwk8sol_a - MATH 202A HOMEWORK 8 Problem 1 .(a) Proof....

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Unformatted text preview: MATH 202A HOMEWORK 8 Problem 1 .(a) Proof. Suppose K is compact in Q I X . Since X is closed for all , Q I X is also Hausdorff and thus K is closed. We just need to show that K = . Let x = ( x ) K . By hint, is continuous, ( K ) X is compact. Since the infinite number of X spaces are noncompact, ( K ) 6 = X for infinitely I . Suppose K 6 = , then U 6 = base element s.t. U K . U = Q U , where U open, U = X for all but finite number . Hence, ( U ) = X for all but finite number . However, since U K , ( U ) ( K ), which contradicts to the above paragraph. (b) Solution. l 2 is a counterexample. Notice the identity map from l 2 with discrete topol- ogy to l 2 with usual l 2 topology. By hint, any discrete topology is locally compact since every point has a single point (itself) open neighborhood, closure of which is compact. This map is continuous. (c) Proof. Let y Y . Since f maps from X onto Y , x X s.t. f ( x ) = y . Since X is locally compact, U , open, containing x and U is compact. Since f is open, y = f ( x ) f ( U ) is open. Hence, f ( U ) is a neighborhood of y . We claim that f ( U ) is compact. We just need to show that f ( U ) = f ( U ) and since f is continuous, U is compact, f ( U ) is compact. First, we show that f ( U ) f ( U ). Since U is compact and f is continuous, f ( U ) is compact. Since Y is Hausdorff, f ( U ) is closed. It is clear that f ( U ) f ( U ) and thus by definition, f ( U ) f ( U ). On the other hand, we show that f ( U ) f ( U ). Since f ( U ) is closed and f is an open map, f- 1 ( f ( U )) is closed in X . It is clear that U f- 1 ( f ( U )) and since f- 1 ( f ( U )) is closed, by definition, we know that U f- 1 ( f ( U )). Therefore, f ( U ) f ( f- 1 ( f ( U ))) f ( U ). (d) Proof. By (c), the projection is continuous, surjective and open and thus we have the conclusion. (e) 1 2 MATH 202A HOMEWORK 8 Proof. By (a), since if infinite number X are compact, then compact sets in the product are nowhere dense. Since U = U , U cannot be compact unless U = ....
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This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.

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hmwk8sol_a - MATH 202A HOMEWORK 8 Problem 1 .(a) Proof....

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