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hmwk8sol_a

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

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MATH 202A HOMEWORK 8 Problem 1 .(a) Proof. Suppose K is compact in α I X α . Since X α is closed for all α , α I X α is also Hausdorff and thus K is closed. We just need to show that K = . Let x = ( x α ) K . By hint, π α is continuous, π α ( K ) X α is compact. Since the infinite number of X α spaces are noncompact, π α ( K ) = X α for infinitely α I . Suppose K = , then U = base element s.t. U K . U = α U α , where U α open, U α = X α for all but finite number α . Hence, π α ( U ) = X α for all but finite number α . However, since U K , π α ( U ) π α ( 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 π α is continuous, surjective and open and thus we have the conclusion. (e) 1

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2 MATH 202A HOMEWORK 8 Proof. By (a), since if infinite number X α are compact, then compact sets in the product are nowhere dense. Since ¯ U = U , ¯ U cannot be compact unless U = . (f) Proof. Let ( x α ) α I X α . Let the index of finitely many not compact X α be α 1 , α 2 , · · · , α n . j [1 , n ], let U α j be open sets in X α j s.t. x α j U α j , U α j compact since X α j is locally compact. Let U = α I U α , where U α = U α j when α = α j and U α = X α otherwise. By Tychonoff Theorem, ¯ U = α I U α is compact.
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hmwk8sol_a - MATH 202A HOMEWORK 8 Problem 1(a Proof Suppose...

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