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hmwk8sol_b

# hmwk8sol_b - Math 202A Homework 8 Roman Vaisberg Problem...

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Math 202A Homework 8 Roman Vaisberg Problem 1. (a) Let X α be a Hausdorff topological space for each α I . Prove that if an infinite number of the coordinate spaces X α are non-compact, then each compact subset of the product X = α I X α is nowhere dense. Proof. Let K be an arbitrary compact subset of X and suppose x = ( x α ) is an interior point of K . Then there exists an open set U in X containing x and contained in K and thus in turn x α I U α U K , where each U α is open in X α and for all but finitely many α , U α = X α . From the statement of the problem there must exist an i I such that X α is non-compact and U α = X α . Letting π i be the projection onto the i -th coordinate we see that X i = π i ( α I U α ) π i ( K ). We also clearly have π i ( K ) X i . However, π i ( K ) must be compact and X i is not, which is a contradiction. Thus int ( K ) = int ( K ) = , where the first equality comes from the fact that X is a product of Hausdorff spaces and thus Hausdorff, and K being a compact subset must then already be closed. Thus K is nowhere dense. (b) Show that a continuous map need not map a locally compact space to a locally compact space. Proof. Consider the identity map from the space R with the discrete topology to R with the lower limit topology. It is obvious that a space with the discrete topology is locally compact as every singleton is open, closed, and compact. Furthermore the identity map is continuous. To prove that R with the lower limit topology is not compact we first show that [0 , 1) is not compact. This is easy since { [0 , 1 - 1 2 n ) } n N is a cover for [0 , 1) which does not have a finite subcover. Thus [ a, b ) is not compact for any real a , b . Next we show that 0 does not have an open neighbourhood whose closure is compact. Let U be a neighbourhood of 0. Then there must exist a basic set [ a, b ) such that 0 [ a, b ) U . But since [ a, b ) is also closed in the lower limit topology (it is easy to see that its complement is open), it would have to be compact if the closure of U were compact. Thus we obtain that the closure of U cannot be compact and R with the lower limit topology is not locally compact. (c) Suppose f : X Y is continuous, open, and surjective. X is locally compact and Y is Hausdorff. Prove that Y is locally compact. Proof. Let y Y and select x X such that f ( x ) = y . Since X is locally compact there exists an open set U containing x such that U is compact. let V = f ( U ). Then V is open. Now observe that f ( U ) is compact and since Y is Hausdorff, f ( U ) is closed. Thus V f ( U ) implies V f ( U ), which implies that V is a closed subset of a compact space and thus compact. This concludes the proof that Y is locally compact. (d) Prove that if a product of Hausdorff spaces is locally compact, then each coordinate space is locally compact.

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hmwk8sol_b - Math 202A Homework 8 Roman Vaisberg Problem...

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