Show that every locally compact Hausdorff space is completely regular

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Problem:

Show that every locally compact Hausdorff space is completely regular.

Proof:

Let X be a locally compact Hausdorff space. One point sets are closed in X since X is hausdorff.

Then there is a space Y such that Y is the one point compactification of X where Y is compact hausdorff.

Let $$x_0$$ $$\in$$ X and B $$\subset$$ X such that $$x_0$$ $$\notin$$ B and B is closed in X. Then the open set U = X - B contains x$$_0$$. Then since U is open in X it is also open in Y by construction of the one point compactification Munkres gives. Then C = Y - U is a closed set in Y that contains B.

Then since Y is compact hausdorff it is also normal. Hence since the set {x$$_0$$} is closed since Y is hausdorff, by the Urysohn lemma we can define a continuous function f: Y -> [a,b] such that f(x$$_0$$) = a and f(x) = b for every x $$\in$$ C.

But since f is continuous we can take the continuous function g on a subspace of the domain of f. Namely, g: X -> [a,b] since X is a subspace of Y. Hence we have a continuous function g that maps g(x$$_0$$) = a and g(B) = b, since B $$\subset$$ C. We can replace [a,b] by [0,1] to satisfy the definition of completely regular.

Hence X is completely regular.

1. $$X$$ embeds into its one-point compactification $$\alpha X$$, which is Hausdorff and compact and thus normal and hence Tychonoff.