It would be very appreciated if someone could review my solution. Thanks!


Show that every locally compact Hausdorff space is completely regular.


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.


It follows from two basic facts:

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

  2. A subspace of a Tychonoff space is again Tychonoff.

Your proof is almost a re-proof of the last fact. If it has been covered in your course, it's better to just quote that instead of redoing its proof in a special case.

  • $\begingroup$ Ok so looks like tychonoff space is another word for a completely regular space from googling around -- munkres doesn't use that term. He does prove a subspace of a completely regular space is completely regular, so I could've used that..... With that said, is my proof correct regardless? $\endgroup$ – H_1317 Apr 17 at 2:04
  • $\begingroup$ @H_1317 yes, it works. But it’s better to use the things you already know. $\endgroup$ – Henno Brandsma Apr 17 at 3:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.