2
$\begingroup$

The following are from Munkres' book on topology:

Theorem 29.2. Let $X$ be a Hausdorff space. Then $X$ is locally compact if and only if given $x$ in $X$, and given a neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ such that $\overline{V}$ is compact and $\overline{V} \subseteq U$.

Lemma 31.1 a. Let $X$ be a topological space. Let one-point sets in $X$ be closed. $X$ is regular if and only if given a point $x$ of $X$ and a neighborhood $U$ of x, there is a neighborhood $V$ of $x$ such that $\overline{V}\subseteq U$.

Doesn't the $(\implies)$ part of theorem 29.2 and the $(\impliedby)$ part of lemma 31.1 a together prove that locally compact Hausdorff spaces are regular? I'm a little bit suspicious because this seems an easy solution, I might be missing something.

$\endgroup$
  • $\begingroup$ Why do I need to show that compact Hausdorff spaces are regular? $\endgroup$ – Kurome May 19 '16 at 14:36
  • $\begingroup$ You’re right: I didn’t read carefully enough and was using a slightly different characterization of local compactness. It’s fine as is. $\endgroup$ – Brian M. Scott May 19 '16 at 14:40
3
$\begingroup$

In fact, X is completely regular: Let X be locally compact Hausdorff. Then, there exists a compact Hausdorff space Y containing X as a subspace. Now since Y is normal, Y is completely regular, which implies that X is completely regular, because it is a subspace of a completely regular space.

$\endgroup$

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.