1
$\begingroup$

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 $\overline{V} \subseteq U$.

Proof: Clearly this new formulation implies local compactness; the set $C = \overline{V}$ is the desired compact set containing a neighborhood of $x$. To prove the converse, suppose $X$ is locally compact, let $x$ be a point of $X$ and let $U$ be a neighborhood of $x$. Take the one-point compactification $Y$ of $X$, and let $C$ be the set $Y - U$. Then $C$ is closed in $Y$, so that $C$ is a compact subspace of $Y$. Apply Lemma 26.4 to choose disjoint open sets $V$ and $W$ containing $x$ and $C$, respectively. Then the closure $\overline{V}$ of $V$ in $Y$ is compact, furthermore, $\overline{V}$ is disjoint from $C$, so that $\overline{V} \subset U$, as desired.

I didn't understand how exactly the fact of $V$ and $W$ are disjoint opent sets containing $x$ and $C$, respectively imply that $\overline{V}$ of $V$ in $Y$ is compact, someone can help me? Thanks in advance!

$\endgroup$
2
$\begingroup$

The set $\overline{V}$ is closed, and since $Y$ is compact, closed subsets of $Y$ are compact by Theorem 26.2 in Munkres.

$\endgroup$
  • $\begingroup$ It's true, thanks, I can not believe I let this go. $\endgroup$ – George Apr 30 '17 at 15:24
1
$\begingroup$

The subspace $\overline{V}$ is compact since $Y$ is a compact space, and since closed subspaces of compact spaces are compact, you have the result.

$\endgroup$

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.