The theorem in Munkres' Topology is

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

One direction is clear, namely the new formulation implies $X$ is locally compact. The converse argument goes: Let $U$ be a neighbordhood 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 $Y$. Then there exist 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. (mostly quoted from Munkres').

How did they get that disjointness (bolded)?

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    $\begingroup$ $W$ is an open set containing $C$, and $V \cap W = \varnothing$. Hence $\overline{V}\cap W = \varnothing$. $\endgroup$ – Daniel Fischer Oct 9 '13 at 19:01
  • $\begingroup$ That almost makes sense intuitively, I will have to prove it quickly... $\endgroup$ – BananaCats Category Theory App Oct 9 '13 at 19:03
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    $\begingroup$ By the way, for the converse argument, do you know why it is possible to take the one-point compactification $Y$ of $X$? I do not understand this because If $X$ is itself compact, then $X$ does not have the one-point compactification. Munkres writes near the top of pg 185 that "X has a one-point compactiication $Y$ if and only if $X$ is locally compact Hausdorff space that is not itself compact. $\endgroup$ – user74261 Jun 21 '15 at 22:15
  • $\begingroup$ @pm021 I think this is just a moment of poor language on Munkres part, due to the fact that there's not really a name for $X\cup\infty$ when X is already compact. Taking this set to be the one point "compactification" seems to work just fine. $\endgroup$ – QTHalfTau Jul 8 '15 at 17:50

Thanks to Daniel Fischer's comment. If $U$, $V$ are two disjoint open sets, then $\overline{U} \cap V = \varnothing$. Proof. If $x \in \overline{U} \cap V$, then every open set around $x$ contains points of $U$ and $V$ contradicting the openness and disjointness of the two sets.


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