Let X be a compact topological Hausdorff space and for $x$ in $X$ let $N$ be an open neighborhood.

It is said that because $X$ is locally compact, there exist neighborhoods $M\subseteq C\subseteq N$ of $x$ with $C$ closed and $M$ open.

I see the existence of $C:$ Locally compact means that for each $x\in X$ each neighborhood contains a compact neighbourhood of $x$. Here, $C$ is closed since $X$ is Hausdorff. Ok!

But why does there exist this $M$? I do not see that.


The existence of $C$ comes from an equivalent condition for local compactness: in a locally compact space $X$, every point $x\in X$ has a local base $\mathcal{B}$ of compact neighborhoods. This implies that for a neighborhood $N$ of $x$, there is some compact neighborhood $C\subseteq N$ of $x$. As $X$ is Hausdorff, $C$ is closed. Then, the existence of an open $M\subseteq C$ comes from the definition of neighborhood: a neighborhood of $x$ is a subset of $X$ that contains an open set containing $x$.

  • $\begingroup$ But N needs not to be an open neighbourhood of x in order to use local compactness to infer the existence of some compact neighbourhood $C\subseteq N$ of x, right? Just some neighbourhood N of x? $\endgroup$ – Rhjg Aug 1 '17 at 21:26
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    $\begingroup$ That is correct. I've edited my answer to include this. $\endgroup$ – Michael Lee Aug 1 '17 at 21:39
  • $\begingroup$ We can take $M=Int (C)$. $\endgroup$ – DanielWainfleet Aug 2 '17 at 2:30
  • $\begingroup$ @DanielWainfleet Yes, but only because $C$ is specifically a neighborhood of $x$, not just a compact set that contains $x$. If $C$ is not necessarily a compact neighborhood of $x$, then there's no saying whether $x$ is in $\operatorname{Int}(C)$. $\endgroup$ – Michael Lee Aug 2 '17 at 2:55

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