0
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $x$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.