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$X$ is Hausdorff space. The following is equivalent.

(a) $X$ is locally compact space.

(b) For every open neighborhood $U$ of $x\in X$ there is a smaller open neighborhood $V$ of $x$ whose closure is compact and is contained in $U$.

I tried the following to show $(a)\Rightarrow(b)$.

Try :

Since $X$ is a locally compact hausdorff space, there exists an open set $W$ of $X$ where $x$ $\in$ $W$ $\subset$ $\overline{W}$ and $\overline{W}$ are compact for a given point $x \in X$.

Then, since $X$ is a regular space, subspace $\overline{W}$ is also a regular space.

Since $U\cap \overline{W}$ is an open set of $\overline{W}$, there exists an open set $O$ of $\overline{W}$ that satisfies $x \in O \subset \overline{O} \subset U\cap \overline{W}$.

Then there exists an open set $G$ of $X$ with $O=G\cap \overline{W}$.

Now, $V=G\cap W$ is an open set of $X$, and $\overline{V}$ is a closed subset of the compact space $\overline{W}$, so $\overline{V}$ is compact.

Q. How do you prove $\overline{V} \subset U$?

It does not look obvious!

Please help me. No matter how worried, it does not improve.

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  • $\begingroup$ I know " math.stackexchange.com/questions/47057/… " is the right way. But is my approach wrong? $\endgroup$ – LeeHanWoong Jan 7 '18 at 4:26
  • $\begingroup$ @RichardClare why? if so, $V=\overline{V}$ ? $\endgroup$ – LeeHanWoong Jan 7 '18 at 4:32
  • $\begingroup$ I did not provide the complete explanation sorry. $\endgroup$ – Richard Clare Jan 7 '18 at 4:34
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    $\begingroup$ Show that if $x \notin U$, then $x \notin \overline{V}$. $\endgroup$ – Richard Clare Jan 7 '18 at 4:35
  • $\begingroup$ @RichardClare It's okay. I am so eager to help you. Do you have any hints before I try? I tried it before, but I do not know where to start. $\endgroup$ – LeeHanWoong Jan 7 '18 at 4:36
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You use in your proof that the space is regular and hence every neighborhood contains a closed neighborhood. But if you know how to prove this, then you can easily prove your claim without that much trouble:

  1. We may assume without loss of generality that $U$ is contained in $W$ (otherwise intersect it with $W$, then find such a $V$ and this very $V$ will do the job for the original $U$).

  2. Let $V$ be the interior of any closed neighborhood of $x$ inside of $U$. Then, since $\bar{V}$ is inside of $\bar{W}$ compact, $\bar{V}$ is also compact. And since $V$ is the interior of a closed neighborhood contained in $U$, $\bar{V}$ is also contained in $U$.

Edit: regarding your original approach, to see the inclusion $\bar{V}\subset U$, you can check that $\bar{V}\subset \bar{O}$ ($\subset U$). To see this inclusion, let $y\in \bar{V}$, i.e. any neighborhood of $y$ meets $V=G\cap W$. But then it also meets $G\cap \bar{W}=O$. Therefore $y\in \bar{O}$.

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  • $\begingroup$ Ah! I understood what you were going to say. I'll try it! Thank you very much! $\endgroup$ – LeeHanWoong Jan 7 '18 at 5:14
  • $\begingroup$ @LeeHanWoong I added the proof of the step that you were missing in your original approach. I hope everything is clear now! $\endgroup$ – Pedro Jan 7 '18 at 5:30

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