# Locally compact at a point implies local basis of open sets whose closure is compact

I'm hoping someone can check my proof of the following problem. I feel like 'check my proof' questions are sort of suboptimal, but as I'm purely a self-studier and in particular the book I'm working from doesn't have solutions, this is sort of my only means of verification, which is always nice to have. So the help is always much appreciated!

"Let $X$ be a space and $x$ a point at which $X$ is locally compact. Prove that there is a local basis $\mathcal{B}$ at $x$ such that $\bar{B}$ is compact for each $B \in \mathcal{B}$."

Answer given below is the original, and a correct, proof.

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Looks good to me. I recently saw a suggestion on Meta that one should post his attempted solution as an answer in questions like these, to avoid having them go apparently unanswered when the attempted proof was successful. Perhaps you'd like to self-answer with this proof? –  Kevin Carlson Oct 8 '12 at 2:39
@KevinCarlson: Thanks for the verification. That works for me. Should I just leave an answer stating that the above proof is correct, and then accept it? I don't see any need to copy the proof into the answer. –  Alex Petzke Oct 8 '12 at 19:13
If it were me, I'd just cut your proof out of the question and paste it as the answer, as this might work best for future MSE-ers looking for answers. Your call. –  Kevin Carlson Oct 8 '12 at 21:04
Sounds good to me. I'll accept it once the site allows me to do so tomorrow. –  Alex Petzke Oct 8 '12 at 22:38

Since $X$ is locally compact at $x$, there is an open set $U$ containing $x$ such that $\overline{U}$ is compact. Let $\{O_\alpha\}_{\alpha \in I}$ be the collection of all open sets containing $x$ and consider the set $\mathcal{B}=\{O_\alpha \cap U \,|\, \alpha \in I\}$. Each $B \in \mathcal{B}$ contains $x$, and if $V$ is an open set containing $x$ then $V=O_\alpha$ for some $\alpha \in I$ and so it contains $O_\alpha \cap U$. Each of these sets is contained in $U$ and thus is contained in $\overline{U}$. Let $B \in \mathcal{B}$. Then $\overline{B}$ is a closed subset of the compact space $\overline{U}$ and so is compact. Therefore, $\mathcal{B}$ is a local basis at $x$ for which $\overline{B}$ is compact for each $B \in \mathcal{B}$.