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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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    Sounds good to me. $I$'ll accept it once the site allows me to do so tomorrow.2012-10-08

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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 $B \in \mathcal{B}$.