Let $X$ be a compact Hausdorff space and $A \subseteq X$ be a closed set. Then $X \setminus A$ is a locally compact Hausdorff space.
That $X \setminus A$ is Hausdorff is quite clear as it is a subspace of a Hausdorff space $X.$ How do I prove that $X \setminus A$ is locally compact? Let $x \in X \setminus A.$ How do I get hold of a compact subset $C_x$ of $X \setminus A$ containing a neighbourhood of $x\ $? Does openness of $X \setminus A$ have something to do with local compactness? I need some help at this stage.
Also I have another question. We know that every locally compact Hausdorff space has a one point compactification. So if $X \setminus A$ is proved to be locally compact what will be it's one point compactification? Any help or suggestion in this regard will be appreciated.
Thanks in advance.
Take $x\in X\setminus A$. Since $X$ is a compact Hausdorff space, it is a normal space. So, since $A$ and $\{x\}$ are closed sets, there are open sets $B$ and $C$ such that $A\subset B$, $x\in C$ and $B\cap C=\emptyset$. But then $C$ is a neighborhood of $x$ and therefore $B^\complement$ is also a neighborhood of $x$. But $B^\complement$ is compact, since it is closed.