Disjoint neighborhoods with compact closures in locally compact Hausdorff spaces Let $X$ be a locally compact Hausdorff space, and $A$ and $B$ be disjoint compact subspaces of $X$. I want to show that $A$ and $B$ have disjoint neighborhoods whose closures are compact. My plan was to use one point compactification. Since $X_{\infty}=X\cup\{\infty\}$ is a compact Hausdorff space, it is a normal space. Then, $A$ and $B$ can be separated by disjoint open sets. The closures of these open sets will compact in $X_{\infty}$. My question is that can we conclude that the closures are also compact in $X$? Or, can you suggest an alternative proof?
 A: Take $X=\mathbb{R}$ and $w<x<y<z\in\mathbb{R}$. The disjoint compact sets $[w,x],[y,z]$ can be separated by the disjoint open intervals $U=(-\infty,a)$ and $V=(b,c)$, where $x<a<b<y$ and $z<c<\infty$. The sets $U,V$ are open in $\mathbb{R}_\infty\cong S^1$ and have disjoint compact closures there. However, $U$ does not have compact closure in $\mathbb{R}$.
Thus it is not true that the closure in $X_\infty$ of a subset of $X$ need have compact closure in $X$. However, this will be true whenever this closure does not contain the point at infinity. That this is so follows from the fact that the closed sets of $X_\infty$ are either compact subsets of $X$, or of the form $A\cup\{\infty\}$, where $A\subseteq X$ is closed.
Thus given disjoint compact $A,B\subseteq X$, let $U,V\subseteq X_\infty$ be disjoint open subsets with $A\subseteq U$ and $B\cup\{\infty\}\subseteq V$. The closure $K$ of $U$ in $X_\infty$ is a compact subset which is disjoint from $B\cup\{\infty\}$, and in particular $K$ a compact neighbourhood of $A$ in $X$ which is disjoint from $B$.
Now repeat the argument to find a compact neighbourhood of $B$ which is disjoint from $K$.
