Show that if \(X\) is a Hausdorff space that is locally compact at the point \(x\), then for each neighborhood \(U\) of \(x\), there is a neighborhood \(V\) of \(x\) such that \(\overline{V}\) is compact and \(\overline{V} \subset U\).
Solution
Let \(U\) be a neighborhood of \(x\). Since \(X\) is locally compact at \(x\), there is a neighborhood \(N\) of \(x\) whose closure \(K=\overline N\) is compact. Replacing \(N\) by \(N\cap U\), we may assume \(N\subset U\), but we still need the closure inside \(U\).
The compact Hausdorff subspace \(K\) is regular. The set \(K\cap U\) is an open neighborhood of \(x\) in \(K\), so there is an open set \(W\) in \(K\) with \[ x\in W\subset \overline W^{\,K}\subset K\cap U. \] Write \(W=K\cap O\) for some open \(O\subset X\), and put \(V=N\cap O\). Then \(V\) is a neighborhood of \(x\). Its closure in \(X\) satisfies \[ \overline V\subset K\cap\overline O\subset \overline W^{\,K}\subset U. \] Also \(\overline V\) is closed in compact \(K\), hence compact. Thus \(V\) has the required properties.