A Hausdorff topological space is known to be locally compact if every point has a compact neighborhood. Prove that every closed subspace of a locally compact Hausdorff space is locally compact.
Solution
Let \(X\) be locally compact Hausdorff, and let \(A\subseteq X\) be closed.
1. The subspace \(A\) is Hausdorff.
Take distinct points \(a,b\in A\). Since \(X\) is Hausdorff, there are disjoint open sets \(U,V\subseteq X\) with \[ a\in U, \qquad b\in V. \] Then \(U\cap A\) and \(V\cap A\) are disjoint open neighborhoods of \(a\) and \(b\) in the subspace topology on \(A\). Thus \(A\) is Hausdorff.
2. Fix \(a\in A\).There is a compact neighborhood \(K\) of \(a\) in \(X\) and an open set \(U\subseteq X\) such that\[ a\in U\subseteq K. \]
Since \(X\) is locally compact, \(a\) has a compact neighborhood \(K\). By the definition of neighborhood, \(K\) contains an open set \(U\) containing \(a\).
3. The set \(K\cap A\) is compact.
2. Therefore \(K\cap A\) is compact.
4. The set \(K\cap A\) is a neighborhood of \(a\) in the subspace \(A\).
By 2, \[ a\in U\subseteq K. \] Hence \[ a\in U\cap A\subseteq K\cap A. \] The set \(U\cap A\) is open in \(A\) by the definition of the subspace topology. Thus \(K\cap A\) contains an open neighborhood of \(a\) in \(A\), so it is a neighborhood of \(a\) in \(A\).
5. Therefore \(A\) is locally compact.
The point \(a\in A\) was arbitrary. By <1>3–<1>4, every point of \(A\) has a compact neighborhood in \(A\). Together with the Hausdorff property from 1, this is precisely local compactness in the sense stated in the problem.