Contents

Background

Definition

(locally compact topological space)

A topological space $X$ is called locally compact if for every point $x \in X$ and every open neighbourhood $U_x \supset \{x\}$ there exists a smaller open neighbourhood $V_x \subset U_x$ whose topological closure is compact and still contained in $U$:

$\{x\} \subset V_x \subset \underset{\text{compact}}{Cl(V_x)} \subset U_x \,.$

Statement and proof

Proposition

(open subspaces of compact Hausdorff spaces are locally compact)

Every open topological subspace $X \underset{\text{open}}{\subset} K$ of a compact Hausdorff space $K$ is a locally compact topological space.

In particular every compact Hausdorff space itself is locally compact.

Proof

Let $x \in X$ be a point and let $U_x \subset X$ an open neighbourhood. We need to produce a small open neighbourhood whose closure is compact and still contained in $U_x$.

By the nature of the subspace topology there exists an open subset $V_x\subset K$ such that $U_x = X \cap V_x$. Since $X$ is assumed to be open, it follows that $U$ is also open as a subset of $K$. Since compact Hausdorff spaces are normal it follows (by this prop.) that there exists a smaller open neighbourhood $W_x \subset K$ whose topological closure is still contained in $U_x$, and since closed subspaces of compact spaces are compact:

$\{x\} \subset W_x \subset \underset{\text{cpt}}{Cl(W_x)} \subset V_x \subset K \,.$

The intersection of this situation with $X$ is the required smaller compact neighbourhood $Cl(W_x) \cap X$:

$\{x\} \subset W_x \cap X \subset \underset{\text{cpt}}{Cl(W_x)} \cap X \subset U_x \subset X \,.$
Remark

Conversely, every locally compact Haudorff space $X$ arises as in prop. , since it may be considered an open subspace in its one-point compactification $X \sqcup \{\infty\}$. See there this example.