Proposition 13

Every locally closed subspace of a locally compact space X is locally compact.

Let A be a locally closed subspace of X; then for each $x \in A$ there is a neighbourhood U of x in X such that $U \cap A$ is closed in U. Let $V \subset U$ be a compact neighbourhood of x in X; $V \cap A = V \cap (U \cap A)$ is closed in V and therefore compact (no. 3, Proposition 3). Since $V \cap A$ is a neighbourhood of x in A, the result is proved (it is clear that A is Hausdorff).

Theorem 1 (no. 1) and Corollary 2 of Theorem 2 (no. 4) do not extend to locally compact spaces which are not compact.

For example, in an infinite discrete space X, the filter consisting of those sets which contain a given point $x \in X$ and have a finite complement has x as its only cluster point but does not converge to x. Since any mapping f of X into a Hausdorff space X' is continuous, the image under f of an arbitrary subset of X (which is closed in X, since X is discrete) will not in general be a closed subset of X'.

The proposition corresponding to Theorem 3 of no. 5 is the following: