Could you help me prove the following fact? I've been trying to prove it and I've searched for a hint in Englking's book, but I haven't come up with anything:
If $X$ is a Hausdorff space and each $x \in X$ has at least one compact neighbourhood, then $X$ is locally compact.
$U$ is a neighbourhood of $x \in X$ $\iff$ $\exists V$ – open $: x \in V \subset U$.
$X$ is locally compact $\iff$ each $x \in X$ has a basis of compact neighbourhoods.
Best Answer
Actually, a stronger result is true: If each point has a compact Hausdorff neighborhood, then $X$ is locally compact.
Let $x\in X$, $U$ a neighborhood of $x$, and $K$ a compact Hausdorff neighborhood of $x$. Since $K$ is regular as a subspace of $X$ and $U\cap K$ is a neighborhood of $x$ in $K,$ there is a neighborhood $C\subseteq U\cap K$ of $x$ in $K$ (and thus also in $X$ since $K$ is a neighborhood of $x$) which is closed in $K$, thus compact. This proves that there are arbitrary small compact neighborhoods around $x$.