Weakly locally compact vs. locally compact

Question

Let locally compact mean that each point of a space has a local basis of compact neighborhoods, while weakly locally compact only requires that each point have a single compact neighborhood. When does one property imply the other?

Answer 1

As you have pointed out, local compactness $\Rightarrow$ weak local compactness, and these notions agree in Hausdorff spaces. There is a kind of converse to this observation. Namely that a locally compact space $X$ is Hausdorff if and only if each of its compact subsets is closed. Note that 'locally compact' cannot be replaced by 'weakly locally compact' in this statement.

Now, the implication itself cannot be reversed, since there are weakly locally compact spaces which are not locally compact. Namely any compact space is weakly locally compact, but that there are even compact spaces which are not locally compact. The one-point compactification $\mathbb{Q}_\infty$ of the rationals is compact but not locally compact.

So here is the positive result: if $X$ is regular and $x\in X$, then the closed neighbourhoods of $x$ form a local base. Since a closed subset of a compact set is itself compact, it follows that if $x$ has some compact neighbourhood, then each of its neighbourhoods contains a compact neighbourhood. Thus we can state;

Suppose $X$ is regular. Then $X$ is weakly locally compact if and only if it is locally compact.

An application is that any regular compact space is locally compact. For instance, if $X$ is regular, then its one-point compactification $X_\infty$ is regular if and only if $X$ is weakly locally-compact.

You can push this a little further in fact. It's true that each regular locally compact space is completely regular. Moreover, in the above you can even go as far as to replace 'regular' with 'preregular'.

Answer 2

It's immediate from the definition that locally compact implies weakly locally compact for any topological space.

Note first that any closed subspace of a compact is itself compact. If the space is Hausdorff, then every compact set $K$ in the space is closed, see How to prove that a compact set in a Hausdorff topological space is closed?.

So then let $K$ be the given compact neighborhood of a point $x$ in a weakly locally compact Hausdorff space, and let $\mathcal B$ be a basis of open neighborhoods of $x$. Then $\mathcal B'=\{cl(B\cap K):B\in\mathcal B\}$ is a basis of closed neighborhoods of $x$. Since each $cl(B\cap K)\subseteq cl(K)=K$ , $\mathcal B'$ is a basis of compact neighborhoods, showing that a weakly locally compact Hausdorff space is locally compact.