Particular Point Space is Weakly Locally Compact

Theorem
Let $T = \left({S, \vartheta_p}\right)$ be a particular point space.

Then $T$ is locally compact.

Proof
We have that $\left\{{p}\right\}$ is compact in $T$.

But by definition $\left\{{p}\right\}$ is open in $T$.

So $p$ is contained in a compact neighborhood, that is, $\left\{{p}\right\}$.

Now let $x \in S: x \ne p$.

Then $\left\{{x, p}\right\}$ is open in $T$.

We have that $\left\{{x, p}\right\}$ has an open cover, trivially, i.e. $\left\{{\left\{{x, p}\right\}}\right\}$ itself.

Hence any open cover of $\left\{{x, p}\right\}$ has a finite subcover: any one set that contains $x$ and $p$ is a cover for $\left\{{x, p}\right\}$.

So $\left\{{x, p}\right\}$ is a neighborhood of $x$ which is compact in $T$.

So $x$ is contained in a compact neighborhood.

Hence the result, by definition of locally compact.