Is every compact space locally compact?

Question

Suppose that $(X,\tau)$ is a topological space.

If $(X,\tau)$ is compact, then $(X,\tau)$ is locally compact.

Does this statement hold for any $(X,\tau)$, or does it only hold when $(X,\tau)$ is Hausdorff?

Answer 1

Because of your question I assume that the definition of compact that you use does not require the space to be Hausdorff. The answer then depends on your definition of locally compact:

If you require every point $x \in X$ to have some compact neighbourhood (weaker condition) then this is always true, because $X$ itself is a compact neighbourhood of $x$.

If you require each point $x \in X$ to have a neighbourhood basis consisting of compact neighbourhoods (stronger condition) then this is not necessarily true.

If $X$ is Hausdorff and compact then $X$ is normal and therefore in particular regular, which (for a Hausdorff space) is equivalent to each point having a neighbourhood basis consisting of closed neighbourhoods. These closed neighborhoods are then also compact, so in this case each point $x \in X$ has a neighbourhood basis cosisting of compact sets, which is why $X$ is locally compact even in the sense of the stronger definition.

But it is not necessary for a compact space $X$ to be Hausdorff to also be locally compact in the sense of the stronger definition: Take any set $X$ together with the indiscrete topology, i.e. $\{\emptyset,X\}$. The resulting space is compact. Then for every $x \in X$ the only possible neighbourhood basis is $\{X\}$, which consists of compact sets.

Answer 2

This is true trivially. A space is locally compact if every point has a compact neighborhood. If the space itself is compact, then it is a compact neighborhood of every point.