# Is every compact space locally compact?

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?

• math.stackexchange.com/questions/727635/… Jan 30, 2016 at 1:01
• What definition of locally compact are you using? Jan 30, 2016 at 1:12
• @Cameron Williams I know two definitions. 1. Every point has a compact neighbourhood. This is trivially satisfied. 2. Every point has a compact neighbourhood base. (local basis) They are equivalent when the space is Hausdorff. Jan 30, 2016 at 1:13
• @HenryW Since one often assumes Hausdorffness (which will then get you .. regularity? normalness? (it's been a while) in the compact setting which is instrumental in the equivalence), you have a result in the positive. Non-Hausdorff spaces are pretty pathological in some ways. Jan 30, 2016 at 1:15

## 2 Answers

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.

• In the last paragraph, are you saying that is an example of a space with is compact, but not Hausdorff and not locally compact? Jun 8, 2016 at 18:21
• If $X$ consists of at least two elements then this is an example for a topological space which is both compact and locally compact but not Hausdorff. More generally, every finite topological space is both compact and locally compact, without necessarily being Hausdorff. So $X$ being Hausdorff is not a necessary condition for the implication ($X$ is compact $\implies$ $X$ is locally compact) to hold. Jun 9, 2016 at 6:21

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.

• Based on the link above, the answer may depend very heavily on the definition of locally compact that OP is working with. Jan 30, 2016 at 1:11
• I must defer to Jendrik's answer. My answer is sufficient for the definition I gave. Jan 30, 2016 at 1:37