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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?

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    $\begingroup$ math.stackexchange.com/questions/727635/… $\endgroup$
    – Maffred
    Jan 30, 2016 at 1:01
  • $\begingroup$ What definition of locally compact are you using? $\endgroup$ Jan 30, 2016 at 1:12
  • $\begingroup$ @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. $\endgroup$ Jan 30, 2016 at 1:13
  • $\begingroup$ @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. $\endgroup$ Jan 30, 2016 at 1:15

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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.

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  • $\begingroup$ In the last paragraph, are you saying that is an example of a space with is compact, but not Hausdorff and not locally compact? $\endgroup$
    – user41728
    Jun 8, 2016 at 18:21
  • $\begingroup$ 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. $\endgroup$ Jun 9, 2016 at 6:21
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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.

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

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