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Since there there several inequivalent ways to define the local compactness, I first give the defintion in Fred H. Croom's Principles of Topology.

Definition: A space $X$ is locally compact at a point $x$ if there is an open set $U$ such that $x\in U$ and $\bar U$ is compact. $X$ is said to be locally compact if it is locally compact at each of its points.

Now, let me write a question in Croom's book:

Question: Let $X$ be a Hausdorff space. Show that $X$ is locally compact if and only if each point of $X$ has a compact neighborhood; i.e., $X$ is locally compact if and only if each point $x\in X$ belongs to the interior of a compact set $K_x$.

I think I can prove this easily.

Proof: "$\Rightarrow$". Suppose $X$ is locally compact. By the definition, for each $x\in X$, there is an open set $U$ such that $x\in U$ and $\bar U$ is compact. So just let $K_x=\bar U$ and we have proved "$\Rightarrow$". Note that we did not use the fact that $X$ is Hausdorff.

"$\Leftarrow$". Suppose each point $x\in X$ belongs to the interior of a compact set $K_x$. Simply set $U=\operatorname{int} K_x$. So $x\in U$ and $\overline{U}$ is closed in $K_x$, which implies that $\bar U$ is compact. So $X$ is locally compact. Again, we did not use $X$ being Hausdorff.

Therefore, it seems that $X$ needs not be Hausdorff. So there should be something wrong with my argument. Please find it. Thank you!

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    $\begingroup$ "$\bar U$ is closed in $K_x$", are you sure about that? $\endgroup$ – David Hartley Aug 16 '18 at 15:12
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    $\begingroup$ If the space is not Hausdorff then $K_x$ might not be closed and $\overline {int (K_x)}$ might not be a subset of $K_x.$ $\endgroup$ – DanielWainfleet Aug 16 '18 at 18:58
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In the $\Leftarrow$ implication you have $U \subseteq K_x$, which implies $\overline{U} \subseteq \overline{K_x}$. You use that $\overline{K_x} = K_x$ which holds as $X$ is Hausdorff (e.g. in an infinite cofinite space $X$, all subsets are compact, and only $X$ and the finite subsets of $X$ are closed..). So then indeed $U$ is as required, as it has compact closure. We really only use the KC property (compact subsets are closed), which is stricly weaker than Hausdorffness.

The $\Rightarrow$ implication always holds, trivially: when $x \in U$ has $\overline{U}$ compact, the latter set is a compact neighbourhood of $x$ by definition.

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You wrote "$\overline{U}$ is closed in $K_x$" and this is where we need the Hausdorff property. In a non-Hausdorff space, compact subsets need not be closed. I cannot think of an explicit counterexample to your problem right now, but consider, for example, the real axis with two zeros, the standard example for a non-Hausdorff manifolds. Let $M$ be the set $\mathbb{R} \times \{0,1\}$ quotient $(x,0) \sim (x,1)$ for $x \ne 0$. Then the compact neighborhood $[-1,1] \times \{0\}$ of $(0,0)$ has interior $(-1,1) \times \{0\}$, but the closure of the latter is actually $[-1,1] \times \{0,1\}$. As I said, this is not a direct counterexample, but should illustrate that the closure of the interior of $K_x$ may become larger than $K_x$, which may cause problems in a more pathological counterexample.

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