equivalent definitions of locally compact for Hausdorff space I think the usual definition of "locally compact" is something like below.
There is a compact neighborhood W of x for any x in the topological space X.
If we assume that X is a Hausdorff space, then the following can be derived. However, I don't know how to prove it, so I would appreciate if you could tell me how to prove it.
There is a relatively compact open neighborhood U of x for any x in the topological space X.
When we take the open neighborhood of x included in W, I think it would be good if the closure of it was included in W...
 A: Statement:
Let $X$ be a Hausdorff topological space. Then X is locally compact if and only
if given $x_0$ in $X$, and given a neighborhood $U$ of $x_0$, there is a neighborhood $V$ of $x_0$ such
that $\overline{V}$ is compact and $\overline{V}\subset U$.
Proof:
the set $\overline{V}:=C$ is the
desired compact set containing a neighborhood of $x_0$. 
To prove the converse, Suppose $X$
is locally compact. 
Let $x_0$ be an arbitrary point of $X$. Since $X$
is locally compact so there exists a compact set $C\subset X$ contains a neighborhood of $x_0$. The $C$ is closed because every compact subspace of a Hausdorff space is closed. Let $U$ be an arbitrary neighborhood of $x_0$. Take the set $A:=C − U$. It is clear that $A$ is a compact subset of $C$.
In other hand, if $A$ is a compact subspace of the Hausdorff space $X$ and $x_0\in X$ is not in $A$, then there exist disjoint open subsets $W$ and $W'$ of $X$ containing $x_0$ and $A$, respectively.
Now if $V:=W\cap int(C)$
Then $\overline{V}\subseteq C$. Every closed subset of a compact set is compact so $\overline{V}$ is compact.  Since $V\subseteq W$ , $W\cap W'=\varnothing$ so $\overline{V}\cap A=\varnothing$, It is clear that $\overline{V}\subset C-A$. Because of $A:=C-U$ we have $\overline{V}\subset U$, as desired.
Your the equivalency follows from the statement simply.
