TITLE: Question regarding the proof of a topological claim
QUESTION [5 upvotes]: The lecturer in the Topology course I'm taking defined the following:
Given a topological space $X$
  we say that:

$X$ is weakly locally compact if for all $x\in X$
there exists a compact nbhd.
$X$ is strongly locally compact if every nbhd of $x$
contains a compact nbhd of $x$
.

We then made the following claim: A weakly locally compact (wlc) Hausdorff space is strongly locally compact. 
Briefly the proof went as follows:

Given $U$ a nbhd of $x\in X$  since $X$ is wlc there is a compact nbhd of $x$
, $C\subseteq X$.
Since $U,C$
are both nbhds of $x$ then $U\cap C$
is also a nbhd of $x$
and thus there is an open set $V\subseteq X$
such that $x\in V\subseteq U\cap C$
.
Since $C$
is a compact Hausdorff space (Hausdorff being hereditary) we know $C$
is regular.
Since regularity is hereditary and $V\subseteq C$
we know $V$
is also regular and thus there is an open set $W\subseteq V$
such that $x\in W\subseteq\overline{W}\subseteq V\subseteq C$ .
Since $C$
is compact and $\overline{W}$
is closed in $C$
we know that $\overline{W}$
is also compact.
Finally $x\in W\subseteq\overline{W}\subseteq V\subseteq U$
and thus $\overline{W}$
is a compact nbhd of $x$
contained in $U$ .

The lecturer then noted that it's important to notice the proof hangs on the fact that the closure of $W$ in $V$ and $C$ is the same. That is since we can only deduce compactness of $\overline{W}$ since it is closed in $C$. However, we used the regularity of $V$ in order to find $W$ and thus the closure of $W$ is relative to the topology in $V$ and not in $C$. He also noted that in fact from the way we carried out the construction the closure of $W$ is the same in all the groups in which it is contained, that is $\overline{W}_{X}=\overline{W}_{C}=\overline{W}_{U}=\overline{W}_{V}$
  (the substring marking closure relative to which space). 
My question is why is it in fact true that $\overline{W}_{X}=\overline{W}_{C}=\overline{W}_{U}=\overline{W}_{V}$ ?

REPLY [3 votes]: There is a useful formula: If $A$ is a subset of $X$ and $W$ is another subset, then $\text{cl}_A(W\cap A)=\text{cl}_X(W\cap A)\cap A$. So let $A$ be the $V$ in your example.
If we take the closure of $W$ with respect to $C$, then $\text{cl}_C(W)=\text{cl}_X(W)$ since $C$ is closed in $X$.
On the other hand $\text{cl}_V(W)=\text{cl}_X(W)\cap V$ which is just $\text{cl}_X(W)$ since $W$ was chosen to have its closure contained in $V$.
Finally, $\text{cl}_U(W)=\text{cl}_X(W)\cap U=\text{cl}_X(W)$ since $\text{cl}(W)\subset U$.
I should also mention that the above formula simplifies one more time if $A$ is open. In this case we have $\text{cl}_A(W\cap A)=\text{cl}_X(W\cap A)\cap A=\text{cl}_X(W)\cap A$. Using this the last two equalities follow automatically since $U$ and $V$ were open.