# Question regarding the proof of a topological claim

The lecturer in the Topology course I'm taking defined the following: Given a topological space $X$ we say that:

1. $X$ is weakly locally compact if for all $x\in X$ there exists a compact nbhd.

2. $X$ 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:

1. Given $U$ a nbhd of $x\in X$ since $X$ is wlc there is a compact nbhd of $x$ , $C\subseteq X$.

2. 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 $x\in V\subseteq U\cap C$ .

3. Since $C$ is a compact Hausdorff space (Hausdorff being hereditary) we know $C$ is regular.

4. 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 $x\in W\subseteq\overline{W}\subseteq V\subseteq C$ .

5. Since $C$ is compact and $\overline{W}$ is closed in $C$ we know that $\overline{W}$ is also compact.

6. 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 $\overline{W}_{X}=\overline{W}_{C}=\overline{W}_{U}=\overline{W}_{V}$ ?

• Note that $C$ is regular, so there is an open $W$ such that $\overline W$ is a subset of $V$, and the closure is taken with respect to $C$. – Stefan Hamcke May 25 '13 at 21:58
• That's true I can take the closure relative to $C$ right away and the proof would be correct without any messing around. That still doesn't explain the claim that $\overline{W}_{X}=\overline{W}_{C}=\overline{W}_{U}=\overline{W}_{V}$ though. – Serpahimz May 25 '13 at 22:02
• @StefanH. I tried taking up your "challenge" to find a similar formula for the interior and ran into a bit of trouble. I couldn't "ping" you in a comment there so I thought I'd ping you here :) – Serpahimz May 26 '13 at 6:40

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.
• I decided to take your suggestion, initially I thought it just might be analogues so that $\mbox{Int}_{A}\left(W\cap A\right)=\mbox{Int}_{X}\left(W\cap A\right)\cap A$ but the same reasoning that works for proving the closure identity doesn't really work here. I then tried going from the definition: $\mbox{Int}_{A}\left(W\cap A\right)=\bigcup\left\{ U\in\mathcal{T}_{A}\,|\, U\subseteq W\cap A\right\} =\bigcup\left\{ U\in\mathcal{T}_{X}\,|\, U\cap A\subseteq W\cap A\right\}$ but I don't really see where I can take this to. Is there actually such a handy formula for the interior? – Serpahimz May 26 '13 at 6:38
• @Serpahimz: There is. To derive it from the previous formula, express $\text{int}_A(W\cap A)$ as $A-\text{cl}_A(A-W)$. According to the formula for the relative closure, this is equal to $A-(\text{cl}_X(A-W)\cap A)=A-\text{cl}_X(A-W)=A\cap\text{int}_X(X-(A-W))=A\cap\text{int}_X(W\cup(X-A))$, so this formula is a bit less intuitive. Again, if $A$ is open, one can prove that $\text{int}_A(W\cap A)=\text{int}_X(X\cap A)\cap A$. – Stefan Hamcke May 26 '13 at 13:33
• @Serpahimz: Of course. Let $x\in\text{cl}_X(W)\cap A$. To show that it is in the relative closure, we need to show that for any open $U\ni x$ the relative neighborhood $U\cap A$ intersects $W\cap A$. But $A$ is open and $x\in A$, so $U\cap A$ is an open neighborhood of $x$, which intersects $W$, thus also $W\cap A$. Hence $x\in\text{cl}_A(W\cap A)$. – Stefan Hamcke May 26 '13 at 17:38
• On second thought the simplification in the interior seems a bit odd since $\mbox{Int}_{X}\left(X\cap A\right)\cap A=\mbox{Int}_{X}\left(A\right)\cap A=\mbox{Int}_{X}\left(A\right)$ so you end up with $\mbox{Int}_{A}\left(W\cap A\right)=\mbox{Int}_{X}\left(A\right)$ – Serpahimz May 26 '13 at 17:45
• @Serpahimz: There's obviously a typo, but giving it a second thought, one can refine it a bit and obtains $\text{int}_A(W\cap A)=\text{int}_X(W\cap A)=\text{int}_X(W)\cap A$. – Stefan Hamcke May 26 '13 at 18:00