This is a follow up to a question I asked earlier. I had previously asked that if a space was regular if and only if any closed set $Z$ was the intersection of all open sets containing $Z$, but the backwards implication turned out to be untrue. I'm now curious to see if this new bi-implication is fully true.

My definition of regular here is that for any $p\in X$ and any closed set $Q$ not containing $p$, there exist open sets $U,V$ such that $p\in U$, $Q\subseteq V$, and $U\cap V=\emptyset$.

I assume $X$ is regular. I think it is clear that $$Z\subseteq\bigcap_{\text{closed}\ N\in\mathscr{N}_Z}N$$

For the other containment, I suppose that $p\not\in Z$. Then since $X$ is regular, there exist open sets $U,V$ such that $p\in U$, $Z\subseteq V$, and $U\cap V=\emptyset$. Then $U\subseteq V^c\subseteq Z^c$, and thus $U^c\supseteq V\supseteq Z$. So $U^c$ is a closed neighborhood of $Z$ which does not contain $p$. Hence $p\not\in\cap\mathcal{N}$, where $\mathcal{N}$ is the family of closed neighborhood of $Z$, and the equality follows.

I feel I have a hole in my reasoning to show the other implication. I suppose now that $Z=\cap\mathcal{N}$, and take some $p\not\in Z$, and so $p\not\in\cap\mathcal{N}$. Anyway, $p\in Z^c$, which is an open neighborhood of $p$. I then take some closed neighborhood $Q$ of $p$ such that $Q\subseteq Z^c$. I then let $U={Q}^{\circ}$, and $V=Q^c$. So both are are open and disjoint. Also, $p\in U$, and $V=Q^c\supseteq (Z^c)^c=Z$, so $X$ is regular.

However, I don't think I'm able to make the assumption that there is some closed neighborhood of $p$ contained in the open neighborhood $Z^c$. Is there some way to use the fact that $p\not\in\cap\mathcal{N}$ to show this is indeed true? Thanks.

EDIT: In case it is not entirely clear, my definition of a neighborhood $N$ of a set $Z$ is that $Z$ is contained in the interior of $N$, so $N$ is a neighborhood of each point $Z$. Then my definition of a neighborhood $M$ of a point $p$ is that $M$ contains an open set containing $p$. $\mathscr{N}_Z$ is my notation for the set of all neighborhoods of $Z$, and $\mathcal{N}$ is a shortcut way of writing the set of $\text{closed}\ N\in\mathscr{N}_Z$.


I agree that you don't know that there is a closed neighborhood of $p$ contained in $Z^c$.

However, this perhaps will work: Since $p$ is not in the intersection $\cap \mathcal{N}$, there is a closed set $V$ with $Z\subseteq V^{\circ}$ and with $p\notin V$. Since $V$ is closed, then $V^c$ is an open set containing $p$; and the interior of $V$ is an open set containing $Z$. Of course, $V^c$ is disjoint from $V^{\circ}$, so don't $V^{\circ}$ and $V^c$ give you open sets, disjoint, one containing $Z$ and the other containing the point $p\notin Z$?

  • $\begingroup$ Thank you Arturo! I feel like I was close, but not close enough, oh well. $\endgroup$ – yunone Nov 24 '10 at 2:17
  • $\begingroup$ You are really fast Arturo. Both typing and thinking! $\endgroup$ – Nuno Nov 24 '10 at 2:19
  • $\begingroup$ @Cromarty: You were indeed very close; your error was staring to closely at $Z$, instead of looking at one those closed neighborhoods thingies. (-; $\endgroup$ – Arturo Magidin Nov 24 '10 at 2:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.