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I'm looking for a reference for a lemma that feels like it should be in the standard reference books, but I haven't been able to find it.

Fix a topology $(T,\mathcal O)$.

Definition: Call $O\in\mathcal O$ regular then $interior(|O|)=O$. In words: $O$ is regular when it is equal to the interior of its closure (see Wikipedia).

Lemma 1: Suppose $p,p'\in T$. Then if $p$ and $p'$ have disjoint open neighbourhoods, then they have disjoint regular open neighbourhoods.
Sketch proof: It would suffice to show that $O$ is contained in a regular open set $R$ disjoint from $O'$, since two applications of this get the required result. We just take $R=interior(|O|)$. Clearly $O\subseteq R$, and $O'$ is disjoint from $|O|$, so we are done.

Question: Can anyone provide references for Lemma 1 in the topological literature, preferably in a standard textbook? I looked in Willard's General topology and also Engelking's General topology and wasn't able to find it.

Thank you.

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  • $\begingroup$ I don't thinks its relevant enough to general topology proofs/exposition for it to be included as a lemma in a textbook like that. $\endgroup$
    – Jakobian
    Jan 7 at 20:06
  • $\begingroup$ If you included a place where this lemma is relevant, it would be helpful for us to find a textbook or reference in which it might be included - so far there is no indication that such application exists. $\endgroup$
    – Jakobian
    Jan 7 at 20:11
  • $\begingroup$ Maybe look in Set Topology by Vaidyanathaswamy, where regular open sets are called "open domains". I recall that he includes more than an average amount of results (some as exercises) on issues related to regular open sets, but I don't have time now to look. $\endgroup$ Jan 7 at 23:43

2 Answers 2

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The result is pretty basic. But if you really want a reference, it's in Handbook of Boolean Algebras, vol. 1, as formula (10) in the proof of Theorem 1.37.

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A similar idea was used in T420, Hausdorff implies semi-Hausdorff of $\pi$-Base and didn't cite a reference as it was seen to be straight-forward from the definitions. Digging a little, I see that this theorem is asserted without proof after Definition 2 of https://doi.org/10.1090/proc/13318 which is probably your best bet for a connection in the published literature. (Digging around for the term "semiregular" (P10) might also reveal something of interest; note that not all Hausdorff spaces are semiregular.)

Note that semi-Hausdorff means that for every pair of points, each is contained in a regular open set missing the other point. You are pointing out something slightly further, that a pair of points being separated by disjoint open sets is equivalent to a pair of points being separated by disjoint regular open sets (and thus, Hausdorff spaces are characterized by the property that points are separated by disjoint regular open sets).

In case it helps a future reader, here is a more careful proof of the result:

Assume $x,y$ are separated by the disjoint open sets $U,V$. Let $U_{r}=int(cl(U))$. Note that $U_{r}\cap V\subseteq cl(U)\cap V$, and since no point of $V$ may be a limit point of $U$ ($V$ it itself a neighborhood of the point missing $U$), we have $cl(U)\cap V=\emptyset$ and therefore $U_{r}\cap V=\emptyset$. Then let $V_{r}=int(cl(V))$. Note that $U_{r}\cap V_{r}\subseteq U_{r}\cap cl(V)$; similarly no point of $U_{r}$ can be a limit point of $V$, so we have $U_{r}\cap cl(V)=\emptyset$ and thus $U_{r}\cap V_{r}=\emptyset$.

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