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Recently I got interested in the following property of topological spaces:

$(X,\mathcal{T})$ satisfies (P) if the following holds:

For any nonempty closed subsets $F$ and $G$ with $F\ne G$, there are closed subsets $F'\subseteq F$ and $G'\subseteq G$ satisfying the following conditions:

  • $F'$ has nonempty interior in $F$,
  • $G'$ has nonempty interior in $G$,
  • $F'\cap G'=\varnothing$.

It is not difficult to show that regularity implies property (P). Moreover, for $T_0$ spaces, property (P) implies $T_1$ (this is also easily seen by letting $x\ne y$ such that $\overline{\{x\}}\subseteq \overline{\{y\}}$, $\overline{\{y\}}\not\subseteq \overline{\{x\}}$, choosing $F=\overline{\{y\}}$ and $G=\overline{\{x\}}$, and deriving a contradiction.).

In view of this situation, I was wondering if this property appears in the literature or if anyone knows any relation to other well-known properties.

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  • $\begingroup$ Notation G for open sets is annoying. Instead of F and G, you may have A and B. $\endgroup$
    – Wlod AA
    Commented Feb 10, 2023 at 17:09
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    $\begingroup$ @WlodAA : Did you mean "notation $G$ for closed sets is annoying"? $\endgroup$ Commented Feb 10, 2023 at 17:18
  • $\begingroup$ @IosifPinelis, yes, of course, "notation 𝐺 for closed sets is annoying" -- sorry for my misprint. $\endgroup$
    – Wlod AA
    Commented Feb 10, 2023 at 17:21
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    $\begingroup$ @WlodAA : I wondered, though, why a standard notation for open sets is $G$ and that for closed sets is $F$. I mean, $F$ precedes $G$ in the alphabet, whereas opens sets are usually introduced before closed ones. $\endgroup$ Commented Feb 10, 2023 at 17:28
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    $\begingroup$ @WlodAA : Interesting. You might want to answer, with bibliographical details, the question at mathoverflow.net/questions/440625/… $\endgroup$ Commented Feb 10, 2023 at 17:36

1 Answer 1

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  1. Hausdorff spaces need not have this property. Consider Bing's Countable connected Hausdorff space (Example 75 in Steen and Seebach's Counterexamples in Topology); it has the property that for every pair of nonempty open sets $U$ and $V$ the closures $\overline U$ and $\overline V$ intersect. Let $U$ and $V$ be nonempty, open, and disjoint, and let $F=\overline U$ and $G=\overline V$. Let $F'$ and $G'$ be closed subsets of $F$ and $G$ respectively with non-empty relative open interior, say $F'\supseteq U'\cap F\neq\emptyset$ and $G'\supseteq G\cap V'\neq\emptyset$ for some open sets $U'$ and $V'$. Then also $U'\cap U$ and $V'\cap V$ are nonempty and $F'$ and $G'$ contain their closures, so $F'\cap G'\neq\emptyset$.
  2. Urysohn spaces do have this property. In Urysohn spaces distinct points have disjoint closed neighbourhoods. So, assume $F\neq G$ and, wlog, take $x\in F\setminus G$ and $y\in G$. Let $U$ and $V$ be open sets with $x\in U$, $y\in V$ and $\overline U\cap\overline V=\emptyset$. Let $F'$ be the closure of $U\cap F$ and let $G'$ be the closure of $V\cap G$. These are as required.
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