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.



Property stronger than $T_1$ and weaker than regularity
1. Hausdorff spaces need not have this property. Consider [Bing's Countable connected Hausdorff space](https://doi.org/10.1090/S0002-9939-1953-0060806-9) (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.


