Is closed set regular-closed in some topological spaces?

Question

The definition of regular closed sets is:

Let $X$ be a topological space and $A\subseteq X$. We say that $A$ is a regular closed if $A=\text{cl}(\text{int}(A))$

We know that the regular closed sets is closed.

Is there a topological space so that the closed sets is regular-closed ?for example in Hausdorff or regular space or .......... .

A trivial class of such spaces are the discrete spaces, where every subset of the space is both open and closed. — Dave L. Renfro, Jan 04 '20 at 16:05
You'd need to consider non-$T_1$ spaces to find other examples. For instance any indiscrete space. — Brian Moehring, Jan 04 '20 at 16:14
Prove that the closure of all singletons ${x}$, for $x\in X$ are clopens that partition $X$. Then, the topology is given by arbitrary unions of those sets. So, topological spaces in which all closed sets are regular closed are those in which the topology consists of arbitrary unions of elements of a partition of $X$. — MoonLightSyzygy, Jan 04 '20 at 16:15

Henno Brandsma · Answer 1 · 2020-01-04T17:09:15.190

If $X$ is $T_1$, the set $\{x\}$ is closed but regular closed iff $x$ is an isolated point.

So the only $T_1$ examples are discrete. The indiscrete/trivial topology is an example that is not $T_1$. In fact Eric Wofsey shows in this answer that the only examples are the ones where the topology is generated by a partition. (All singletons for the discrete case, a single $\{X\}$ in the indiscrete case).

As a side note, there is a nice (but weird) class of spaces called extremally disconnected where the closure of an open set is again open, so regular closed sets are clopen.

If a space $X$ is regular then the regular open sets form a base for its topology, and so by duality the regular closed sets form a "closed base" (every closed set is an intersection of regular closed sets).

Is closed set regular-closed in some topological spaces?

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