For which topological spaces does the closure commute with the interior?

Question

For which topological spaces $X$ is it the case that for all $Y \subseteq X$, $\mathrm{cl}(\mathrm{int}(Y)) = \mathrm{int}(\mathrm{cl}(Y))$?

If one replaces equality with inclusion (i.e., $\mathrm{cl}(\mathrm{int}(Y)) \subseteq \mathrm{int}(\mathrm{cl}(Y))$), one gets the extremally disconnected spaces. So, such spaces should be special kinds of extremally disconnected spaces.

Any discrete space clearly satisfies the condition. Indiscrete spaces with at least two points do not satisfy the condition, however, as for any nonempty proper subset, the closure of the interior is empty, while the interior of the closure is the whole space.

Consider the Sierpinski space $(\{0,1\},\{\emptyset, \{1\}, \{0,1\}\})$. Let's check the condition for each of the four subsets:

1. $\mathrm{cl}(\mathrm{int}(\emptyset)) = \mathrm{cl}(\emptyset) = \emptyset$ and $\mathrm{int}(\mathrm{cl}(\emptyset)) = \mathrm{int}(\emptyset) = \emptyset$
2. $\mathrm{cl}(\mathrm{int}(\{0\})) = \mathrm{cl}(\emptyset) = \emptyset$ and $\mathrm{int}(\mathrm{cl}(\{0\})) = \mathrm{int}(\{0\}) = \emptyset$
3. $\mathrm{cl}(\mathrm{int}(\{1\})) = \mathrm{cl}(\{1\}) = \{0,1\}$ and $\mathrm{int}(\mathrm{cl}(\{1\})) = \mathrm{int}(\{0,1\}) = \{0,1\}$
4. $\mathrm{cl}(\mathrm{int}(\{0,1\})) = \mathrm{cl}(\{0,1\}) = \{0,1\}$ and $\mathrm{int}(\mathrm{cl}(\{0,1\})) = \mathrm{int}(\{0,1\}) = \{0,1\}$

So, the Sierpinski space is a nondiscrete space satisfying the condition.

More generally, any extremally disconnected space that is also a door space (i.e., every subset is open, closed, or both) satisfies the condition. Indeed, if $Y$ is open, then $\mathrm{cl}(Y)$ is also open, and so both sides are equal to $\mathrm{cl}(Y)$, while if $Y$ is closed, then $\mathrm{int}(Y)$ is also closed, and so both sides are equal to $\mathrm{int}(Y)$. If $Y$ is clopen, then both sides are equal to $Y$. The Sierpinski space is an example of an extremally disconnected door space, and of course, discrete spaces are also extremally disconnected door spaces.

The disjoint union of a family of spaces $(X_i)$ satisfies the condition if and only if each $X_i$ individually satisfies the condition. So, the disjoint union of the Sierpinski space with itself (i.e., $\{0,1,2,3\}$ with the open sets being the subsets that do not contain $0$ without also containing $1$, nor contain $2$ without also containing $3$) satisfies the condition but is not a door space, as $\{0,3\}$ and $\{1,2\}$ are neither open nor closed.

Answer 1

Since "$\mathrm{cl}(\mathrm{int}(A)) \subseteq \mathrm{int}(\mathrm{cl}(A))$ for all subsets $A$" has already been characterized by the OP, I'm concentrating on the other inclusion.

A subset $A$ of a toplogical space is called $\delta$-set, if $\mathrm{int}(\mathrm{cl}(A)) \subseteq \mathrm{cl}(\mathrm{int}(A))$.

It is well-known that the following are equivalent for the subset $A$:

1. $A$ is $\delta$-set.
2. There exist an open set $U$ and a nowhere dense set $N$ such that $A = U \cup N$.
3. The boundary $\delta A$ of $A$ is nowhere dense.

See here for the above, and further information about $\delta$-sets.

Call a space deinde, if each dense subset has dense interior.

This is just my name for it. I don't know, if this property is explicitly named somewhere. Anyway, I think it's a well-treated "factor" of submaximality, namely, a space is submaximal, iff it is nodec and deinde (see here for submaximal and nodec).

With this notation, the following are equivalent for the topological space $X$:

1. $X$ is deinde.
2. Each subset with empty interior is nowhere dense.
3. Each subset is $\delta$-set.

PROOF. "(1 $\Rightarrow$ 2):" If $\mathrm{int}(A) = \emptyset$, then $X \setminus A$ is dense, hence $\mathrm{int} (X \setminus A)$ is dense, hence $\mathrm{int} (X \setminus (\mathrm{int} (X \setminus A))) = \emptyset$, hence $\mathrm{int}(\mathrm{cl}(A)) = \emptyset$.
"(2 $\Rightarrow$ 3):" Let $A$ be a subset of $X$. Obviously, $\mathrm{int} (A \setminus \mathrm{int}(A)) = \emptyset$, hence $A \setminus \mathrm{int}(A)$ is nowhere dense.
"(3 $\Rightarrow$ 1)" is obvious.

Hence, the spaces the OP is looking for, are exactly the extremally disconnected deinde spaces. For related spaces see this paper of van Douwen.