Geometric Intuition of regular open sets

Question

In Oxtoby's Measure and Category book, he has the following definition at page $20.$

A regular open set is a set that is equal to the interior of its closure.

If I am not mistaken, by definition, $A = int (\overline{A})$ if and only if $A$ is regular open.

Question: What is its geometric intuition of the definition?

It seems that open set may not be regular open, for example, $A=(1,2)\cup(2,3).$

Also, closed set is not regular open, as we have $int(\overline{A}) = int({A})$ is open and $A$ is closed.

Answer 1

The geometric intuition is that it has no “skinny holes” (holes without interior).

The closure fills the isolated holes, and the interior then makes it open again.

Note that for general sets, the operation also removes "skinny islands" (skinny holes of the complement; removed during forming the interior), but open sets don't have them to begin with.

Also note that the operation maps an arbitrary set to a regular open set.

Answer 2

A regular open set $U\subseteq X$ is a pseudocomplement of the open set $V=X\setminus\overline U.$ This means that the sets $U$ and $V$ are a disjoint partition of $X$ apart from a nowhere dense set, their common boundary $\overline U\setminus U=\overline V\setminus V.$

A regular open set is an open set that gives a nice partition of $X$ in this way. The geometrical intuition is that $U$ is a nice representative of this kind of partition. Note you can recover $U$ from $V,$ and vice versa. You could equally well represent the partition by the regular closed set $\overline U,$ and get a similar theory.

This is the geometrical consequence of the regular open sets forming a Boolean algebra.