Closure of opening of closure in $\mathbb R^2$

Question

My question is somehow related to Closure of the interior of another closure

However, I go a bit further. I have a closed set $X\subseteq \mathbb R^2$ and $Y:=\operatorname{cl}\operatorname{int} X$.

I would like to know whether $Y=\operatorname{cl}\operatorname{int} Y$.

(Here, $\operatorname{cl}$ is the closure and $\operatorname{int}$ is the interior.)

Answer 1

Let $(X, τ)$ be a topological space. We have the following:
a.) If $F$ is closed, then $int(cl(int(F)))=int(F)$.
b.) From (a) you can deduce that for any open set $U$, $cl(int(cl(U)))=cl(U)$
if you set $U:=int(X)$ you have the desired result.

$Int(ClIntX) ⊂ ClIntX \Rightarrow Cl(IntClIntX) ⊂ Cl(ClIntX) = ClIntX,$
$Cl(IntX) ⊃ IntX \Rightarrow ClInt(ClIntX) ⊃ ClInt(IntX) = ClIntX$

Answer 2

Suppose $U$ is open, ie $U=U^\circ$. We have $U \subset \overline{U}$ and $\overline{U}^\circ \subset \overline{U}$. Since $U$ is open and contained in $\overline{U}$, we have $U \subset \overline{U}^\circ \subset \overline{U}$. Taking closure of both sides shows $\overline{\overline{U}^\circ} = \overline{U}$.

Now let $U = X^\circ$, $Y = \overline{U}$. Then we have $Y = \overline{Y^\circ}$.