Is the interior of the closure of the interior of the closure of a set equal to the interior of the closure of that set? Let $S$ be a subset of a topological space. I want to prove or disprove the following claim:
$\left(\overline{\left( \overline{S} \right)^\circ}\right)^\circ=\left( \overline{S} \right)^\circ$
Setting $A=\left( \overline{S} \right)^\circ$, we have: 
$A=\left( \overline{A} \right)^\circ$.
I know counterexamples where $A$ is open and this does not hold (for example: $(-1,0)\cup(0,1) $ in R), but I cannot find $S$ such that $A=\left( \overline{S} \right)^\circ$.
Thus, I guess the statement is true, and I am trying to prove it.
I proved that $A\subseteq\left( \overline{A} \right)^\circ$, but I did not manage to proof the other implication yet.
 A: Quoting myself from my note here.
As $(\overline{A})^\circ$ is open and a subset of $\overline{(\overline{A})^\circ}$ trivially, by maximality of interior we have 
indeed $$ (\overline{A})^\circ \subseteq (\overline{(\overline{A})^\circ})^\circ$$
Also $(\overline{A})^\circ \subseteq \overline{A}$, (the interior of a set is a subset of it) this implies (taking the closure on both sides  using $\overline{A}$ is closed already) that 
$\overline{(\overline{A})^\circ} \subseteq \overline{A}$, and then taking the interior on both sides (which preserves the inclusion) gives 
$$(\overline{(\overline{A})^\circ})^\circ \subseteq \overline{A}^\circ$$ so we have equality $$(\overline{(\overline{A})^\circ})^\circ = \overline{A}^\circ$$
A: A slightly more streamlined proof.

*

*Taking the closure or interior both preserve inclusions in the sense that $B\subseteq C$ implies $B^\circ\subseteq C^\circ$ and $\bar B\subseteq \bar C$.


*One has $B^\circ \subset B\subset \bar B$.
This means that in any combination of taking the interior or closure, we can add the operation $(\ )^\circ$, resp $\overline{(\ )}$, and make the resulting space smaller, resp bigger.  Thus,
$$
\overline{A^\circ}
= \overline{\bigl(A^\circ\bigr)^\circ}
\subset \overline{\overline{A^\circ}^\circ}
\subset \overline{\overline{A^\circ}}
= \overline{A^\circ}
$$
A: I will use the notations $\operatorname{Cl}(U)$ for the closure of a subset $U$, and $\operatorname{Int}(U)$ for the interior of a subset $U$. So you claim $\operatorname{Int}(\operatorname{Cl}(\operatorname{Int}(\operatorname{Cl}(S)))) = \operatorname{Int}(\operatorname{Cl}(S))$. Instead, we will show $\operatorname{Cl}(\operatorname{Int}(\operatorname{Cl}(\operatorname{Int}(A)))) = \operatorname{Cl}(\operatorname{Int}(A))$ (as this is equivalent: just take $A = $ complement of $S$).
$\subset$  Let $B = \operatorname{Cl}(\operatorname{Int}(A))$.  The interior of $B$ is the largest open subset inside of $B$, so $\operatorname{Int}(B) \subset B$.  \operatorname{Cl}osure preserves subsets, so $\operatorname{Cl}(\operatorname{Int}(B)) \subset \operatorname{Cl}(B) = B$.  Note this direction makes no use of the topological properties of $B$ apart from being closed.
$\supset$ Is $B$ the smallest closed set containing $\operatorname{Int}(B)$?  Suppose, to prove by contradiction, that $\operatorname{Int}(B) \subsetneq C \subsetneq B$ with $C$ closed.  By writing this statement more explicitly, we have $\operatorname{Int}(\operatorname{Cl}(\operatorname{Int}(A))) \subsetneq C \subsetneq \operatorname{Cl}(\operatorname{Int}(A))$.  We know any set is a subset of its closure and interior preserves open sets, so $\operatorname{Int}(A) \subset \operatorname{Int}(\operatorname{Cl}(\operatorname{Int}(A))) \subsetneq C \subsetneq \operatorname{Cl}(\operatorname{Int}(A))$.  But this last statement is tantamount to saying that there is a closed subset strictly between $\operatorname{Int}(A)$ and $\operatorname{Cl}(\operatorname{Int}(A))$, which is a contradiction of the definition of $\operatorname{Cl}(\operatorname{Int}(A))$.
