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.
Best Answer
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 $$(\overline{(\overline{A})^\circ})^\circ = \overline{A}^\circ$$