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.

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    $\begingroup$ It is also true that Cl(Int(Cl(Int(A)=Cl(Int(A). $\endgroup$ May 13 '17 at 17:05
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    $\begingroup$ @DanielWainfleet That's also in the note I linked to. $\endgroup$ May 13 '17 at 18:05
  • $\begingroup$ Note: This is a counterpart to the claim posted at math.stackexchange.com/questions/2269110/… , which says the same thing but with the roles of "closure" and "interior" swapped. (Of course, it is equivalent to that claim, because if you can replace all subsets by their complements, then closures and interiors trade places.) $\endgroup$ May 23 '19 at 14:22

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 $$(\overline{(\overline{A})^\circ})^\circ = \overline{A}^\circ$$

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    $\begingroup$ Thank you very much for the clear answer and for the link! $\endgroup$
    – A-B-izi
    May 13 '17 at 15:28

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 $\operatorname{Cl}(\operatorname{Int}(A))$.


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