TITLE: Idempotence of the interior of the closure
QUESTION [5 upvotes]: I'm reading the Complex Analysis text by Ahlfors. I'm stuck on exercise 5 on chapter 3:
Prove that $\overline{\overline{\overline{{\overline{X}}^c}^c}^c}^c=\overline{{\overline{X}}^c}^c$. 
I've manged to rephrase the question as $ Int(Cl(Int(Cl(X))))=Int(Cl(X))$.
I've proven one inclusion like so: $Int(Cl(X)) \subseteq Cl(Int(Cl(X)))$, taking interior of both sides gives $LHS \supseteq RHS$. How can I prove the other inclusion? Is there a way of proving both at once?
Thank you!
P.S. in the book, topological spaces haven't been defined yet. So it might be true for metric spaces only.

REPLY [4 votes]: Interior of the closure is the topological operation that models double-negation in intuitionistic logic.   Negation is modeled by interior of the complement, and intuitionistic negation satisfies $N = N^3$ and thus $N^k = N^{k+2}$ for all positive $k$.  Double negation $D = N^2$ therefore satisfies $D = D^2 = D^3 = \dots$.  
The problem in Ahlfors is about $D^2 = D^4$, which trivially follows from the above.  So it is enough to prove two simpler facts, that can be restated in the language of topology with no reference to any logic:

if $D(s)$ is interior of the closure of $s$, and $N(s)$ is interior of the complement of $s$, then $D = N^2 \quad$ (explicitly, $D(s)=N(N(s))$ for all sets $s$)
$N = N^3$

The rest is just associativity of function composition.  
The logic interpretation of the idempotence of $D$ is that $D$ is the operator of projection from intuitionistic onto classical logic.