Idempotence of the interior of the closure 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.
 A: Provided that you know the following:
For any set $A$, $Int(A)\subset A \subset Cl(A)$ and for any sets $A \subset B$, $Int(A) \subset Int(B)$ and $Cl(A) \subset Cl(B)$ 
As you said, $Int(Cl(A))\subset Cl(Int(Cl(A)))$ and taking interiors gives $Int(Int(Cl(A))) \subset Int(Cl(Int(Cl(A))))$ so $Int(Cl(A)) \subset Int(Cl(Int(Cl(A))))$.
For the other direction, note that $Int(Cl(A)) \subset Cl(A)$ which gives $Cl(Int(Cl(A))) \subset Cl(Cl(A))=Cl(A)$ and this $Int(Cl(Int(Cl(A)))) \subset Int(Cl(A))$.
This is in fact true for any topological space, as is the related fact:
$Cl(Int(Cl(Int(A)))) = Cl(Int(A))$. These two statements tell us that for a given set $A$ we cant obtain a maximum of $7$ sets (including $A$) by taking interiors and closures. A good exercise is to find some $A \subset \mathbb{R}$ where all of the $7$ possible sets are distinct!
A: If we denote by $\mathcal O(X)$ the set of open sets and by $\mathcal C(X)$ the set of closed sets, then $Cl$ is an inclusion-preserving map from $\mathcal O$ to $\mathcal C$ while $Int$ is such a map from $\mathcal C$ to $\mathcal O$. These maps satisfy the properties $U\subseteq Int(Cl(U))$ and $Cl(Int(A))\subseteq A$. It follows that $Cl(U)\subseteq Cl(Int(Cl(U)))\subseteq Cl(U)$ and $Int(A)\subseteq Int(Cl(Int(A)))\subseteq Int(A)$ for all open $U$ and closed $A$. Since $Cl(Y)$ is closed and $Int(Y)$ is open for any set $Y$ in $X$, the equalities follow. 
In the language of category theory this says that $Cl$ is a left adjoint functor to $Int$. Don't worry if you do not know what that means. In an abstract setting the inclusion $U\subseteq Int(Cl(U))$ is replaced by morphisms $U\to F(G(U))$, and the posets $\mathcal O$ and $\mathcal  C$ become general categories with functors $F,G$ instead of order-preserving maps $Cl$ and $Int$. Having a adjoint-functor pair is then a nice relation between $F,G$ implying certain existence and uniqueness statements, which are radically simplified if the categories are just posets like in your case. For example, you could also have proven the equality using the fact that $Cl(U)\subseteq A\iff U\subseteq Int(A)$.
A: 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.
