Hint on Kuratowski 14-Set Theorem Proof I am trying to prove Kuratowski 14-Set Theorem and in doing so I have arrived at the following claim (for the sake of simplicity I shall denote the closure by $Cl()$ and the complementation by $C()$):

If $B$ is the closure of an open set, then: $$Cl(C(Cl(C(B))))=B$$

I would like to know if this result is true (it seems to be so by looking at fairly complicated examples on the real line) and if so, I would like just some general hint toward its proof!
 A: For any set $U$ we have $X/clU=int(X/U)$. Then for  $\, U=B^{c}$ we get
$X/cl(B^{c})=int(X/B^{c})=intB$. So all we have to prove is that,
$cl(intB)=B$. Clearly $\,\,cl(intB)\subseteq\,B$. Now we will show that
$B\,\subseteq\,cl(intB)$. Assume that $x\in\,B$ and $x\notin\,cl(intB)$.
That implies that there is a $S(x,\epsilon)$ such that $S(x,\epsilon)\bigcap\,cl(intB)=\varnothing$.
But $cl(intB)=\bigcap F$ where $F$ closed such that $F\supseteq int(B)$.
Therefore $S(x,\epsilon)\bigcap\,F =\varnothing$ for all such $F$.
But $B$ is one of these $F$, because $B\supseteq\,int(B)$. So $S(x,\epsilon)\bigcap\,B=\varnothing$
which is a clear contradiction since $x\in\,B$. Thus $cl(intB)=B$ and the result is proved!!
A: The result is true.
Since $B$ is the closure of an open set, we have $B = Cl(I(X))$ for some $X$, where $I()$ denotes the interior of a set.
So this amounts to showing:
$Cl\circ C\circ Cl\circ C\circ Cl\circ I = Cl \circ I$.
Now some hints for you:
(1) Show in fact we have $I = C \circ Cl \circ C$.
(2) Show indeed $Cl\circ C\circ Cl\circ C\circ Cl\circ C \circ Cl \circ C = Cl \circ C \circ Cl \circ C$, which gives your claim. To make this easier to read: $( Cl \circ C)^4 = (Cl \circ C)^2$.
To help you see why (2) is true, use (1) and also:
(3) $Cl\circ I \circ Cl \circ I = Cl\circ I$. (Show these)
In summary, first convince yourself algebraically that using (1) and (3) gives (2), which implies your claim. Then prove (1) and (3), which can be done by using the definitions of closure (and adherent points) and interior (and interior points).
A: Notice that $U \subseteq B$ is an open set if and only if  $U^c \supseteq B^c$ is closed. So we have
\begin{equation}\operatorname{Cl}(B^c)^c = \left(\bigcap_{\substack{F: F \text{ is closed,}} \\ \substack{\text{and } B^c \subseteq F}} \displaystyle{F} \right)^{\displaystyle{c}}=\bigcup_{\substack{F: F \text{ is closed,}} \\ \substack{\text{and } B^c \subseteq F}} \displaystyle{F^c}=\bigcup_{\substack{F^c: F^c \text{ is open,}} \\ \substack{\text{and } F^c \subseteq B}} \displaystyle{U}=\operatorname{Int}(B).\tag{1}\end{equation}
Now suppose $B=\operatorname{Cl}(U)$ for some open set  $U$. Since $U \subseteq \operatorname{Int}(B) \subseteq B$, we have
\begin{equation}B=\operatorname{Cl}(U) \subseteq \operatorname{Cl}\left(\operatorname{Int}(B)\right) \subseteq \operatorname{Cl}(B)=B.\tag{2}\end{equation}
We combine $(1)$ and $(2)$ to see that we now have
$$\operatorname{Cl}\left(\operatorname{Cl}(B^c)^c\right) = \operatorname{Cl}\left(\operatorname{Int}(B)\right)=B.$$
