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My question is somehow related to Closure of the interior of another closure

However, I go a bit further. I have a closed set $X\subseteq \mathbb R^2$ and $Y:=\operatorname{cl}\operatorname{int} X$.

I would like to know whether $Y=\operatorname{cl}\operatorname{int} Y$.

(Here, $\operatorname{cl}$ is the closure and $\operatorname{int}$ is the interior.)

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2 Answers 2

up vote 2 down vote accepted

Let $(X, τ)$ be a topological space. We have the following:
a.) If $F$ is closed, then $\operatorname{int}(\operatorname{cl}(\operatorname{int}(F)))=\operatorname{int}(F)$.
b.) From (a) you can deduce that for any open set $U$, $\operatorname{cl}(\operatorname{int}(\operatorname{cl}(U)))=\operatorname{cl}(U)$
if you set $U:=\operatorname{int}(X)$ you have the desired result.

$\operatorname{Int}(\operatorname{Cl}\operatorname{Int}X) ⊂ \operatorname{Cl}\operatorname{Int}X \Rightarrow \operatorname{Cl}(\operatorname{Int}\operatorname{Cl}\operatorname{Int}X) ⊂ \operatorname{Cl}(\operatorname{Cl}\operatorname{Int}X) = \operatorname{Cl}\operatorname{Int}X,$
$\operatorname{Cl}(\operatorname{Int}X) ⊃ \operatorname{Int}X \Rightarrow \operatorname{Cl}\operatorname{Int}(\operatorname{Cl}\operatorname{Int}X) ⊃ \operatorname{Cl}\operatorname{Int}(\operatorname{Int}X) = \operatorname{Cl}\operatorname{Int}X$

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If you had a good reference to your statement that could be used in a scientific article, I would be 100% happy to accept your answer! Thanks! –  tohecz Dec 27 '12 at 22:54
    
I supplied the proof using a "subset" argument. If you google this subject you will find a PDF that discusses this exact problem and they provide an identical result. link –  Rustyn Dec 27 '12 at 23:02
    
Thanks again! It seems it is quite trivial, so I'll just claim it as a fact, and if the referee complains, I'll include a proof. –  tohecz Dec 27 '12 at 23:07
    
@tohecz Yeah that sounds good. –  Rustyn Dec 27 '12 at 23:09

Suppose $U$ is open, ie $U=U^\circ$. We have $U \subset \overline{U}$ and $\overline{U}^\circ \subset \overline{U}$. Since $U$ is open and contained in $\overline{U}$, we have $U \subset \overline{U}^\circ \subset \overline{U}$. Taking closure of both sides shows $\overline{\overline{U}^\circ} = \overline{U}$.

Now let $U = X^\circ$, $Y = \overline{U}$. Then we have $Y = \overline{Y^\circ}$.

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But then $\emptyset=\operatorname{cl}\operatorname{int}\emptyset$. Or have I overseen something? –  tohecz Dec 27 '12 at 22:40
    
?????? If $X=\{x\}$ then $Y:=clint(X)=\emptyset$, and then $clint(Y)=\emptyset$=$Y$?????? –  Rustyn Dec 27 '12 at 22:40
    
You say "not in general". However, your $X$ surely satisfies my conjecture, because $Y=\emptyset$ and whence $Y=\operatorname{cl}\operatorname{int}Y$. –  tohecz Dec 27 '12 at 22:44
    
I misread the question earlier. I have answered the question as asked now! –  copper.hat Dec 28 '12 at 0:39

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