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As a part of a problem I'm working on, I think that I need to show that for any set A in any topological space,

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

where the bar denotes closure and the notation $A^o$ denotes the interior of a set.

I have already convinced myself that in general, one cannot assume that $(\overline{A^o})^o = A^o$ since it does not work on the set $(1,2)\cup(2,3)$ in the reals.

But since $(\overline{A^o})^o \subset \overline{A^o}$ by basic properties of the interior, its closure must also be a subset: $ \overline{(\overline{A^o})^o} \subset \overline{A^o}$.

The inclusion in the other direction is messing with me -- it could simply be the excess of repeating symbols. I'm betting it's really simple, even just hinting at me what property to use on which set should be enough. Any suggestions?

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    $\begingroup$ Does one of $A^o$ and $(\overline{A^o})^o$ contain the other? $\endgroup$ May 6, 2017 at 22:04
  • $\begingroup$ I guess that since $A^o \subset \overline{A^o}$ then the interior of both sets maintains the inclusion? That gets me there, thanks! $\endgroup$
    – Opal E
    May 6, 2017 at 22:08
  • $\begingroup$ See my note at.yorku.ca/p/a/c/a/24.htm for full proofs. Identity p) is what you want. But don't skip the notation part. $\endgroup$ May 6, 2017 at 22:12
  • $\begingroup$ That's the problem I'm working on -- I'll wait to check my answer there until I'm done with the second sequence of sets! Thanks, though! $\endgroup$
    – Opal E
    May 6, 2017 at 22:13
  • $\begingroup$ @DanielFischer your comment got me what I needed & is much clearer than the other answer -- if you add it in the answers, I'll be happy to accept it. $\endgroup$
    – Opal E
    May 7, 2017 at 20:22

2 Answers 2

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$A^o$ is an open subset contained in $\overline{A^o}$, so we have

$$A^o \subset \bigl(\overline{A^o}\bigr)^o.$$

Taking the closure yields $\overline{A^o} \subset \overline{\bigl(\overline{A^o}\bigr)^o}$. Together with the inclusion $\overline{\bigl(\overline{A^o}\bigr)^o} \subset \overline{A^o}$ that you got from $\bigl(\overline{A^o}\bigr)^o \subset \overline{A^o}$, this yields equality.

By a similar argument, or by taking complements and using this result, one sees that also $A \mapsto \bigl(\overline{A}\bigr)^o$ is an idempotent operator.

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cl int cl int A = cl int A.
Proof.
cl int cl int A subset cl cl int A = cl int A
= cl int int A subset cl int cl int A.

Likewise the dual statement
int cl int cl A = int cl A
which also comes by taking the complement of both
sides of the previous result and replacing A^c with A.

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