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?

  • 1
    $\begingroup$ Does one of $A^o$ and $(\overline{A^o})^o$ contain the other? $\endgroup$ – Daniel Fischer May 6 '17 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 '17 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. $\endgroup$ – Henno Brandsma May 6 '17 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 '17 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 '17 at 20:22

$A^o$ 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.


cl int cl int A = cl int A.
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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.