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I'm asking, in a topological space, does a set $A$'s interior equals to $A$'s closure's interior? Or to ask that do all the sets that have the same closure also have the same interior?

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    I seem to remember one can produce **seven** (but never more) different sets starting from some $A\subset\mathbb R$ and applying repeatedly either the closure or the interior operation.2012-04-26
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    @Didier Interesting result. Is there a reference for this result? I'd be happy to know. A little googling tells me these kinds of results are called Kuratowski-type results after his 14 sets theorem, about which I know only now. Thank you for mentioning this result here. Regards,2012-04-26
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    And open set is called *regular open* or *open domain* if $U=\operatorname{Int}\overline U$. So you questions asks about the sets such that their interior is a regular open set. You can find some basic stuff about regular open sets in many texts on general topology. Also the book Givant, Halmos: Introduction to Boolan Algebras has a [chapter](http://books.google.com/books?id=ORILyf8sF2sC&pg=PA66) on regular open sets; they give quite detailed proofs of several basic facts on regular open sets there.2012-04-26
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    @Kannapan: The 7-sets result I mentioned in a previous comment is attributed to T.A. Chapman in the introduction of [this paper](http://www.latrobe.edu.au/mathstats/department/algebra-research-group/Papers/GJ_Kuratowski.pdf) on the Kuratowski closure-complement theorem, for which the maximal number of different sets becomes 14.2012-04-29

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David's answer is more than sufficient, but I just wanted to point out it's not even true if $A$ is open to begin with. If $A = (0,1)\cup (1,2)$, then $\overline{A} = [0,2]$ so $(\overline{A})^o = (0,2) \neq A$.

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    Good answers both-+1 for both you and David. : )2012-04-26
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No, consider the rationals in $\Bbb R$ with the usual topology.