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Assume you've got an arbitrary topological space $X$. Now let $I$ be the set of the interiors of all closed subsets of $X$. And now assume you give me $I$, but don't tell me what $X$ is. Can I reconstruct the topology from $I$ alone? If it is not always possible, does there exist any commonly assumed additional feature of topological spaces so that if I restrict $X$ to topological spaces with that feature, the reconstruction is possible?

Note that while all members of $I$ are open by construction, generally not all open sets of $X$ will be in $I$. For example on $\mathbb R$, the set $(-1,0)\cup(0,1)$ is open, but not the interior of a closed set.

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    What do you mean by "reconstruct the topology from $I$ alone"?2012-09-16
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    The sets in $I$ are called *regular open sets* (or *open domains* in Engelking). In Steen and Seebach's *Counterexamples in topology* Hausdorff spaces whose regular open sets form a (sub)base are called *semiregular* which is a separation property strictly between $T_2$ and $T_3$.2012-09-16
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    @ArthurFischer It probably means take the topology generated by $I$.2012-09-16
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    A [semiregular space](http://en.wikipedia.org/wiki/Semiregular_space) is a topological space whose regular open sets (sets that equal the interiors of their closures) form a base.2012-09-16
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    @ArthurFischer: With "reconstruct the topology from $I$ alone" I mean to derive the topological structure of $X$ given nothing but $I$.2012-09-16
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    @Matt: No, it means to re-derive the topology of the original space $X$.2012-09-16
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    @t.b., MartinSleziak: Thanks for the information. So if I know that $X$ is semiregular, I can reconstruct it from $I$ because $I$ is a base. From Wikipedia I get that it is indeed a quite common condition. However, if I don't know in advance that $X$ is semiregular, could I still reconstruct it? (And somewhat related: Could I figure out from $I$ whether the space is semiregular?)2012-09-16
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    What do you mean by re-derive? Are you looking for two distinct topologies on the same space that give rise to the same collection $I$ of open sets?2012-09-16
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    @t.b.: Yes, basically that's it. If no two topologies had the same set of regular open sets, then by knowing $I$ I'd be able to derive the topology which gave rise to $I$. In the mean time Chris Eagle already gave a counter example, so the answer to that question is indeed "No".2012-09-16
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    I see. What if you make this a bit more challenging by adding the Hausdorff condition?2012-09-16
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    Well, that's somewhat covered in the second question: Although not explicitly stated, I'm of course interested in a condition that's as weak as possible. From the comments of you and Martin Sleziak I know that being semiregular suffices, but Hausdorff would of course be better (BTW, what would be an example of a non-semiregular Hausdorff space?). Also, maybe an even weaker condition suffices.2012-09-16
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    You can use [Austin Mohr](http://math.stackexchange.com/users/11245/)'s handy [spacebook](http://austinmohr.com/home/?page_id=146) to get a list of examples from Steen and Seebach. This yields e.g. [Bing's irrational slope topology](http://dx.doi.org/10.1090/S0002-9939-1953-0060806-9) among many others.2012-09-16
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    @t.b. Thanks for the links.2012-09-16

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This isn't possible in general. For example, if $X$ is an infinite set, then the trivial topology and the cofinite topology on $X$ have the same interiors-of-closed-sets.