# Topology from interiors of closed sets?

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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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 $T_3$. –  t.b. Sep 16 '12 at 14:28
A semiregular space is a topological space whose regular open sets (sets that equal the interiors of their closures) form a base. –  Martin Sleziak Sep 16 '12 at 14:31
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? –  t.b. Sep 16 '12 at 14:41
@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". –  celtschk Sep 16 '12 at 14:53
You can use Austin Mohr's handy spacebook to get a list of examples from Steen and Seebach. This yields e.g. Bing's irrational slope topology among many others. –  t.b. Sep 16 '12 at 15:25

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.