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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.

celtschk
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  • What do you mean by "reconstruct the topology from $I$ alone"? – user642796 Sep 16 '12 at 14:18
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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
  • @ArthurFischer It probably means take the topology generated by $I$. – Matt Sep 16 '12 at 14:29
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    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
  • @ArthurFischer: With "reconstruct the topology from $I$ alone" I mean to derive the topological structure of $X$ given nothing but $I$. – celtschk Sep 16 '12 at 14:32
  • @Matt: No, it means to re-derive the topology of the original space $X$. – celtschk Sep 16 '12 at 14:33
  • @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?) – celtschk Sep 16 '12 at 14:41
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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? – t.b. Sep 16 '12 at 14:41
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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". – celtschk Sep 16 '12 at 14:53
  • I see. What if you make this a bit more challenging by adding the Hausdorff condition? – t.b. Sep 16 '12 at 14:58
  • 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. – celtschk Sep 16 '12 at 15:14
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    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
  • @t.b. Thanks for the links. – celtschk Sep 16 '12 at 16:00

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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.

Chris Eagle
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