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An open subset U of a space X is regular if it equals the interior of its closure, as we learn from the Wikipedia glossary of topology. Furthermore, the regular open subsets of a space (any space) form a complete Boolean algebra.

I'm coming to this from logic and algebra, with not much background in topology. I can't figure out which topologies have "interesting" collections of regular open sets. For example, in the trivial topology and the discrete topology, every open set is regular, if I'm not mistaken. Those are not "interesting" topologies. I assume there are other topologies in which the space X and the null set are the only regular open sets. If so, those aren't "interesting" either, at least not with regard to their regular open ets.

I believe what I'm looking for is topologies in which every open set is regular, other than the ones I've just described. Thanks for any help.

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up vote 10 down vote accepted

Well, for starters, if your space is $T_1$ (so that one-point sets are closed), then it must be discrete.

Proof: Suppose $X$ is $T_1$ and every open subset of $X$ is regular. Let $y \in X$. Then $X \backslash \{y\}$ is open. Now $X \backslash \{y\}$ must also be closed. For if not, then its closure is necessarily $X$, whose interior is $X$, not $X \backslash \{y\}$, contradicting regularity. Since $X \backslash \{y\}$ is closed, $\{y\}$ is open.

This suggests to me that there are not going to be very many interesting topologies with this property.

Addendum: Another possibility would be to consider spaces for which the regular open sets form a basis for the topology. In some sense, this ensures that there are "enough" regular open sets. Such spaces include $\mathbb{R}^n$, all topological manifolds, and all Banach spaces. I can't offhand think of an example of a space without this property; can anyone? If there are interesting necessary and/or sufficient conditions for this property, better still.

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The Zariski topology of an irreducible variety has the property that non-empty open sets are dense. –  t.b. Jul 19 '11 at 1:06
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Steen and Seebach in Counterexamples in topology call a Hausdorff space in which the regular open sets form a base semi-regular. That property lies somewhere between $T_2$ and $T_3$, see page 16. –  t.b. Jul 19 '11 at 1:57
    
Nate and Theo: Thanks for your replies. I have one follow-on question. What about the Cantor space? This has a countable basis of clopen sets, and every clopen set is regular open. But I can't quite see if this entails that it has a basis of regular open sets. I'm out of my depth here. My interest is in logic, where the Cantor space keeps popping up, is why I ask. –  MikeC Jul 19 '11 at 4:38
    
@Michael: I'm not sure I follow you. Yes, the Cantor space has a (countable) basis of clopen sets. What should be of interest to you is also the Wikiedia page on zero-dimensional spaces –  t.b. Jul 19 '11 at 5:02

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