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.


2 Answers 2


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.

  • 4
    $\begingroup$ The Zariski topology of an irreducible variety has the property that non-empty open sets are dense. $\endgroup$
    – t.b.
    Commented Jul 19, 2011 at 1:06
  • 1
    $\begingroup$ 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. $\endgroup$
    – t.b.
    Commented Jul 19, 2011 at 1:57
  • $\begingroup$ 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. $\endgroup$
    – MikeC
    Commented Jul 19, 2011 at 4:38
  • $\begingroup$ @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 $\endgroup$
    – t.b.
    Commented Jul 19, 2011 at 5:02

It turns out that every open set is regular $\iff$ every open set is clopen. (Hence, for example, a non-discrete space satisfying this condition is obtained by putting the topology $\{\varnothing,\{1\},\{2,3\},\{1,2,3\}\}$ on the set $\{1,2,3\}$.)

($\Leftarrow$) The proof in this direction is obvious.

($\Rightarrow$) Suppose $X$ is a topological space such that every open set is regular open. Let $A\subset X$. By hypothesis, $iA=ibiA$ where $i$ is the interior operator and $b$ is the closure operator. Since $A$ is arbitrary, this implies that $i=ibi$ on all of $2^X$.

The result now follows from Theorem 2.1 in Gardner and Jackson's 2008 paper, The Kuratowski Closure-Complement Theorem (hereafter referred to as GJ). The authors show that any topological space satisfying the relation $ibi=i$ must be a partition space (the standard definition of which says that an equivalence relation exists such that the equivalence classes form a basis for the topology, but this is equivalent to saying that all open sets are clopen). $\,\square$

Note. Theorem 2.1 classifies all topological spaces according to their Kuratowski monoid, namely the monoid of operators generated by the complement and closure operators. Page 15 of GJ presents the following Hasse diagram for the six different Kuratowski monoids that are possible:

enter image description here

ED stands for extremally disconnected and OU stands for open unresolvable. Interestingly enough, the authors claim that a point-set argument using the additive property $b(A\cup B)=bA\cup bB$ of topological closure is required to prove that there are no topological spaces with Kuratowski monoids strictly between ED and Partition. The proofs of the other six edges are each based on the poset structure of the 14-element monoid labeled Kuratowski.

  • $\begingroup$ Your first example doesn't give a topology consisting of clopen sets. $\endgroup$ Commented Sep 13, 2021 at 9:59
  • $\begingroup$ @HolyMoly thanks I fixed it. $\endgroup$ Commented Sep 13, 2021 at 14:21

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