Suppose that the family $\mathrm{RO}(X)$ of regular open subsets of $(X,\mathscr{O})$ is a basis of $X$. Let the density of $\mathrm{RO}(X)$ (considered as boolean algebra) be $\aleph_0$.

Does $X$ have to be second-countable? If not, what if we add regularity of $X$ (both $T_1$ and $T_3$ separation axioms)? If answers to both questions are negative, what is the maximal cardinality (relative to $|X|$) of the set of points not covered by dense countable $D\subseteq\mathrm{RO}(X)$?


The answer is no, not necessarily.

For a counterexample, consider the Sorgenfrey line, which is the topology on $\mathbb{R}$ with basis consisting of the half-open intervals $[a,b)$. These are each clopen and hence regular open in that topology. Furthermore, every such nonempty open interval contains a nonempty half-open interval $[p,q)$ where $p,q\in\mathbb{Q}$, and so the regular open algebra is $\aleph_0$-dense as a Boolean algebra. The Sorgenfrey line is normal but not second-countable. Thus, it serves as a counterexample to your question.

For the last part of the question, this gives an example where the cardinality of the excess not covered has size continuum.

  • $\begingroup$ Joel, thanks a lot. Is there any non-trivial condition put directly on $RO(X)$ (again, considered as boolean algebra) that could force existence of countable base? (by trivial I mean e.g., $RO(X)$ is countable). $\endgroup$ – Mad Hatter Feb 28 '18 at 20:28
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    $\begingroup$ I'm not sure. I think RO(X) can never be countable (unless finite), since there are no countably infinite complete Boolean algebras. So RO(X) will be size continuum when it is countably-dense and infinite. What the example in my answer shows is that RO(X) up to isomorphism does not determine the topology, since there is only one countably-dense complete Boolean algebra up to isomorphism, but they can come from non-homeomorphic topological spaces. $\endgroup$ – Joel David Hamkins Feb 28 '18 at 20:36
  • $\begingroup$ Right, it indeed seems impossible to ensure existence of countable base via countability properties of RO(X). Thanks one more time. $\endgroup$ – Mad Hatter Feb 28 '18 at 21:03

Joel has already shown you that none of the standard separation axioms is enough to force a positive answer to your question (by the way, the Sorgenfrey line is even perfectly normal).

A notable case for which your question has a positive answer is that of a topological group.

Define the $\pi$-weight of $X$ ($\pi w(X)$) to be the least cardinality of a family $\mathcal{P}$ of non-empty open subsets of $X$ such that for every non-empty open set $U \subset X$ there is $P \in \mathcal{P}$ such that $P \subset U$. It is clear that if $RO(X)$ is a base for $X$ then the density of $RO(X)$ coincides with the $\pi$-weight of $X$. This happens, in particular, if $X$ is a regular space. Now topological groups are regular, and the $\pi$-weight and the weight of a topological group coincide.

(for a proof of that see, for example, Arhangel’skii, Alexander; Tkachenko, Mikhail, Topological groups and related structures, Atlantis Studies in Mathematics 1. Hackensack, NJ: World Scientific; Paris: Atlantis Press (ISBN 978-90-78677-06-2/hbk). xiv, 781 p. (2008). ZBL1323.22001.)

Another case in which $\pi w(X)=w(X)$ is when $X$ is a metric space (because the $\pi$-weight is bounded below by the density and bounded above by the weight, and for metric spaces, the weight and the density coincide).

  • $\begingroup$ So $\pi$-weight is the density of the lattice $\Omega(X)$ of all open subsets of $X$? $\endgroup$ – Mad Hatter Mar 3 '18 at 12:19
  • $\begingroup$ Yes, that's exactly what it is. $\endgroup$ – Santi Spadaro Mar 9 '18 at 11:00

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