# Does $\aleph_0$-density of regular open algebra entail existence of countable basis?

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.

• 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). Feb 28 '18 at 20:28
• 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. Feb 28 '18 at 20:36
• Right, it indeed seems impossible to ensure existence of countable base via countability properties of RO(X). Thanks one more time. 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).

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