Description of atomless complete Boolean algebras with a countable $\pi$-base

Question

Recall that a subset $A$ of a Boolean algebra $B$ is a $\pi$-base if for every $b>0$ there is $a\in A$ with $0<a\le b$. For example, the definition of atomicity says that atoms constitute a $\pi$-base. In particular, it is easy to see that an atomic Boolean algebras has a countable $\pi$-base iff it is isomorphic to $2^{\mathbb{N}}$ or $2^{n}$. I have a suspicion that the list of complete atomless Boolean algebras with a countable $\pi$-base is also short (perhaps it is unique?). The only example that I know are the regular open sets of metrizable compacts.

What are complete atomless Boolean algebras with a countable $\pi$-base?

Answer 1

To make the answers concrete: the Boolean algebra of clopen subsets of the Cantor set is the unique countable atomless Boolean algebra. Its completion is the regular-open algebra of the Cantor set, which is isomorphic to the regular-open algebra of the unit interval (the map that identifies end points of complementary interval is an irreducible map from the Cantor set onto the unit interval).

And so, yes, the Stone space of the completion is the absolute of the Cantor set, the unit interval, $\dots$, every compact metrizable space without isolated points.

Answer 2

Yes. The complete atomless Boolean algebra with a countable $\pi$-base is unique up to isomorphism. For a proof, we observe that if $B$ has a countable $\pi$-base $A$, then the Boolean algebra $C$ generated by $A$ is a countable atomless Boolean algebra.

But one can use a back-and-forth algorithm to show that all countable atomless Boolean algebras are isomorphic. And the Boolean algebra $B$ is the completion of the Boolean algebra $C$, so since Boolean algebras have only one Boolean completion up-to-isomorphism, the Boolean algebra $B$ is unique up-to-isomorphism.