Countable subsets of complete Boolean algebras

Question

Suppose that $B$ is a complete Boolean algebra and let $S$ be a countably infinite subset of $B$. Let $C$ be the smallest complete Boolean subalgebra of $B$ that contains $S$. Is $C$ isomorphic to $\wp(\mathbb{N})$, the power set of natural numbers?

I think that the answer should be positive as $S$ generates a countable Boolean algebra on its own, so it embeds into $\wp(\mathbb{N})$. Taking completion should be in a sense coherent between $\wp(\mathbb{N})$ and $C$ but it may well be wishful thinking.

Answer 1

This is quite false, as demonstrated by the following theorem of Solovay.

Theorem. Let $X$ be an infinite set with the discrete topology, and consider the product topology on $X^\omega$. Then the regular open algebra $\operatorname{RO}(X^\omega)$ has a countable set of complete generators, and has cardinality at least $|X|$.

(This theorem can be found in S. Koppelberg's Handbook of Boolean Algebras, vol.1 (J.D. Monk, ed.), p.191.)

So there is no bound on the cardinality of a complete Boolean algebra with a countable set of complete generators.