Separability of the Stone space of a free sigma-algebra

Question

let $A$ be the free $\sigma$-algebra on $\omega_1$ free $\sigma$-generators and $X$ its Stone space. Is $X$ separable, i.e. does $X$ contain a countable dense subset?

Answer 1

Of course not. Let $\{b_0, b_1, b_2,\ldots\}$ be a countable subset of the generators.

Define for $b$ in a Boolean ($\sigma $-algebra $b^{(0)} = b, b^{(1)}=b’$ (the complement of $b$).

For any $f \in \{0,1\}^\omega$ we can thus define $$b_f = \bigwedge_{n \in \omega} b_n^{(f(n))}$$ in our free $\sigma$-algebra and note that $f \neq g$ implies $b_f \land b_g =0$.

It follows that the corresponding clopen sets in the Stone space are also disjoint. And so that space has continuum many pairwise disjoint non-empty open sets and as a dense set has to intersect all of them, no dense set of size smaller than continuum can exist. In particular, it’s not a separable space.