Why algebra generated by any set $A$ is of the form $\cup_{i=1}^{m}\cap_{j=1}^{n_i} A_{ij}$?

Question

Perhaps this is a duplicate question, I was trying to find a proof that an algebra generated by a countable collection is itself countable. As in

Algebra generated by countable family of sets is countable?

and

algebra generated by a countable set is countable

But none of these questions gave complete proof, I just couldn't understand the hint and in the second question, the OP mentioned

"algebra generated by any set $A$ is of the form $\cup_{i=1}^{m}\cap_{j=1}^{n_i} A_{ij}$ where $A_{ij}$ or $A_{ij}^c$ is in $A$."

Why is that? Also, can you please give a proof of countably generated algebra is also countable (even a short one) based on this fact?

Answer 1

Check that all sets of that form do form an algebra $\mathcal{A}$ and it contains $A$ and any algebra containing $A$ must contain all these sets, so $\mathcal{A}$ is minimal.

And any member of it is described by a finite sequence of finite sequence of integers, really: Find a bijection $f:\mathbb{N} \to A$ and encode $A_{ij} \in A$ as $l$ when we use $A_{ij}=f(l)$ and as $-l$ when we use $A_{ij}^\complement$.

Show by set theory that the set of all finite sequence sets is countable.