TITLE: Non-isomorphic countable Boolean algebras
QUESTION [6 upvotes]: I'm trying to solve the next exercise:
Construct a sequence $\mathcal{B}_0,\mathcal{B}_1, \ldots$ of countable Boolean algebras such that for all $m \neq n$ then $\mathcal{B}_m \ncong \mathcal{B}_n$.

I know that two countable atomless Boolean algebras are isomorphic, so I guess it has something to do with the number of atoms?! But what are examples of these countable Boolean algebras and how can I construct them?

REPLY [2 votes]: Take a countable atomless Boolean algebra $A$ and let $B_n$ be the finite Boolean algebra with $n$ atoms (and $2^n$ elements).
Then the product $A\times B_n$ is a countable Boolean algebra with $n$ atoms, and all the $A\times B_n$ are non-isomorphic.