Any two countable atomless Boolean algebras are isomorphic

Question

How to prove that Any two countable atomless Boolean algebras are isomorphic. This is an exercise of Jech - Set theory book for which I have some difficulty.

Answer 1

Sketch: Given two countable atomless Boolean algebras $A$ and $B$, you can construct an isomorphism between them by transfinite induction.

Start by mapping $0$ to $0$ and $1$ to $1$. Then proceed in stages:

In an odd-numbered stage, choose the first element of $A$ that still hasn't been mapped, and map it to the first element of $B$ that compares in the right way to all the finitely many elements that have already been mapped. (This is where you use the fact that $B$ is atomless, which guarantees that it will contain at least one such element). Then add the necessary mappings to make the domain of the isomorphism so far closed under joins, meets, and complements.

Even-numbered stages are the same, except with the roles of $A$ and $B$ swapped.

After $\omega$ steps, the mapping converges to one where every element of $A$ and every element of $B$ has been mapped.

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You may recognize here the structure of the standard argument that any two countable dense linear orders with/without first or last elements are isomorphic. "Dense" and "atomless" serve roughly similar purposes here.