Boolean algebras without atoms Why is the theory of Boolean algebras without atoms $\omega$-categoric?
 A: It isn't true generally that all atomless Boolean algebras of the same cardinality are isomorphic, so we don't expect $\kappa$-categoricity, and the diversity of such Boolean algebras give rise to all the various distinct forcing notions.
But in the countable case, it turns out that there is just one atomless Boolean algebra up to isomorphism. There are evidently a variety of proofs.

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*Here is a 1972 article by Abian that gives a brief topological proof, as well as a detailed proof for the case of atomless Boolean rings.


*This book by Givant seems to have an explanation of the proof using the back-and-forth technique, which I believe is probably how you would prefer to understand it. (Here is a link to [the Google Books version][3], where you can see a complete and fully detailed proof with exercises afterward.)
[3]: https://books.google.com/books?id=ORILyf8sF2sC&pg=PA134&lpg=PA134&dq=countable+atomless+Boolean+algebra&source=bl&ots=YL2jQrXiS9&sig=PTWu4vLDZO2TqSgHj_ksKjZ6oFU&hl=en&ei=fYHRTYCqKILe0QGK86yODg&sa=X&oi=book_result&ct=result#v=onepage&q=countable atomless Boolean algebra&f=false
A: An answer using back-and-forth is as follows:
Let $A=\{a_n:n\in \mathbb N \}$ and $B=\{b_n:n\in \mathbb N \}$ with $a_0=0=b_0$ and $a_1=1=b_1$.
Put $\pi(a_i)=b_i$ for $i=0,1$.
Suppose we have defined $\pi$ for the subalgebra generated by $a_0,\ldots,a_n$, in a way that $\pi$ restricted to this subalgebra is an embedding into $B$. Let $x_0,\ldots,x_m$ be the atoms of $\langle a_0,\ldots,a_n \rangle$.
If $a_{n+1}\in \langle a_0,\ldots,a_n \rangle$, we are done.
Now suppose $a_{n+1}\notin \langle a_0,\ldots,a_n \rangle$, let $y_0,\ldots,y_k$ be the atoms of $\langle a_0,\ldots,a_{n+1} \rangle$, then for each $i\leq m$ we can find $y_{i,0},\ldots,y_{i,n_i}\in \{y_0,\ldots,y_k\}$  with $y_{i,0}+\cdots+y_{i,n_i}=x_i$. Then $\{y_{i,j}:i\leq m, j\leq n_i\}=\{y_0,\ldots,y_k\};$ as $\Sigma\{x_0,\ldots,x_n\}=1$.
Since $B$ is atomless, for each $i\leq m$ we can find $w_{i,0},\ldots,w_{i,n_i}\in B$ pairwise disjoint such that $w_{i,0}+\cdots+w_{i,n_i}=\pi(x_i)$, then $\{w_{i,j}:i\leq m, j\leq n_i\}$ is a set of pairwise disjoint elements of $B$ whose sum is $1$, hence if we define $\pi(y_{i,j})=w_{i,j}$ for all $i\leq m$ and $j\leq m_i$, $\pi$ is an embedding into $B$ and $\pi$ is an extension of our previous definition of $\pi$.
For the "forth" step interchange the roles of $a$ and $b$.
