If $(B\_1, \cdot\_1, +\_1, -\_1)$ is a complete atomic Boolean algebra (where the induced partial order is $\leq\_A$), and $(B\_2, \cdot\_2, +\_2, -\_2)$ is a complete atomic algebra (where the induced partial order is $\leq\_B$), do the class of functions from $B\_1 \to B\_2$ form a complete atomic Boolean algebra where the ordering $\leq\_{[B\_1 \to B\_2]}$ is such that $f \leq\_{[B\_1 \to B\_2]} g$ if and only if, for all $b \in B\_1$, $f(b) \leq\_{B2}g(b)$?



A complete Boolean algebra on a function space
If you want to define an algebra of arbitrary subsets of $B\_1$, then indeed the function space from $B\_1$ to $2$ is what you want, and the boolean algebra structure of $B\_1$ need play no role (though it of course plays a role in defining $\sigma$).


So Andreas Blass's comment gives what you need: The algebra in question is the product of $|B\_1|$ copies of $B\_2$ and thus is a complete atomic Boolean algebra.


