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 $b \in B_1$, $f(b) \leq_{B2}g(b)$?
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1$\begingroup$ As it stands, the Boolean algebra structure of $B_1$ plays no role. Is that intentional? If so, then the answer is affirmative, because you're just forming the product of $|B_1|$ copies of $B_2$. $\endgroup$– Andreas BlassMay 16, 2021 at 18:19
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$\begingroup$ I'm not entirely sure. Basically, I'm trying to replicate the effect of a certain operator $\sigma$ on subsets of $B_1$. Where $A \subseteq B_1$, the operator is defined as $\sigma (A) = \begin{cases} \sqcup A& \text{if } \sqcup A \in A \\ \# & \text{otherwise}\end{cases}$. Ignoring $\#$ (which stands for the undefined object), I have been wondering how to code create an operator which does the same thing as the $\sigma$ operator, but which applies to the characteristic function of $A$, so that $\sqcup$ applies to the characteristic function of $A$. $\endgroup$– user65526May 16, 2021 at 20:14
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$\begingroup$ I was therefore wondering whether we can code up this operation by considering $B_1$ and letting $B_2$ be 2, and then defining a complete, atomic Boolean algebra on the function space between $B_1$ and $B_2$. $\endgroup$– user65526May 16, 2021 at 20:14
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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.