Is this a complete and/or atomic subalgebra of $2^{2^S}$? For some infinite set $S$, let
$W:=\mathcal{P}(S)$
$B:=\mathcal{P}(W)$
$F:= \{p\in B: \exists s\in S\text{ s.t. }p=\{w\in W:s\in w\}\text{ or }p=\{w\in W:s\not\in w\}\}$
$A:= \{p \in B: \forall X\subseteq F(\bigcap X\neq\emptyset\text{ and }\bigcap X\subseteq p \Rightarrow \exists Y\subset X\text{ s.t. }\bigcap Y\subseteq p)\}$.
(We might think of $S$ as a set of independent possible events, $W$ as the set of possibilities (one for each set of events, in which all and only those events obtain), $B$ as the set of propositions (with a proposition identified with the set of possibilities in which it is true), $F$ as the set of fundamental propositions (those saying that some given event either does or does not obtain), and $A$ as the set of amorphous propositions (those which, when true, lack any minimal basis among the fundamental propositions which implies their truth.))
My question is:
Is $A\cup\{W,\emptyset\}$ a Boolean subalgebra of $B$ (under the natural set-theoretic operations)? If so, is it complete and/or atomic?
 A: No.  For instance, take a countably infinite proper subset $T\subset S$, and topologize $\mathcal{P}(T)$ by identifying it with $\{0,1\}^T$ with the product topology.  Say two elements of $\mathcal{P}(T)$ are equivalent if their symmetric difference is finite.  By transfinite recursion, you can construct a subset $p\subset\mathcal{P}(T)$ which is a union of equivalence classes such that $p$ intersects every uncountable closed subset of $\mathcal{P}(T)$ but does not contain any uncountable closed subset of $\mathcal{P}(T)$.  (Sketch: there are $\mathfrak{c}$ uncountable closet subsets and each has cardinality $\mathfrak{c}$, so you can go through them one by one and put an equivalence class which intersects each one in $p$ and another such equivalence class in the complement of $p$.)
I claim then that $p\in A$.  Indeed, suppose $X\subseteq F$, $\bigcap X$ is nonempty, and $\bigcap X\subseteq p$.  If $\bigcap X$ is finite, then $X$ must contain either $\{w\in W:s\in w\}$ or $\{w\in W:s\not\in w\}$ for all but finitely many $s\in S$.  In particular, there is some such element of $X$ for which $s\in T$; let $Y$ be $X$ with that element removed.  Then $\bigcap Y$ will still be contained in $p$ since $p$ is a union of equivalence classes.  If $\bigcap X$ is infinite, then it is an uncountable closed subset of $\mathcal{P}(T)$, so it cannot be contained in $p$.
So $p\in A$, and by the same reasoning, $\mathcal{P}(T)\setminus p\in A$.  However, the union $p\cup(\mathcal{P}(T)\setminus p)=\mathcal{P}(T)$ is not in $A\cup\{W,\emptyset\}$.
