TITLE: Reduced products of complete Boolean algebras
QUESTION [6 upvotes]: I expect that complete Boolean algebras are not closed under reduced products modulo $\kappa$-complete filters, for any regular cardinal $\kappa$. Is it true? And, is a reference for this?

REPLY [8 votes]: If you write $2$ for the $2$-element Boolean algebra, and $F$ for the filter of cobounded sets in $\kappa$, then $2^\kappa/F$ is just $P(\kappa)/[\kappa]^{<\kappa}$, which is certainly not complete. 
Here is an ad hoc proof that it is not complete, but I am sure that deeper reasons could be found. If it were complete, you could find a maximal antichain $ (b_i:i\in \kappa)$.  Let $c_i:= \bigvee_{j\ge i} b_j$.  The elements $c_i$ are equivalence classes of sets, equivalent mod $F$.  Choose representatives $C_i$ in $c_i$.  Let $C'_i:= C_i \cap \bigcap_{k<i} C_k$, then $C'_i =_F C_i$, and the $C'_i$ are  actually decreasing. Now find a pseudo-intersection $C$ of the $C'_i$ (by induction), of size $\kappa$. Then $[C]_F$ contradicts the maximality of the $b_i$.