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?
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$\begingroup$ I guess that $\kappa$ is not only the completeness of the filter, but also the number of factors in the product, correct? $\endgroup$– GoldsternMar 10, 2020 at 9:45
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1$\begingroup$ Finite Boolan algebras are complete. Is the $2$-element Boolean algebra a good test case? Are such reduced powers of the $2$-element Boolean algebra complete? $\endgroup$– bofMar 10, 2020 at 10:00
1 Answer
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 $b_i$.
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$\begingroup$ Goldstern Thank you for your answer. The number of factors could be larger than $\kappa$ but this is even better. $\endgroup$ Mar 10, 2020 at 11:28