1
$\begingroup$

A filter $U$ over a boolean algebra $A$ (isomorphic to a powerset algebra) "preserves" a join $a = \bigcup_{i\in I}a_i$, if $a\in U$ implies $a_i\in U$ for some $i\in I$. A join $a$ is infinite if $I$ is. There exist ultrafilters preserving countable sets of infinite joins and, moreover, for an arbitrary non-zero element $e \in A$, there is such an ultrafilter containing $e$.

The question is: if we have given a subset $S\subset A$ with finite intersection property (each non-empty finite subset $T\subseteq S$ has a non-zero meet), and a countable set of infinite joins in $A$, does there exist an ultrafilter containing $S$ and preserving these joins? If the general answer is negative, are there any additional conditions on $A$ or $S$ that ensure the existence of such an ultrafilter?

$\endgroup$
0
$\begingroup$

This is certainly not true in general. For instance, if you take a single infinite join $1=\bigcup a_i$ where the $a_i$ are disjoint, and let $S$ consist of the complements of all the $a_i$, then no proper filter containing $S$ can preserve this join. More generally, an obvious necessary condition is that for each join $a=\bigcup a_i$ which you want to preserve, the filter generated by $S$ cannot contain every $a\setminus a_i$. (Conversely, it is similarly obviously both necessary and sufficient that it be possible simultaneously to choose an $i$ for each join such that $S$ together with all the elements $\neg(a\setminus a_i)$ still has the finite intersection problem. I don't see any way to simplify that condition in general, though.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.