Is there a Boolean algebra without $\sigma$-complete ultrafilters?

I am looking for an example (in ZFC) of a Boolean algebra $A$ such that no ultrafilter on $A$ is $\sigma$-complete. Since every principal ultrafilter is complete, there has to be no principal ultrafilter on $A$, so $A$ has to be atomless.

Alternatively, can one show (in ZFC) that every Boolean algebra has a $\sigma$-complete ultrafilter?

• What about an atomless boolean algebra in which no strictly decreasing sequence has an infimum? $\mathcal{P}(\omega)/ fin$ is a classic example. – Jonathan Apr 24 '18 at 17:55
• @Jonathan It works perfectly, thank you. I did not realized that an ultrafilter may violate $\sigma$-completeness by containing a sequence $(a_n)_{n\in\omega}$ such that $\bigwedge a_n$ does not exist in $A$, not just that $\bigwedge a_n$ is not in ultrafilter. What happens if I moreover ask that $A$ is $\sigma$-complete? – Peter Elias Apr 24 '18 at 19:15
• Well, it is not so clear what the definition of "$\sigma$-complete ultrafilter" should be in an algebra that is not $\sigma$-complete. What definition do you have in mind? (It seems to me the most natural definition would be that if $C$ is a countable subset of the ultrafilter, then there exists an element of the ultrafilter that is less than or equal to every element of $C$.) – Eric Wofsey Apr 24 '18 at 19:23
• @EricWofsey Yes, that definition looks very natural. However, I am mostly interested in the case of a $\sigma$-complete algebra. Actually I am trying to find out whether $\sigma$-complete ultrafilters (of a $\sigma$-complete algebra) may serve as points in some sort of representation of that algebra (in a Stone or Loomis-Sikorski style). – Peter Elias Apr 24 '18 at 19:40

Less trivially, a simple counterexample is any countable atomless Boolean algebra $B$ (in fact, up to isomorphism, there is only one nontrivial such algebra, the clopen algebra of the Cantor set). If $U$ is any ultrafilter in $B$, note that $\bigwedge U=0$ (if $u\in B$ were any nonzero lower bound of $U$, then it would have to be an atom generating $U$, but $B$ is atomless). Since $U$ is countable, this implies it is not $\sigma$-complete.
[As I commented, there are multiple definitions of what "$\sigma$-complete ultrafilter" might mean in a Boolean algebra that is not itself $\sigma$-complete, but this example will fail all possible definitions, since the meet $\bigwedge U$ does exist in $B$.]
If you want a nontrivial example of a $\sigma$-algebra (or even a complete Boolean algebra) with no $\sigma$-complete ultrafilter, you can take the completion $C$ of a countable atomless Boolean algebra $B$ (explicitly, $C$ would be the regular open algebra of the Cantor set). If $U$ is an ultrafilter in $C$, then $U\cap B$ is an ultrafilter in $B$. Then as above, $\bigwedge U\cap B=0$ in $B$. But the inclusion $B\to C$ preserves all joins and meets that exist in $B$, so $\bigwedge U\cap B=0$ in $C$ as well. Since $U\cap B$ is a countable subset of $U$, this shows $U$ is not $\sigma$-complete.