ZFC proves every infinite Boolean algebra has infinitely many ultrafilters. If every ultrafilter over $\omega$ is principal, then $\mathcal{P}(\omega)/\mathrm{fin}$ has no ultrafilter.

Is it consistent with ZF that there is an infinite Boolean algebra with a unique ultrafilter? Thanks for any help.

  • $\begingroup$ I seem a bit lost, perhaps confused about definitions. Isn't the unit interval $[0,1]$ (with $\min$ and $\max$) an infinite Boolean algebra with a unique ultrafilter (as $[0,1]$ is a linear order)? (And, perhaps you mean "Is it consistent...?" rather than "It is consistent...?".) $\endgroup$ – Mirko May 13 '17 at 0:54
  • $\begingroup$ @Mirko It's not a Boolean algebra - what's the complement of ${1\over 2}$? $\endgroup$ – Noah Schweber May 13 '17 at 1:13
  • $\begingroup$ @mirko Thank you for letting me know a grammatical error. But... does your algebra has a negation operation? $\endgroup$ – Hanul Jeon May 13 '17 at 1:13
  • $\begingroup$ @NoahSchweber obviously I am confusing a Boolean algebra with a lattice. It seems the result (about ZFC) stated in the question is well-known, but not to me, any reference? $\endgroup$ – Mirko May 13 '17 at 1:20

Take the Boolean algebra $B=\mathcal{P}(\omega)/\mathrm{fin}\times \{0,1\}$. An ultrafilter on $B$ has the form either $U\times\{0,1\}$ or $\mathcal{P}(\omega)/\mathrm{fin}\times U$ where $U$ is an ultrafilter on the respective coordinate. So if $\mathcal{P}(\omega)/\mathrm{fin}$ has no ultrafilters, then the only ultrafilter on $B$ is $\mathcal{P}(\omega)/\mathrm{fin}\times\{1\}$.


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