It is a simple consequence of AD that there are no non-principal ultrafilters on $\omega$: for $U$ an ultrafilter on $\omega$, consider the game $G_U$ where players I and II play natural numbers $x_0$ < $y_0$ < $x_1$ < $y_1$ < . . .

Let $$R_I=\lbrace 0, 1, 2, . . . , x_0\rbrace\cup\lbrace y_0+1, y_0+2, . . , x_1\rbrace\cup . . .$$ and $$R_{II}=\lbrace x_0+1, x_0+2, . . . , y_0\rbrace\cup\lbrace x_1+1, x_1+2, . . , y_1\rbrace\cup . . .$$ Then either $R_I$ or $R_{II}$ is in $U$. Say that I wins if $R_I\in U$, and II wins otherwise. Then if $U$ were non-principal, neither player can have a winning strategy, by a strategy-stealing argument.

Bringing this proof down into the AC world, we have that, for instance, projective determinacy implies that no non-principal ultrafilter on $\omega$ is projective. The hypothesis here can't be removed completely: assuming $V=L$, there are projective non-principal ultrafilters.

My question is whether this reverses. Does "every projective ultrafilter is principal" imply PD over ZFC? If not, then what are the consequences of this claim - and what is the consistency strength of ZFC+"every projective ultrafilter is principal?"


The answer is no. The strength of the theory "ZFC+ there is no projective nonprincipal ultrafilter" is at most that of an inaccessible cardinal, which is strictly weaker than PD.

The reason is that if there is an inaccessible cardinal, then in the Solovay model, the $L(\mathbb{R})$ of the Levy collapse $V[G]$, every set is Lebesgue measurable. It follows that every projective set in $V[G]$ is Lebesgue measurable, and consequently, there can be no nonprincipal projective ultrafilters in $V[G]$, since the existence of a nonprincipal ultrafilter implies the existence of a non-measurable projective set, namely, the ultrafilter itself, as I explain in my answer to a remark of Connes.

Since an inaccessible cardinal has weaker consistency strength than PD, which is equiconsistent with infinitely many Woodin cardinals the assertion that every real is in an inner model with an arbitrarily large finite number of Woodin cardinals (see Andres's comment), we cannot make a model of PD this way (unless our theories are inconsistent).

Perhaps there may be a way to get rid of the inaccessible, by using the property of Baire instead of Lebesgue measurability. [Update:] And indeed, by a result of Shelah, it is equiconsistent with ZFC that every projective set has the property of Baire, and in that model, there can be no projective nonprincipal ultrafilters, since they would need to be meager and this leads to a contradiction. (See comments by Asaf and Andreas.) So the theory "ZFC + there is no projective nonprincipal ultrafilter" is equiconsistent with ZFC.

  • $\begingroup$ Oh drats. I was going to say that you can get rid of large cardinals using BP, but then I saw that you already mentioned that. $\endgroup$ – Asaf Karagila Oct 15 '12 at 18:29
  • $\begingroup$ Do you have an easy argument that ultrafilters do not have BP? $\endgroup$ – Joel David Hamkins Oct 15 '12 at 18:34
  • 1
    $\begingroup$ I don't recall the arguments in details, but we did prove that last year in the descriptive set theory course I attended. This is in the same vein as their non-measurability. If it had BP then it would have to be meager; but it cannot be meager because the complement is continuous (or Baire?) so it would contradict the fact that $2^\omega$ is not the union of two disjoint meager sets. $\endgroup$ – Asaf Karagila Oct 15 '12 at 18:37
  • 1
    $\begingroup$ If a filter $F$ on $\omega$ contains all the cofinite sets and has the Baire property, then (it is meager and), by a theorem of Talagrand, there is a partition of $\omega$ into a sequence of finite intervals such that every set in $F$ intersects all but finitely many of the intervals. Equivalently, there's a finite-to-one map sending $F$ to the cofinite filter. So $F$ is very far from being an ultrafilter. $\endgroup$ – Andreas Blass Oct 15 '12 at 22:15
  • $\begingroup$ I guess just knowing it is meager already implies directly that it is not an ultrafilter, since the bit-flipping map preserves meager-ness, and so would make the whole of Cantor space the union of two meager sets. $\endgroup$ – Joel David Hamkins Oct 15 '12 at 23:49

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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