Determinacy and definable ultrafilters - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:54:48Z http://mathoverflow.net/feeds/question/109739 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109739/determinacy-and-definable-ultrafilters Determinacy and definable ultrafilters Noah S 2012-10-15T17:54:28Z 2012-10-16T00:08:51Z <p>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 $x_0$ &lt; $y_0$ &lt; $x_1$ &lt; $y_1$ &lt; . . .</p> <p>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.</p> <p>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.</p> <p>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?" </p> http://mathoverflow.net/questions/109739/determinacy-and-definable-ultrafilters/109742#109742 Answer by Joel David Hamkins for Determinacy and definable ultrafilters Joel David Hamkins 2012-10-15T18:07:20Z 2012-10-16T00:08:51Z <p>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. </p> <p>The reason is that if there is an inaccessible cardinal, then in the <a href="http://en.wikipedia.org/wiki/Solovay_model" rel="nofollow">Solovay model</a>, 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 href="http://mathoverflow.net/questions/57072/a-remark-of-connes/57108#57108" rel="nofollow">a remark of Connes</a>.</p> <p>Since an inaccessible cardinal has weaker consistency strength than PD, which is equiconsistent with <strike>infinitely many Woodin cardinals</strike> 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). </p> <p>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. </p>