Exact consistency-strength of "all projective sets are Ramsey" - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:52:01Z http://mathoverflow.net/feeds/question/76280 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76280/exact-consistency-strength-of-all-projective-sets-are-ramsey Exact consistency-strength of "all projective sets are Ramsey" David Schrittesser 2011-09-24T18:01:23Z 2011-09-28T15:27:36Z <p>I wonder if the exact consistency strength of <strong>"All projective sets have the Ramsey property"</strong> is still open. In Solovay's model, all sets have the Ramsey property, so the consistency strength of this is below an inaccessible. As far as I know, there are some implications under forcing axioms. And all sets up to a certain level of the projective hierarchy having the Ramsey property is still do-able from just ZFC. But I can't seem to find anything which shows Solovays inaccessible is really necessary in case of the Ramsey property (like Raisonnier/Shelahs result for Lebesgue measure), or oppositely, that you can get a model where "All projective sets have the Ramsey property" from just ZFC (like for the Baire property).</p> http://mathoverflow.net/questions/76280/exact-consistency-strength-of-all-projective-sets-are-ramsey/76284#76284 Answer by Andres Caicedo for Exact consistency-strength of "all projective sets are Ramsey" Andres Caicedo 2011-09-24T19:46:03Z 2011-09-24T19:46:03Z <p>Hi David. </p> <p>This is still open, and I don't know of any strategy that would result in a model with the property for all projective sets but not in a model with the property for all sets in \$L({\mathbb R})\$.</p> <p>Carlos Di Prisco has worked on this problem, you may want to contact him. Other than Carlos, the person to contact is Andrey Bovykin. He has been working on a project involving Harvey Friedman's "Boolean relation theory" whose goal is to establish the consistency strength of the Ramsey property. Unfortunately, I do not have any additional details on what his approach involves, and am curious as well.</p> http://mathoverflow.net/questions/76280/exact-consistency-strength-of-all-projective-sets-are-ramsey/76650#76650 Answer by Todd Eisworth for Exact consistency-strength of "all projective sets are Ramsey" Todd Eisworth 2011-09-28T15:27:36Z 2011-09-28T15:27:36Z <p>Hi David,</p> <p>This is my first foray onto MathOverflow as well, so this answer is an experiment to see if I can get things to work, rather than an attempt to convey a lot of serious information.</p> <p>As Andres said, the problem is still open as far as I know. I worked on this with Shelah a bit in the late 90s and we generated many things that led nowhere. Some related material:</p> <p>1) Roslanowski and Shelah investigated "sweetness" and "sourness" (properties of ccc posets motivated by the constructions in Shelah's "Can you take Solovay's Inaccessible Away") in a series of quite technical papers early in the 2000s. This was partially motivated by the problem of getting all nicely definable sets to be Ramsey without using an inaccessible.</p> <p>2) CH + "every set of real in L(R) is Ramsey with respect to every Ramsey ultrafilter" is equiconsistent with the existence of Mahlo cardinals. Mathias got the consistency result assuming a Mahlo, and the other direction is in a paper of mine from 1999 or so. The trick I used didn't seem to shed any light on whether or not the inaccessible is needed when we drop the reference to ultrafilters.</p> <p>Best,</p> <p>Todd</p>