Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I wonder if the exact consistency strength of "All projective sets have the Ramsey property" 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).

share|improve this question
1  
Welcome to mathoverflow, David! –  Joel David Hamkins Sep 25 '11 at 1:03
    
Hi Joel! Thanks! I have the impression I will like it here. See you next week, I guess. –  David Schrittesser Sep 25 '11 at 21:10
add comment

2 Answers

Hi David,

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.

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:

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.

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.

Best,

Todd

share|improve this answer
3  
Hi Todd, Nice to see you here. –  Andres Caicedo Sep 28 '11 at 15:42
    
Good to be here, though it took me a little while to get enough reputation points to comment! ;) –  Todd Eisworth Sep 28 '11 at 19:39
    
Thanks, Todd, great answer! –  David Schrittesser Sep 28 '11 at 20:00
    
Have a few more reputation points on me ... now you owe me a pint! –  Simon Thomas Sep 28 '11 at 21:52
1  
Heh! Now I've got to fight the urge to post a billion questions just for the thrill of "talking" to other set theorists... –  Todd Eisworth Sep 28 '11 at 22:38
add comment

Hi David.

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 $L({\mathbb R})$.

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.

share|improve this answer
1  
You mention a strategy to get a model with the property for all projective sets but not for all sets in $L(\mathbb{R})$. Is there such a strategy for the better understood cases of having Lebesgue-measure or the Baire property? –  David Schrittesser Sep 24 '11 at 20:04
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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