Which forcings preserve (some) determinacy? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T16:05:13Z http://mathoverflow.net/feeds/question/109647 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109647/which-forcings-preserve-some-determinacy Which forcings preserve (some) determinacy? Noah S 2012-10-14T19:56:50Z 2012-10-15T01:19:05Z <p>The question is exactly as in the title. I'm interested in general in all questions of the form "which forcings preserve property P?" for any P, but determinacy assumptions occupy a special place in my interests. More specifically, what results are known of the form "the forcings which preserve $\Gamma$-determinacy are exactly the following: . . ." for $\Gamma$ some reasonably natural pointclass? I am, for the purposes of this question, taking my base set to be $\mathbb{N}$ (that is, all payoff sets are subsets of $\mathbb{N}^\mathbb{N}$).</p> <p>What I've been able to figure out so far (which is not very much):</p> <ul> <li><p>Continuum-closed forcings preserve all determinacy assumptions. This is just because continuum-closed forcings add no new sets of reals - hence no new payoff sets, no new points in old payoff sets, and no new strategies.</p></li> <li><p>Countably closed forcings preserve PD (projective determinacy). A countably closed forcing adds no new reals, and hence preserves the truth of analytic formulas with parameters (=definitions of projective sets), and since no new reals are added, no new strategies are added either. (As Andreas points out below, "countably closed" can be replaced by "adds no new reals.")</p></li> <li><p>Countable closure is <em>not</em> enough to guarantee preservation of AD. The usual construction of a non-determined game can be reformulated as a countably closed forcing construction over a model of ZF; and even if the ground model satisfies AD, the generic extension will have a non-determined game.</p></li> </ul> <p>I have tried to figure out whether either of these results reverse, but I've had no success here. The way I would attempt to phrase such a reversal would be something like the following:</p> <p>(*) If $\mathbb{P}$ is some poset without property $P$, then there is a transitive model of set theory $W$ containing $\mathbb{P}$ and satisfying $\Gamma$-determinacy such that forcing with $\mathbb{P}$ over $W$ does not preserve $\Gamma$-determinacy.</p> <p>[EDIT: As Francois points out, this isn't a good way to phrase a reversal statement, and it's not clear what a good way would be. So as an additional question, how can this idea be phrased in a non-silly way? Or is there good reason to believe that this can't be done?]</p> <p>So, in addition to the main question, I have the following subquestions:</p> <p>(i) Are any results along the lines of (*) known?</p> <p>(ii) What methods seem like they could be useful for proving results along the lines of (*)?</p> <p>(iii) For that matter, is my reasoning in the bullet points above correct? It seems straightforward enough, but I've been very wrong about these sorts of things before.</p> <p>Thanks in advance!</p> <p>[EDIT: I forgot to mention this initially, but for the purposes of this question I'm assuming the consistency of arbitrary large cardinals, although I am very interested in how much large cardinal strength any answers require.]</p> http://mathoverflow.net/questions/109647/which-forcings-preserve-some-determinacy/109664#109664 Answer by Joel David Hamkins for Which forcings preserve (some) determinacy? Joel David Hamkins 2012-10-15T00:51:21Z 2012-10-15T00:51:21Z <p>Here is a way to answer for projective determinacy. The basic situation is that if there are sufficient large cardinals, then projective determinacy is indestructible by any kind of forcing. </p> <p>First, if it is consistent with ZFC that there are infinitely many strong cardinals, then it is consistent with ZFC that $\Gamma$-determinacy is exactly preserved by all forcing notions, for any projective class $\Gamma$. The reason is that Kai Hauser proved (see his <a href="http://www.google.com/url?sa=t&amp;rct=j&amp;q=projective%20absoluteness&amp;source=web&amp;cd=4&amp;cad=rja&amp;ved=0CDQQFjAD&amp;url=http%3A%2F%2Fwww.math.cas.cz%2F~jech%2Flibrary%2Fhauser%2Fconsistency.ps&amp;ei=f1l7UIWlMNG-0QH0q4DgAg&amp;usg=AFQjCNG-2jSdMsmK5pvgn8OSzdW7yjXV3A" rel="nofollow">habilitation</a>) that the existence of infinitely many strong cardinals is equiconsistent with projective absoluteness, which means that any given projective assertion is absolute to any forcing extension. Since $\Gamma$-determinacy is projective whenever $\Gamma$ is, this means that under projective absoluteness, $\Gamma$-determinacy is exactly preserved to all forcing extensions.</p> <p>Secondly, if there is a proper class of Woodin cardinals, then not only does $\text{AD}^{L(\mathbb{R})}$ hold, but the theory of $L(\mathbb{R})$ is absolute by forcing, which means that PD will continue to hold in all forcing extensions, since this is expressible as a part of the theory of $L(\mathbb{R})$.</p> <p>This would seem to pour some cold water on any nontrivial version of $\ast$ in the presence of large cardinals. </p>