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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 $\mathbb{N}^\mathbb{N}$).

What I've been able to figure out so far (which is not very much):

  • 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.

  • 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.")

  • Countable closure is not 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.

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:

(*) 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.

[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?]

So, in addition to the main question, I have the following subquestions:

(i) Are any results along the lines of (*) known?

(ii) What methods seem like they could be useful for proving results along the lines of (*)?

(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.

Thanks in advance!

[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.]

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Statements like $(\ast)$ are always tricky to formulate. I think you need some more qualifiers on the transitive model $W$. As is, we could take a large $W$ where $\mathbb{P}$ is countable and then everything falls apart. But that's not really an answer since things probably fell apart in $W$ before forcing with $\mathbb{P}$. Working inside a fixed universe $V$ would help but then the assumptions of $(\ast)$ are likely to need more large cardinal power than what you intended. –  François G. Dorais Oct 14 '12 at 20:54
    
Good points. One thing that might be doable is to ask when a theorem of the following form is provable in ZFC (+ large cardinals?): "$Det(\Gamma)\implies$ "if $\mathbb{P}$ is any poset such that $\Vdash_\mathbb{P} $"$Det(\Gamma)$", then $\mathbb{P}$ has property $P$." This is expressible in the language of ZFC, since forcing is definable, as long as $\Gamma$ is a sufficiently nice pointclass. Would this work? –  Noah S Oct 14 '12 at 21:57
    
(Sorry, I screwed up my quotation marks: 'has property $P$."' should be 'has property $P$.""') –  Noah S Oct 14 '12 at 21:58
    
What you wrote about countably closed forcing applies more generally to any forcing that doesn't add reals (since that's all you used from the "countably closed" hypothesis) --- and there are forcings that don't add reals but aren't countably closed, for example the standard forcing to add a club subset in a statinoary, co-stationary subset of $\omega_1$. –  Andreas Blass Oct 15 '12 at 1:05
    
@Andreas, good point. Edited. –  Noah S Oct 15 '12 at 1:18

1 Answer 1

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.

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 habilitation) 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.

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 $L(\mathbb{R})$.

This would seem to pour some cold water on any nontrivial version of $\ast$ in the presence of large cardinals.

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Hi Joel. Yes, the "interesting region" lies between $\omega$ Woodins and the sharp for $\omega$ Woodins. More precisely: If there is such a (fully iterable) sharp, AD holds in $L(\mathbb R)$, and the theory of $L(\mathbb R)$ is invariant under forcing. Otherwise, it is not, and if AD holds in $L(\mathbb R)$, there is an extension where it fails. –  Andres Caicedo Oct 15 '12 at 0:56
    
Thanks for the additional information. This way of thinking does not seem to extend much beyond PD, since as the OP notes, AD is never indestructible by all forcing. Any ideas for an answer in the more general case? –  Joel David Hamkins Oct 15 '12 at 1:01
    
Quick question: what is a "projective class"? –  Noah S Oct 15 '12 at 1:17
    
I just meant a class like $\Sigma^1_5$, a class of projective sets of a uniform type. –  Joel David Hamkins Oct 15 '12 at 1:25
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I'm not sure, Joel. Cohen forcing preserves determinacy, and this generalizes (essentially, you can obtain generic embeddings between the respective $L(\mathbb R)$s), but I do not know of a characterization of the preserving posets, or anything of the sort. –  Andres Caicedo Oct 15 '12 at 2:11

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