# Which forcings preserve (some) determinacy?

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.

[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.]

• 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. 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? Oct 14 '12 at 21:57
• (Sorry, I screwed up my quotation marks: 'has property $P$."' should be 'has property $P$.""') 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 $\omega_1$. Oct 15 '12 at 1:05
• @Andreas, good point. Edited. Oct 15 '12 at 1:18

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.

• 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. 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? Oct 15 '12 at 1:01
• Quick question: what is a "projective class"? Oct 15 '12 at 1:17
• 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. Oct 15 '12 at 2:11
• @AndrésE.Caicedo Yes, that's true (that issue cropped up in the comments below my answer as well!). Jul 24 '20 at 21:10

Bumping to mention/advertise a recent development:

Today, William Chan and Stephen Jackson posted this paper to the arxiv. They proved that a broad class of forcings (under reasonable hypotheses) always kill determinacy. Specifically, they showed

Theorem 3.2: ZF + AD proves that every well-orderable forcing of cardinality $$<\Theta$$ forces $$\neg$$AD

and

Theorem 5.6: ZF + AD + "$$\Theta$$ is regular" proves that every nontrivial forcing which is a surjective image of $$\mathbb{R}$$ forces $$\neg$$AD, as does ZF + AD$$^+$$ + $$\neg$$AD$$_\mathbb{R}$$ + V=L$$(\mathcal{P}(\mathbb{R}))$$.

• Note that we generally don't have $$L(\mathbb{R})^V[g]=L(\mathbb{R})^{V[g]}$$ (which would of course contradict these results).

Below I'll give a tissue-thin, hopefully-not-too-inaccurate description of their paper.

The arguments in the paper generalize the existing proofs of "determinacy-destruction" for a wide class of forcings under various assumptions (specifically: Cohen forcing, $$Col(\omega_1,\omega_2)$$, and $$Col(\omega,\omega_1)$$) from prior analysis by Ikegami and Trang:

Ikegami and Trang initiated the study of the preservation of AD under forcing. They showed that many forcings, such as Cohen forcing, can never preserve AD. They also showed that if one is working with natural models of AD, i.e. models satisfying ZF + AD+ + V = L$$(\mathcal{P}(\mathbb{R}))$$, then any forcing which preserve AD must preserve $$\Theta$$, where $$\Theta$$ is the supremum of the ordinals which are surjective images of $$\mathbb{R}$$. They also showed that the consistency of ZF + AD+ + $$\Theta>\Theta_0$$ implies the consistency of ZF + AD and there is a forcing which preserve AD and increases $$\Theta$$. Thus necessarily this forcing must disturb $$\mathcal{P}(\mathbb{R})$$ by adding a new set of reals

(Unfortunately, it looks like this work hasn't appeared yet, and Ikegami and Trang don't appear in the bibliography. My quick impression, though, is that the Chan/Jackson paper subsumes their results.)

Chan and Jackson isolate a common combinatorial aspect of these examples: the ground club property at $$\kappa$$ (Definition 2.9), that any new club in $$\kappa$$ contains a ground club in $$\kappa$$. The ground club property provides a connection between Ramsey properties and reals (slightly rephrased):

Lemma 2.10: If $$V[G]$$ is a generic extension via a forcing with the ground club property at $$\kappa$$ and $$V[G]\models\kappa\rightarrow(\kappa)^\omega_2$$, then $$\mathbb{R}^V=\mathbb{R}^{V[G]}$$.

From this they quickly deduce Theorem 3.2 above. (A different argument, joint with Goldberg, improves the theorem by replacing "cardinality $$<\Theta$$" with "adds a real" - Corollary 3.5.)

This lemma is also fundamental to the proof of Theorem 5.6, but the argument there is much more complicated, and relies on an analysis of the behavior of $$\Theta$$ after forcing, building off of a previous result by Ikegami and Trang. In particular, they show that (in ZF + AD) any nontrivial forcing which is a surjective image of $$\mathbb{R}$$ and preserves AD must add a real (Fact 4.4), and so must preserve $$\Theta$$ (Lemma 4.3 - which surprised me quite a bit).

I hope I haven't misrepresented anything here; regardless of my understanding, this is certainly a very relevant paper.

• How is this possible? Suppose $c$ is a Cohen generic. Then $L(\mathbb R)[c] = L(\mathbb R)^{V[c]}$, since every new real is definable from a ground model real and $c$. Mar 19 '19 at 17:05
• @MonroeEskew Am I misreading their corollary 3.5? (I'm just the messenger, after all.) Also, are you sure about the equality you've stated? $\mathbb{R}^V\in L(\mathbb{R})^V[c]$, but it's not obvious to me that $\mathbb{R}^V\in L(\mathbb{R})^{V[c]}$. Mar 19 '19 at 17:07
• Oh I see, you're right. $L(\mathbb R)[c]$ is not the same as $L(\mathbb R \cup \{ c \})$. Mar 19 '19 at 17:10