# When does “sufficient genericity” actually suffice?

Fix a forcing notion $$\mathbb{P}$$. Say that a formula $$\varphi(x)$$ with parameters is $$\mathbb{P}$$-enforceable if there is some countable set $$\mathcal{D}$$ of dense sets in $$\mathbb{P}$$ such that for every $$\mathcal{D}$$-generic filter $$G\subseteq\mathbb{P}$$ there is some $$p\in G$$ such that for every $$\mathcal{D}$$-generic filter $$H\subseteq \mathbb{P}$$ with $$p\in H$$ we have $$\varphi(G)\iff\varphi(H).$$ Note that this all takes place within $$V$$ itself, and since $$\mathcal{D}$$ is required to be countable enforceability isn't trivial.

In general, we don't always have that $$\varphi$$ is $$\mathbb{P}$$-enforceable$$^*$$ - however, for reasonably natural $$\varphi$$ and $$\mathbb{P}$$ it does tend to be the case that $$\varphi$$ is $$\mathbb{P}$$-enforceable. I'm interested in the particular case when $$\mathbb{P}$$ is Cohen forcing $$\operatorname{Fin}(\omega, 2)$$ and $$\varphi$$ is projective with only real parameters. I've heard it stated repeatedly that the usual suspects guarantee tameness, but I've never seen a citation for this and the proof isn't immediate to me. So my question is:

Is it actually the case that, assuming reasonable large cardinals, every projective formula $$\varphi$$ with only real parameters is $$\mathbb{P}$$-enforceable? If so, what's a citation for this fact?

$$^*$$Here's one counterexample - indeed, where $$\varphi$$ is parameter-free and projective. Assume $$V=L$$, take $$\mathbb{P}$$ to be the usual Cohen forcing, let $$a$$ be a projective bijection from $$\mathbb{R}$$ to $$\omega_1$$, and let $$b$$ be a bijection between the set of countable sets of dense subsets of $$\mathbb{P}$$ and $$\omega_1$$. We can define by recursion a projective injection $$f:\alpha\rightarrow\mathbb{R}$$ such that $$f(\alpha)$$ is $$b^{-1}(\beta)$$-generic for all $$\beta\le\alpha$$. Now let $$\varphi(x)$$ be the projective formula "$$x\in \operatorname{ran}(f)$$ and $$f^{-1}(x)$$ is a limit ordinal." Of course, this sort of nonsense relies on wild projective sets, so large cardinals rule it out.

## 1 Answer

Assume $$\text{AD}^{L(\mathbb R)}$$. Let $$A$$ be projective. Since $$A$$ is $${}^\infty$$Borel in $$L(\mathbb R)$$, there is a set of ordinals $$S$$ and a formula $$\psi$$ such that $$\omega_1$$ is strongly inaccessible in $$L[S]$$ and for all reals $$x$$, $$x\in A$$ if and only if $$L[S,x]\vDash \psi(S,x).$$ Let $$\mathcal D$$ be the set of dense subsets of Cohen forcing that belong to $$L[S]$$. Suppose $$g$$ is $$\mathcal D$$ -generic. Let $$p\in g$$ decide whether $$L[S,g]\vDash \psi(S,g).$$ Then for any $$\mathcal D$$-generic $$h$$ containing $$p$$, $$L[S,h]\vDash \psi(S,h)$$ if and only if $$L[S,g]\vDash \psi(S,g).$$ Hence $$h\in A$$ if and only if $$g\in A$$. Thus $$A$$ is enforceable.

EDIT: One can prove in ZFC that all $${\bf \Pi}^1_1$$ sets are enforceable. Suppose $$A$$ is $${\bf \Pi}^1_1$$. Let $$T$$ be a tree on $$\omega\times \omega$$ such that $$x\in A$$ if and only if $$T_x$$ is wellfounded. Let $$M$$ be a countable transitive model of enough set theory containing $$T$$. Suppose $$g$$ is Cohen generic over $$M$$ and $$g\notin A$$. Then $$T_g$$ is illfounded, and hence $$T_g$$ is illfounded in $$M[g]$$: otherwise $$T_g$$ would be ranked in $$M[g]$$, and hence wellfounded in $$V$$, a contradiction. It follows that some $$p\in g$$ forces relative to $$M$$ that $$T_g$$ is illfounded in $$M[g]$$, so every $$M$$-generic Cohen real $$h$$ containing $$p$$ is such that $$h\notin A$$. Similarly, if $$g\in A$$, one can find a $$p\in g$$ that enforces this for $$M$$-generics. Thus $$A$$ is enforceable (taking $$\mathcal D$$ equal to the set of dense sets in $$M$$).

EDIT 2: I think I had too much coffee yesterday. A set $$A$$ is enforceable (for Cohen forcing) if and only if $$A$$ has the Baire property. The reason is that the comeager filter on $$\omega^\omega$$ is generated by sets of the form $$\{g\in \omega^\omega : g\text{ is }\mathcal D\text{-generic}\}$$ for $$\mathcal D$$ a countable collection of dense subsets of Cohen forcing. Thus $$A$$ is enforceable if and only if $$A$$ and $$\omega^\omega - A$$ are open on a comeager set, which is easily equivalent to $$A$$ having the Baire property. This subsumes all the other things I said.

• I guess you may want to add that $\mathcal D$ is countable, so $g$ actually exists (in $V$). – Andrés E. Caicedo Dec 15 '19 at 19:17
• Awesome! (And this is more high-powered than I was expecting; I feel less bad now that I couldn't figure it out myself. :P) A side question: Is every $\Pi^1_1$ formula enforceable in ZFC? (The counterexample above is $\Pi^1_2$, if my counting is right, so this would be optimal.) In ZFC alone, while every $\Pi^1_1$ set is $^\infty$Borel $\omega_1$ need not be inaccessible to reals, so at a glance your argument doesn't lift. – Noah Schweber Dec 15 '19 at 19:23
• Another question: do you know a citation for this result? (I'd like to cite it in a paper I'm working on.) If not, do you mind if I include this argument (with attribution of course)? – Noah Schweber Dec 15 '19 at 19:24
• I'm not sure I've seen this result exactly, but this kind of argument comes from Solovay's proof of the consistency of ZF + DC + all sets Lebesgue measurable. No problem if you include the argument, of course. – Gabe Goldberg Dec 16 '19 at 1:56
• I updated the answer with a much simpler solution. – Gabe Goldberg Dec 16 '19 at 19:26