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.


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.

  • $\begingroup$ I guess you may want to add that $\mathcal D$ is countable, so $g$ actually exists (in $V$). $\endgroup$ – Andrés E. Caicedo Dec 15 '19 at 19:17
  • $\begingroup$ 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. $\endgroup$ – Noah Schweber Dec 15 '19 at 19:23
  • $\begingroup$ 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)? $\endgroup$ – Noah Schweber Dec 15 '19 at 19:24
  • $\begingroup$ 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. $\endgroup$ – Gabe Goldberg Dec 16 '19 at 1:56
  • $\begingroup$ I updated the answer with a much simpler solution. $\endgroup$ – Gabe Goldberg Dec 16 '19 at 19:26

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