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:



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> 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?
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$^\*$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.



When does "sufficient genericity" actually suffice?
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.


