Consider the following proof of the assertion in the title:
Say $A=\{x\in 2^\omega: \phi(x,y)\}$, where $\phi$ is a $\Sigma_2^1$ formula and $y\in 2^\omega$.
By Shoenfield's absoluteness,
$$x\in A \iff L(x,y)\models \phi(x,y)$$
Because $\omega_1$ is inaccessible to the reals, $\mathcal{P}(\mathbb{P})\cap L(z)$ is countable for every $z\in 2^\omega$, where $\mathbb{P}=\text{Fn}(\omega,2)$ is Cohen forcing. Therefore, the set
$$\{g\in 2^\omega: g\text{ is Cohen generic over }L(z)\}$$
is comeager. Now, for every Cohen generic $g$ over $L(y)$,
$$
\begin{aligned}
g\in A&\iff L(y,g)\models \phi(g,y)\\
&\iff \exists n<\omega \left[g\upharpoonright n \Vdash^{L(y)}\phi(\dot g,\check y)\right]\\
&\iff g\in \bigcup\left\{N_s\cap 2^\omega: s\Vdash^{L(y)} \phi(\dot g,\check y)\right\}
\end{aligned}
$$
and this set is open.
Now, the title of the question is a theorem of ZFC. However, when looking at the proof, I'm confused, because we're $L(x,y)$ and $L(y,g)$ are proper classes, and we're dealing with $L(x,y)\models \phi(x,y)$ for all $\Sigma_2^1$ formulas at once. And we can't write $N\models \psi$ as a property of the set $\psi$ (assuming we've formalized first order logic in ZFC) when $N$ is a proper class by Tarski's Undefinability of Truth. We can metalinguistically define $\psi^N$ one formula at a time, but this does not seem to be the spirit of the proof. Moreover, one can define $\mathbf{\Sigma}_2^1$ without any logic (say, as projections of co-analytic sets).
Is there a flaw in my reasoning? How should one reconcile these points?
Best Answer
To interpret a $\Sigma^1_2$ formula in $L(x,y)$, you don't need a notion of truth in the whole proper class $L(x,y)$. Since the formula involves quantification only over reals, you can just interpret it in the set of reals of $L(x,y)$.