numerical formulas over $V$ and $L$

Question

Let $\phi$ be a numerical formula, i.e., all variables range over naturals. Then we know that $\phi^{V} = \phi^{L}$, where $V$ is the universe and $L$ the constructible sets. Our previous equality holds because for a finite ordinal $\alpha$ we have $V_{\alpha} = L_{\alpha}$. However, this is no longer true for bigger ordinals, like $\omega$. I would like to extend this to consider formulas of the form $\forall x \subseteq \omega\, \varphi(x)$, where all quantifiers over $\varphi$ range over natural numbers. What do I mean by extend? To find some $A$ such that $\phi^{L[A]} = \phi^V$ provided that $\phi$ is a formula of the form $\forall x \subseteq \omega\, \varphi(x)$.

Can you give me a hint? Any help is appreciated it.

Answer 1

Actually, this still works:

(Mostowski absoluteness) Every $\Pi^1_1$ sentence is absolute between $V$ and $L$ (even merely assuming $V\models\mathsf{ZF}$).

This is a consequence of the fact that $(i)$ $V$ and $L$ have the same ordinals, $(ii)$ a $\Pi^1_1$ fact is witnessed by wellfoundedness of an associated partial order, and $(iii)$ $\mathsf{ZF}$ proves that a partial order is well-founded iff it admits a ranking function into some ordinal.

Even this can be strengthened: a cute extension of the same argument leads to Shoenfield absoluteness, which applies to all $\Pi^1_2$ sentences. And there things stop, since "Every real is constructible" is $\Pi^1_3$.

• Actually things don't really stop there since there are absoluteness results conditional on large cardinals which - necessarily - contradict $\mathsf{V=L}$. This is a bit technical and is probably best ignored for now, but if you're interested see the discussion here.