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Suppose that $\phi(x)$ is a $\Delta^{1}_{2}$-formula (without parameters) and let $A:=\{x\subseteq\omega:\phi(x)\}$. It is clear that, e.g. if there are Cohen-generics over $L$, then $A$ cannot be the set of constructible reals, roughly because $\Delta^{1}_{2}$ cannot distinguish between sufficiently high Cohen-generics in $L$ and Cohen-generics over $L$. I wonder whether this can be strengthenend in the following way:

Given appropriate largeness assumptions (existence of generic filters, large cardinals...), at least one of $A$ and $\mathfrak{P}(\omega)\setminus A$ contains real numbers of all degrees of constructibility. In other words, can $\Delta^{1}_{2}$ "separate" degrees of constructibility in a nontrivial way?

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    $\begingroup$ Some weak positive comments: Under appropriate large cardinal assumptions, $\mathcal{P}(\omega)^L$ is countable and hence $L$-Cohen generics exist. More interestingly, suppose there are appropriate large cardinals and both $A$ and $A^c$ are each uncountable. Then since $\Delta^1_2$ sets have the perfect set property (because of the large cardinal assumption), let $P,Q$ be perfect subsets of $A,A^c$ respectively. Then we conclude that both $A$ and $A^c$ have reals of all sufficiently high constructibility degrees - namely, above reals coding $P$ and $Q$ respectively. $\endgroup$ Oct 16, 2018 at 11:55
  • $\begingroup$ I had posted an answer earlier, but I had forgotten about the difference between $\Delta^1_2$ and ``absolutely $\Delta^1_2$." Namely: it could be that $A \subseteq 2^\omega$ is $\Delta^1_2$, but in forcing extensions, the $\Sigma^1_2$ and $\Pi^1_2$-definitions of $A$ differ... $\endgroup$ Oct 19, 2018 at 21:13
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    $\begingroup$ @DouglasUlrich In $L$, I think there is a $\Delta_2^1$ set of reals so that it and its complement does not have a perfect subset. In $L$, the canonical wellordering is $\Sigma_1$. By diagonalizing against perfect trees and putting all reals into $A$ or $A$ complement, I think you make such a set. $\endgroup$
    – William
    Oct 19, 2018 at 21:15
  • $\begingroup$ @William yes, sorry. $\endgroup$ Oct 19, 2018 at 21:16

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Expanding on Douglas Ulrich's answer:

Note that this question is only meaningful when there are nonconstructible reals since otherwise there is only one constructibility degree.

If there is a nonconstructible real and $A$ is $\Delta_2^1$ set, then either $A$ or $\mathbb{R} \setminus A$ is a $\Sigma_2^1$ set of reals with a nonconstructible element. Without loss of generality, suppose $A$ has the nonconstructible real. By the Mansfield-Solovay theorem (using the $\omega_1$-Suslin representation via the Shoenfield Tree $S$ for $A$ which belongs to $L$), there is a constructible tree $T$ so that $[T] \subseteq A$.

Every constructible tree is an $L$-pointed tree. So $[T]$ has every $L$-degree.

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    $\begingroup$ This is also simpler than my argument. And it might be better to say "patching" rather than expanding :) $\endgroup$ Oct 19, 2018 at 21:46

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