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If $A$ is a set of reals such that $A \cap L[x] \in L[x]$ for each real $x$, is there a real $z$ such that $A \cap L[x] \in OD_z^{L[x]}$ for a cone of $x$?

This can be proved under the Axiom of Determinacy using a game of roughly the same complexity as $A$.

Just assuming boldface ${\bf \Delta}^1_2$ determinacy, if $A\cap L[x] \in OD^{L[x]}_z$ for a cone of $x$ then it can be shown that $A$ itself is $OD_z$. So perhaps one can find a counterexample by finding a homogeneous forcing that does not add reals but adds a set $A$ of reals with $A \cap L[x] \in L[x]$ for each real $x$.

(The assumption of ${\bf \Delta}^1_2$ determinacy is necessary here because starting with $V=L$ it is possible to force a new set $A$ with $A\cap L[x] \in OD^{L[x]}$ for each real $x$ by a homogeneous forcing that does not add reals.)

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It's definitely interesting. Do you have a counterexample showing that a parameter $z$ is required? That is, a set $A$ that is amenable to every $L[x]$, but not necessarily being OD there? –  Joel David Hamkins Jul 18 '12 at 1:48
    
I don't know that the parameter $z$ is required, but I put it in there because that's the version of the statement that I know how to prove under AD. –  Trevor Wilson Jul 18 '12 at 6:26
    
Acutally I think that taking $A = \{z\}$ where $z$ is a real that is not OD would typically show that a real parameter (e.g., $z$ itself) is required. –  Trevor Wilson Jul 18 '12 at 20:12
    
Oh, of course ! –  Joel David Hamkins Jul 19 '12 at 0:11
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