Let $\alpha$ be an ordinal, and let $a\subseteq\alpha$ such that $\alpha$ is countable in $L[a]$. Moreover, let $\beta>\alpha$ be an ordinal such that, in $L[a]$, $\alpha$ and $\beta$ have the same cardinality and such that $\beta$ is "reasonably closed", say $L_{\beta}[a]\models\text{ZF}^{-}$. Denote by $P_{\alpha}$ the forcing of finite partial functions from $\alpha$ to $\{0,1\}$, ordered by reverse inclusion.

Does it follow that there is a subset $x\subseteq\alpha$ that is $P_{\alpha}$-generic over $L_{\beta}[a]$ (i.e. generic with respect to all dense subsets in $L_{\beta}[a]$) and such that $L[x]=L[a]$?

In other words, can such "locally generic" sets encode everything?

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    $\begingroup$ I think that I don't understand the question, or that some assumptions are missing: if we take $\alpha = \beta$ to be a regular uncountable cardinal in $L$ then every maximal antichain in $P_\alpha$ in $L$ belongs also to $L_{\beta}[a]$. Thus, $L_\beta[a]$-generic is the same as $L$ generic. If $a$ cannot be added by the Cohen forcing then clearly $a\notin L[x]$ for any $L_\beta[a]$-generic filter $x$. $\endgroup$ – Yair Hayut Oct 8 '18 at 19:47
  • $\begingroup$ You are right, the question was missing the assumption that $\alpha$ is countable in $L[a]$. Sorry for that. $\endgroup$ – M Carl Oct 9 '18 at 9:52

No. Suppose $a$ is a random real over $L$. Then $a$ is random over any $L_\alpha$ satisfying $ZF^-$ plus "$\mathbb R$ exists." If there were $x$ as in your question, then it would be $Add(\omega,\alpha)$-generic over $L_\beta$. But by the well-known orthogonality of random and Cohen forcing, $L[x]$ thinks there are no random reals over $L$, and $L[a]$ thinks there are no Cohen reals over $L$.

We could also do this assuming $\alpha$ is an $L$-cardinal which is countable in $L[a]$, but the above argument also covers many $\alpha$ which are countable in $L$.

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