Encoding sets in locally generic sets

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?

• 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 $x$. – Yair Hayut Oct 8 '18 at 19:47
• You are right, the question was missing the assumption that $\alpha$ is countable in $L[a]$. Sorry for that. – 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 $$L$$.