If all reals are generic, is the set of reals generic? Let $W\subseteq V$ be two models of $\sf ZFC$ with the same ordinals. Is the following situation consistent:


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*For every $x\in\Bbb R^V$ there is some $P_x\in W$ such that for some $G\subseteq P_x$ which is $W$-generic, $x\in W[G]$.

*There is no $P\in W$ and $G\subseteq P$ that is $W$-generic such that $\Bbb R^{W[G]}=\Bbb R^V$.
Namely, each real is [set-]generic over $W$, but the set of reals is not.
This sort of situation of course immediately exclude the case that $V$ is a generic extension of $W$; but also things like when $V=L[r]$ is obtained by coding $W$ into a real $r$.
(We may assume that $\sf CH$ holds in $V$, otherwise we can force it without adding real numbers.)
 A: I met Woodin recently and asked him that. He came up with a solution, modulo some technical assumption which Ashutosh showed to be consistent (although admittedly, not the same suggestion that Woodin had for solving this issue). With his kind permission, I am posting this solution here.


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*$W$ is a model of $\sf ZFC+GCH+$"There are $\aleph_1$ ccc forcings which add independent reals" (call these forcings $\Bbb P_\alpha$).

*$V_1$ is a class generic extension of $W$ in which a proper class of cardinals were collapsed while preserving $\sf ZFC$ (e.g. collapse all $\aleph_{\alpha\cdot\omega+3}$ to $\aleph_{\alpha\cdot\omega+2}$).

*$V_2$ is coding $V_1$ into a subset of $\omega_1$ without adding reals over $W$, so $V_2=W[A]$ where $A\subseteq\omega_1$.

*Finally, $V$ is the finite support product of $\Bbb P_\alpha$ for $\alpha\in A$ over $V_2$.
Since from $W$ to $V_2$ we didn't add any reals, and every real added to $V$ came from a countable part of the product (which is in $W$), it follows that every real number is $W$-generic for some suitable part of the product. But if you had a $W$-generic $G$ (for a set forcing) such that $W[G]$ and $V$ had the same reals, you would be able to extract $A$ and therefore compute the class generic for the now-collapsed cardinals.
