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Let $W\subseteq V$ be two models of $\sf ZFC$ with the same ordinals. Is the following situation consistent:

  1. 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, $x\in W[G]$.

  2. 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.)

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  • $\begingroup$ What happens if we consider the following: Let $(x_i: i<c^V)$ enumerate $\mathbb{R}^V,$ and for each $i$ let $P_i=P_{x_i}.$ What if we let $P$ be the product of $P_i$'s, $i < c^V$? $\endgroup$ Jul 10, 2014 at 8:58
  • $\begingroup$ Then you've added all the reals. But perhaps many more. And it's not clear to me why $\Bbb R^V$ is a set in that generic extension (or why this $P$ is even in $W$, actually). $\endgroup$
    – Asaf Karagila
    Jul 10, 2014 at 9:08
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    $\begingroup$ Under your hypotheses, we can assume without loss of generality that all the $P_x$'s are the same, namely, the collapse forcing $\text{Coll}(\omega,\delta)$ of some fixed $\delta$ to $\omega$. This is because any forcing notion embeds into this, for sufficiently large $\delta$, and we may simply take $\delta$ large enough to work for all $x$. $\endgroup$ Jul 10, 2014 at 12:07
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    $\begingroup$ @MonroeEskew I make the claim only for countable models. This follows as a consequence of theorem 34 in my paper, "Set-theoretic geology" (jdh.hamkins.org/set-theoreticgeology), which says that one can get above any countable sequence of successive forcing extensions of a countable model $W$, if the forcing has bounded size in $W$. There is also a more hands-on proof for this case, which simplifies some of the generalities of that proof. $\endgroup$ Jul 10, 2014 at 18:42
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    $\begingroup$ Another note: We can always take $V$ to be $W[x]$ for some $x \subset \omega_1.$ To see this, consider the given $V$. Force over it to force $GCH$. The resulting generic extension adds no new reals. Force over the extension by Jensen's coding to code every thing into a subset $x$ of $\omega_1.$ Again we do not add new reals and the final model is of the form $L[x]=W[x].$ So $V$ and $W[x]$ have the same reals and we can replace $V$ by $W[x].$ $\endgroup$ Jul 11, 2014 at 5:25

1 Answer 1

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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.

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

  2. $V_1$ 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 $\aleph_{\alpha\cdot\omega+2}$).

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

  4. 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.

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  • $\begingroup$ That's a nice argument. $\endgroup$ Oct 21, 2015 at 12:19
  • $\begingroup$ What was Woodin's example for $W$ in 1? $\endgroup$
    – Ashutosh
    Oct 25, 2015 at 21:13
  • $\begingroup$ @Ashutosh: $L$ with selective ultrafilters. $\endgroup$
    – Asaf Karagila
    Oct 25, 2015 at 21:30
  • $\begingroup$ What are the $P_i$'s? $\endgroup$
    – Ashutosh
    Oct 25, 2015 at 21:55
  • $\begingroup$ @Ashutosh: Shooting fast sets through the selective ultrafilters. $\endgroup$
    – Asaf Karagila
    Oct 26, 2015 at 5:46

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