7
$\begingroup$

The usual argument for the minimality result for Sacks forcing uses choice.

Theorem (Sacks): Let $s \subseteq \mathbb S_\kappa$ be generic for the forcing to add a Sacks subset to $\kappa$, where $\kappa$ is either $\omega$ or inaccessible. Then for any set of ordinals $A \in V[s]$, either $A \in V$ or $s \in V[A]$.

The only proof of which I know for this result starts by considering $p \in \mathbb S_\kappa$ forcing that $\dot A \not \in V$. We find $q \le p$ and a set of ordinals $\{\xi_t : t \text{ a splitting node of } q\}$ so that $q \upharpoonright {t {}^\frown 0}$ and $q \upharpoonright {t {}^\frown 1}$ decide $\xi_t \in \dot A$ differently. From these and $A$ we can reconstruct $s$. We build $q$ as an intersection of a fusion sequence $\{p_\alpha\}$. Given $p_\alpha$, to get $p_{\alpha+1}$ we consider all the $\alpha$th splitting nodes $t$ of $p_\alpha$. For each such $t$, we pick an ordinal $\xi$ so that $p_\alpha \upharpoonright t$ does not decide $\xi \in \dot A$. We then pick $r_0 \le p_\alpha \upharpoonright t {}^\frown 0$ and $r_1 \le p_\alpha \upharpoonright t {}^\frown 1$ which decide $\xi \in \dot A$ differently. We amalgamate all these $r_i$ for all splitting nodes of $p_\alpha$ to produce $p_{\alpha+1} \le_{\alpha+1} p_\alpha$.

Picking an ordinal $\xi$ is of course unproblematic. This is not the case for picking the conditions $r_0$ and $r_1$. Here we need to appeal to choice as there are lots and lots of appropriate conditions below $p_\alpha \upharpoonright t {}^\frown i$.

My question is whether choice is necessary for the minimality result.

Question: Are there $M \models \mathsf{ZF} + \neg \mathsf{AC}$ and $s \subseteq \mathbb S_\kappa^M$ generic for $M$ so that $M[s]$ is not minimal over $M$? That is, are there $M$, $s$ as above and $A \in M[s]$ a set of ordinals so that $M \subsetneq M[A] \subsetneq M[s]$?

$\endgroup$
2
  • 1
    $\begingroup$ My uneducated guess, is that the answer is most likely probably maybe yes. $\endgroup$
    – Asaf Karagila
    Nov 4, 2015 at 22:17
  • 1
    $\begingroup$ Although, coming to think about it, maybe for sets of ordinals you can in fact prove it. Because sets of ordinals mean nothing in arbitrary models of ZF... $\endgroup$
    – Asaf Karagila
    Nov 5, 2015 at 0:40

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.