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Suppose $\mathbb{P}$ is a definable proper forcing (for instance Sacks forcing). Let $\alpha$ be some ordinal. Let $\mathbb{P}_\alpha$ be the countable support iteration of $\mathbb{P}$ of length $\alpha$.

It is well known that $\mathbb{P}_\alpha$ is also proper. Hence for any countable elementary structure $M \prec H_\Theta$ for sufficiently large $\Theta$ such that $\mathbb{P}_\alpha \in M$ and any $p \in \mathbb{P}_\alpha \cap M$, there exists some $q \leq_{\mathbb{P}_\alpha} p$ which is a $(M, \mathbb{P}_\alpha)$-generic condition.

Let $X = \alpha \cap M$. The question is whether a $(M, \mathbb{P}_\alpha)$ generic condition $q \leq_{\mathbb{P}_\alpha} p$ can be found such that $\text{supp}(q) \subseteq X$.

Thanks for any information that can be provided.

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  • $\begingroup$ Probably you want to assume not only that $\mathbb{P}$ is definable and proper, but also that it is necessarily proper, in the sense that the poset defined by that definition is proper in the intermediate models arising during the course of the iteration; and probably you also intend that the definition of $\mathbb{P}$ and $\mathbb{P}_\alpha$ are absolute from $V$ to $H_\theta$. $\endgroup$
    – Joel David Hamkins
    Aug 15, 2014 at 14:05
  • $\begingroup$ Otherwise (without the "necessarily proper" part), the iteration $\mathbb{P}_\alpha$ will not be proper, and there may be no generic conditions at all. $\endgroup$
    – Joel David Hamkins
    Aug 15, 2014 at 14:28
  • $\begingroup$ @JoelDavidHamkins Yes, I definitely need that each partial iteration forces the name for $\mathbb{P}$ to be proper. Thanks. $\endgroup$
    – William
    Aug 15, 2014 at 15:55

2 Answers 2

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I may be missing something here, because if I've got the question right then the definability doesn't come in to play:

The answer is yes, and the reason is implicit in the proof that $\mathbb{P}_\alpha$ is proper. Shelah's proof of Theorem 3.2 on page 109 of Proper and Improper forcing mentions this. In particular, (*) on the bottom of page 109 points out when building the generic conditions, "we could add $dom(r)\cap [i, j) = N\cap [i, j))$'' which decodes into what you are looking for in the case where $i = 0$ and $j = \alpha$.

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  • $\begingroup$ By definable, I meant something along the line that it would be meaningful to iterate the "version" of $\mathbb{P}$ from the previous model. For example, if $V[G]$ is the Sacks extension, I would like to force again with the perfects trees in $V[G]$. Nevertheless, I think your answer is exactly what I was looking for. Thanks. $\endgroup$
    – William
    Aug 15, 2014 at 15:52
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See also my recent paper Understanding preservation theorems, Chapter VI of Proper and Improper Forcing, I for what I've been told is the clearest exposition of iterated proper forcing.

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    $\begingroup$ Chaz, welcome to MathOverflow! $\endgroup$
    – Joel David Hamkins
    Aug 19, 2014 at 18:52

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