I'm trying to get a better grasp of iterated forcing, and I ran across the following problem:


0) Let $P\_\alpha$ be posets in a c.t.m. $M$, $\alpha<\beta$, and for each $\alpha$ let $G\_\alpha$ be $P\_\alpha$ generic. Let $P$ be the finite support iteration of the $P\_\alpha$. Then is there necessarily a $G$ which is $P$-generic such that for all $\alpha$, $G\_\alpha\in M[G]$?


My guess for how to do this would be transfinite induction, but I run into some problems when I try that. Specifically, the successor step makes sense, but the limit step and the base case $(\beta=2)$ don't.


[For what it's worth, here's the successor step. Suppose the answer to (0) is "yes" for all $\beta<\delta$, $\delta\ge 2$. Let $P\_\alpha$ be posets for $\alpha<\delta+1$, let $G\_\alpha$ be $P\_\alpha$-generic, let $P$ be their finite support iteration, and let $P'$ be the finite support iteration of the $P\_\alpha$ for $\alpha<\delta$. Then $P$ is equivalent (in fact, isomorphic) to $P'\times P\_\delta$. By the induction hypothesis, we have some $G'$ which is $P'$-generic such that for all $\alpha<\delta$, $G\_\alpha\in M[G']$. Applying the induction hypothesis again (since $\delta>1$) we have some $G$ which is $P'\times P\_\delta$-generic such that $G', G\_\delta\in M[G]$. But then $M[G']\subseteq M[G]$, so $G\_\alpha\in M[G]$ for all $\alpha<\delta$.] 


Related questions:


1) What happens if we restrict to c.c.c. posets?


2) What happens if we specify $\beta=2$?


3) What happens if we try to generalize to other kinds of supports?


I have a feeling I'm missing something obvious, but I've thought about this for a while now with no success.



A question about iterated forcing
I'm not sure I really understand the question, because of the mixture of iteration and product, but I believe the following negative answer is independent of such issues because it uses $\beta=2$ and takes both factors (or iterands) to be Cohen forcing (so the product forcing is equivalent to the iteration). Recall that a real (regarded, as usual in such contexts, as an element of $2^\omega$) is Cohen-generic over $M$ if and only if it belongs to every dense $G\_\delta$ set coded in $M$. Since $M$ is countable, the set of Cohen-generic reals over $M$ is comeager. Now choose some real $z$ that codes a well-ordering of $\omega$ longer than the height of $M$. Since both $\{x\in 2^\omega:x\text{ Cohen over }M\}$ and its translate $\{x\in 2^\omega:x\oplus z\text{ Cohen over }M\}$ (where $\oplus$ is pointwise addition mod 2) are comeager, they have an element $x$ in common. Take $G\_0$ to be the generic filter coding $x$ and take $G\_1$ to be the generic filter coding $x\oplus z$. No generic extension of $M$ (by $P\_0\times P\_1$ or by any other forcing) can contain both of these $G\_\alpha$'s. [Proof: Such a model would contain both $x$ and $x\oplus z$, and would therefore contain $z$. Being a model of ZF, it would have to contain the ordinal isomorphic to the ordering of $\omega$ coded by $z$. But the ordinals of a forcing extension of $M$ are the same as those of $M$, so this contradicts the choice of $z$.] 


