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Suppose $M \subseteq N$ are models of ZFC such that $(ORD^\omega)^M = (ORD^\omega)^N$. Let $\langle P_n : n \in \omega \rangle$ be a sequence of countably closed partial orders in $M$, and let $\langle G_n : n \in \omega \rangle$ be a sequence of filters in $N$ such that for each $n$, $G_n$ is $P_n$-generic over $M[G_0,...,G_{n-1}]$. Is $\Pi_{n \in \omega} G_n$ generic for $\Pi_{n \in \omega} P_n$ over $M$?

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Related interesting question at mathoverflow.net/questions/38666/…. –  Joel David Hamkins Nov 24 '12 at 21:56

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up vote 5 down vote accepted

This is a very nice problem, but unfortunately the answer can be negative.

Let me describe a counterexample. Consider the forcing to add $\omega$ many Cohen subsets of $\omega_1$. So $P_n=\text{Add}(\omega_1,1)$ adds one Cohen subset to $\omega_1$ and the (full support) product $\Pi_n P_n$ is $\text{Add}(\omega_1,\omega)$. Suppose that $G\subset\Pi_n P_n$ is $M$-generic for the product forcing, and consider the two models $M\subset N=M[G]$. Since the forcing is countably closed, it adds no new $\omega$-sequences over $M$.

We may think of $G$ as filling in a $\omega\times\omega_1$ matrix with $0$s and $1$s. Generically, there will be many all-zero rows, that is, rows having zeros all the way across, so that $G(n,\alpha)=0$ for all $n$, where this is the $\alpha^{th}$ row.

Let us define $G^\ast$ to be just like $G$ in every column, except that in any such all-zero row in $G$, we change the first bit to a $1$ in the first column in $G^\ast$, leaving the rest of the row all $0$s. This operation ensures that $G^\ast$ has no all-zero rows, and thus ensures that $G^\ast$ is definitely not $M$-generic for the product forcing. But meanwhile, I claim that this operation does not affect the $M$-genericity of any finite number of the columns of $G^\ast$. For this, it is an elementary exercise to see that for any dense set $D$ for the forcing in the first $n$ columns, there is a dense set $E$ in the full product, such that the operation applied to conditions in $E$ gives a condition in $D$. So generically, $G$ is such that $G^\ast$ will have its first $n$ factors in $D$.

Thus, every finitely many factors of $G^\ast$ are $M$-generic, but the whole product $G^\ast$ is not $M$-generic; so it is a counterexample.

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Another way to think about it. Let $G$ be generic for the full product. Now replace the first column $G_0$ with the sequence of $0$s and $1$s that indicate whether the whole rest of the row in $G$ is $0$ or $1$. This new $G_0$ is $M$-generic for $\text{Add}(\omega_1,1)$, and mutually generic with any finite number of the columns $G_i$, but not with the whole matrix $G$. –  Joel David Hamkins Nov 25 '12 at 12:18
    
I meant that $G_0$ would have a $0$ digit iff the entire rest of that row is $0$ in $G$. This is $M$-generic for Cohen forcing, and is mutually generic with any finitely many $G_n$, but not with the full product. –  Joel David Hamkins Nov 25 '12 at 15:57
    
Nice answer to a nice problem. But I am not sure I agree with "unfortunately". –  Goldstern Nov 25 '12 at 18:15

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