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 $M$?



Infinite products of forcings
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.


