MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
show/hide this revision's text 2 improved formatting

Take the steps $Q_\alpha$ to be Cohen forcing, and begin by taking a generic filter $G$ for the usual finite-support product. So $G$ codes an $\omega$-sequence of Cohen reals $x_n\in 2^\omega$. Fix some $z\in 2^\omega$ coding an ordinal larger than the height of $M$. Define $y_n\in 2^\omega$ to be exactly the same as $x_n$ except that $y_n(0)=z(n)$. Since only one entry in $x_n$ has been changed, it is clear that $y_n$ is Cohen-generic over $M$ and that, for each natural number $n$, $\langle y_k:k<n\rangle$ is generic for $P_0*\dots*P_{n-1}$; let $G_n$ be the corresponding generic filter in $P_0*\dots*P_{n-1}$. The sequence `$\langle G_n:n\in\omega\rangle$ is not in any extension of $M$ with the same ordinals, because it encodes $z$.
show/hide this revision's text 1

Take the steps $Q_\alpha$ to be Cohen forcing, and begin by taking a generic filter $G$ for the usual finite-support product. So $G$ codes an $\omega$-sequence of Cohen reals $x_n\in 2^\omega$. Fix some$z\in 2^\omega$coding an ordinal larger than the height of $M$. Define$y_n\in 2^\omega$ to be exactly the same as $x_n$ except that $y_n(0)=z(n)$. Since only one entry in $x_n$ has been changed, it is clear that $y_n$ is Cohen-generic over $M$ and that, for each natural number $n$, $\langle y_k:k<n\rangle$ is generic for $P_0*\dots*P_{n-1}$; let $G_n$ be the corresponding generic filter in $P_0*\dots*P_{n-1}$. The sequence `$\langle G_n:n\in\omega\rangle$ is not in any extension of $M$ with the same ordinals, because it encodes $z$.