Let $M_1$ be the canonical inner model with one Woodin cardinal $\delta$. Now suppose that $\mathbb{P}$ is a forcing notion of size $< \delta$, which preserves $\omega_1$ and that $G$ is a generic filter for the forcing. Note that, assuming $M_1^{\sharp}$ exists, there exists a formula which defines $M_1$ in the generic extension properly. Suppose that in $M_1[G]$ the $(M_1)$-initial segment $(M_1)_{\alpha}$ has size $\aleph_1$.

My question is whether it is possible, for unboundedly many $\alpha < \aleph_2^{M_1[G]}$ to find an elementary continuous chain $(N_i \, : \, i \in \omega_1)$ of elementary submodels of $(M_1)_{\alpha}$ such that $\bigcup_{i < \omega_1} N_i = (M_1)_{\alpha}$, and such that for every model $N_i$ in the chain, its transitive collapse is an initial segment of $M_1$ again?

Note that the statement is true if $\mathbb{P}$ is just the trivial forcing, however I would be interested in the case when we collapse cardinals down to $\omega_1$.

  • $\begingroup$ Why is $M_1$ definable in $M_1[G]$? $\endgroup$ Nov 3, 2015 at 17:03
  • $\begingroup$ @Yizheng: Ground model definability. Or do you mean definable without parameters? $\endgroup$
    – Asaf Karagila
    Nov 3, 2015 at 17:50
  • $\begingroup$ @AsafKaragila Yes, the latter. The point is that to identify $M_1$ as such you need its sharp, which of course is not in $M_1[G]$, so you need to define $M_1$ some other way. (Presumably as a version of $K$, though this would also need to be clarified.) $\endgroup$ Nov 3, 2015 at 20:06
  • $\begingroup$ @Yizheng@Andres: Thanks for pointing that out. If I assume the existence of $M_1^{\sharp}$, then it should be definable right? $\endgroup$ Nov 3, 2015 at 20:59
  • 3
    $\begingroup$ M_1 is definable in this extension. it is K up to the woodin and then L above it. $\endgroup$
    – Grigor
    Dec 28, 2015 at 9:30

1 Answer 1


i think the answer is yes.

let $M=M_1$. let $\lambda=\omega_2^M$ and say $G$ collapses $\omega_2$ to $\omega_1$. let $f\colon\omega_1\to\lambda$ be an increasing cofinal function. let construct a continuous chain $(M_i: i<\omega_1)$ of submodels of $M|\lambda$ such that $f(i)$ is in $M_i$.

let $N_i$ be the transitive collapse of $M_i$. then $N_i$ is an initial segment of $M$, this follows from comparison.

let then $M_{\omega_1}$ be the direct limit of $N_i$. there is a map $\pi\colon M_{\omega_1}\to M_i|\lambda$. this map has the range of $f$ in its range and its critical point is bigger than $\omega_1$, so it is identity.

you can do this with any $\gamma<\omega_3^M$. fix $f\colon\omega_1\to\lambda$ and $g\colon\omega_1\to\gamma$ that are increasing and cofinal. then construct a chain that exhaust $f$ and $g$. similar argument would show that $M_{\omega_1}$ is $M|\gamma$.


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