I'm trying to show that if a premouse $\mathcal M$ is 1-small then it's also tame.

Definition. $\mathcal M$ is 1-small if for every extender $E$ on the $\mathcal M$-sequence, $\mathcal J^{\mathcal M}_{\text{crit }E}\models\text{ there is no Woodin cardinal}$.

Definition. $\mathcal M$ is tame if for every extender $E$ on the $\mathcal M$-sequence, $\mathcal J^{\mathcal M}_{\text{lh }E}\models\forall\delta\geq\text{crit } E:\delta\text{ is not Woodin}$.

The argument as I see it is that, assuming $\mathcal M$ isn't tame with $E$ and $\delta$ witnessing it, we form the $E$-ultrapower, note that $\delta$ is still a Woodin cardinal in the ultrapower, so that the ultrapower thinks that there is a Woodin cardinal between $0$ and $i_E(\text{crit }E)$, so $\mathcal M$ thinks that there is a Woodin between $0$ and $\text{crit }E$, so $\mathcal J^{\mathcal M}_{\text{crit }E}\models\text{there is a Woodin}$, thereby showing that $\mathcal M$ isn't 1-small.

So, my main question is why $\delta$ is still Woodin in the ultrapower. Couldn't we wind up with subsets of $\delta$ that can't be reflected by any $\kappa<\delta$ in the ultrapower?

This argument is from Proposition 2.31 in Löwe-Steel's "An introduction to core model theory", but it seems like they assume that the model thinking that there is a Woodin between some $\alpha$ and $\beta$ is equivalent to the $\beta$-initial segment thinking that there is a Woodin above $\alpha$. Is this true?

Thank you in advance!


1 Answer 1


The Woodiness of $\delta$ in $\mathcal{J}^{\mathcal{M}}_{lh(E)}$ is witnessed by a bunch of extenders, which are either on the $\vec{E}$ sequence of $\mathcal{M}$ or are definable from elements of $\vec{E}$. As $\mathcal{M}$ is a premouse we know that its extender sequence $\vec{E}$ satisfies the coherence condition, i.e.

if $i: \mathcal{J}^{\mathcal{M}}_{lh(E)} \rightarrow Ult(\mathcal{J}^{\mathcal{M}}_{lh(E)}, E)$ denotes the ultrapower embedding for the extender $E$, and let $\kappa$ be its critical point, then $i(\vec{E} \upharpoonright \kappa) \upharpoonright \alpha = \vec{E} \upharpoonright \alpha$, which has as a consequence that also the ultrapower $Ult(\mathcal{J}^{\mathcal{M}}_{lh(E)}, E)$ has all the necessary extenders to witness the Woodiness of $\delta$ in it.

  • $\begingroup$ So, the reason why the extenders you mention are sufficient to capture all subsets of $V_\delta$ in the ultrapower is because $\text{lh }E$ is a cardinal of the ultrapower, so that any subset of $V_\delta$ lies in $\mathcal J_{\text{lh }E}^{\text{Ult}}$, which by coherence is just $\mathcal J_{\text{lh }E}^{\mathcal M}$, so we've already taken care of that subset. Is this correct? $\endgroup$ Jul 5, 2016 at 10:02
  • $\begingroup$ Yes that's right $\endgroup$ Jul 5, 2016 at 12:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.