# Preservation of Woodinness when it overlaps the active extender

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

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.

• 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? Jul 5, 2016 at 10:02
• Yes that's right Jul 5, 2016 at 12:40