# $M$-amenable ultrafilters on $\kappa$ are $\kappa$-powerset preserving

Let $M$ be a transitive model of $\operatorname{ZFC-}$ and let $$j \colon M \rightarrow N$$ be elementary with $\operatorname{crit}(j) = \kappa \in \operatorname{wfp}(N)$. Let $U_j$ be defined by $$X \in U_j :\Leftrightarrow X \in \mathcal P(\kappa)^M \text{ and } \kappa \in j(X).$$

We say that $j$ is $\kappa$-pp ($\kappa$-powerset preserving) iff $\mathcal P(\kappa)^M = \mathcal P(\kappa)^N$. And we say that $U_j$ is $M$-amenable iff for all $f \in ^\kappa \mathcal P(\kappa) \cap M \colon \{ \alpha < \kappa \mid f(\alpha) \in U_j \} \in M$.

I'd like to see that $U_j$ to be $M$-amenable implies $j$ to be $\kappa$-pp.

My thoughts:

If $N = \operatorname{ult}(M; U_j)$, this is easy: Let $X \in \mathcal P(\kappa)^N$ and fix $h \in ^\kappa \kappa \cap M$ with $X = [h]$. Then for all $\alpha < \kappa \colon$ \begin{align} \alpha \in X & \Leftrightarrow [c_\alpha] \widetilde \in [h] \\ & \Leftrightarrow \{ \xi < \kappa \mid \alpha \in h(\xi) \} \in U_j \end{align} Thus defining $f \colon \kappa \rightarrow \mathcal P(\kappa), \alpha \mapsto \{\xi < \kappa \mid \alpha \in h(\xi)\}$ yields $X = \{ \alpha < \kappa \mid f(\alpha) \in U_j \}$. As $f \in M$ (provided $\kappa \times \mathcal P(\kappa)^M \in M$), the $M$-amenability of $U_j$ yields $X \in M$.

So far, I don't see how to adopt the above to the situation where $N \neq \operatorname{ult}(M; U_j)$.

I think that your claim is false. I will try to construct a counterexample. Suppose that $\kappa$ is measurable and $U$ is a normal measure on $\kappa$. Let $j:V\to M$ be the ultrapower by $U$. First, we build $M\prec H_{\kappa^+}$ such that $U'=U\cap M$ is $M$-amenable. Let $M_0$ be any transitive elementary substructure of $H_{\kappa^+}$ of size $\kappa$. Let $M_1$ be a transitive elementary substructure of $H_{\kappa^+}$ which contains $M_0$ and $U\cap M_0$ as elements. Continuing in this way construct a sequence $\langle M_n\mid n<\omega\rangle$ and let $M=\bigcup_{n<\omega}M_n$. It is easy to see that $U'=U\cap M$ is $M$-amenable. Consider the restriction $j:M\to j(M)$. Note that by elementarity $j(M)\prec H_{j(\kappa)^+}^{M}$. Find some $j(M)\prec N\prec H_{j(\kappa)^+}^{M}$ such that $N$ has a subset of $\kappa$ not in $M$. Note that $j:M\to N$ is elementary and clearly the ultrafilter for $M$ generated from $j$ is precisely $U'$, which is $M$-amenable by construction.
• Thanks, Asaf! It's all Neil Barton's fault with his $KM$ questions :). Nov 25, 2015 at 19:33