Product Measure $\mathcal{V}$ of normal measure $\mathcal{U}$ s.t $\mathcal{U}\nleq_{RK} \mathcal{V}$

Question

Let $\kappa$ be measurable and let $\mathcal{U}$ be some measure on it we define a measure on $\kappa\times\kappa$

$$ A\in \mathcal{V} \iff \{\alpha \ \big |\ \{\beta\ \big|\ (\alpha,\beta)\in A\}\in \mathcal{U}\} \in \mathcal{U}$$

show that $\mathcal{V}$ is a $\kappa$-complete ultrafilter and prove that $ \mathcal{U} \leq_{RK} \mathcal{V}$ and $\mathcal{V} \nleq_{RK} \mathcal{U}$. (where $\leq_{RK}$ is defined in a previous question)

I'm stuck in the last part of proving $\mathcal{V} \nleq_{RK} \mathcal{U}$, we were given a hint first to show that

$$ \kappa =[\pi_1]_\mathcal{V} <[\pi_2]_\mathcal{V}<j_\mathcal{V}(\kappa)$$

which I managed to do. Then we were supposed to show that $ j_\mathcal{U}(\kappa) \leq [\pi_2]_\mathcal{V}$, but I got suck at how to desribe $j_\mathcal{U}(\kappa)$ inside the ultrapower by $\mathcal{V}$?

Moreover given that $ j_\mathcal{U}(\kappa) \leq [\pi_2]_\mathcal{V}$ how do I get to a contradiction? In the previous question we proved that being in order is equivalent to having an elementary embedding $k\circ j_\mathcal{U} = j_\mathcal{V}$, I understand that this somehow mixes the order but I can't prove it.

Any help would be appriciated.

Answer 1

On the one hand, assume that $[f]_\mathcal{U}<j_\mathcal{U}(\kappa)$. You can see that $[f\circ \pi_1]_\mathcal{V}< [\pi_2]_\mathcal{V}$ (since $\{(\alpha,\beta)\in \kappa^2 \mid f(\alpha)<\beta\}\in\mathcal{V}$), so $[f]_\mathcal{U}\le k([f]_\mathcal{U})<[\pi_2]_\mathcal{V}$. Thus $j_\mathcal{U}(\kappa)\le [\pi_2]_\mathcal{V}.$ (Note that the idea of the proof is similar to that of $\kappa\le [\mathrm{Id}]_\mathcal{U}$.)

On the other hand, it holds in general that if $\mathcal{U}\le_{RK}\mathcal{V}$ and $\mathcal{V}\le_{RK}\mathcal{U}$ then they generate the same ultrapower: if $\mathcal{U}\le_{RK}\mathcal{V}$ and $\mathcal{V}\le_{RK} \mathcal{U}$, then we have maps $k:\operatorname{Ult}(V,\mathcal{U})\to \operatorname{Ult}(V,\mathcal{V})$ and $k':\operatorname{Ult}(V,\mathcal{V})\to \operatorname{Ult}(V,\mathcal{U})$, which are given by witnesses $s$ and $t$ of $\mathcal{U}\le_{RK} \mathcal{V}$ and $\mathcal{V}\le_{RK} \mathcal{U}$ respectively. (That is, for example, $k([f]_\mathcal{U})=[f\circ s]_\mathcal{V}$.) They also satisfy

• $k\circ j_\mathcal{U}=j_\mathcal{V}$, and
• $k'\circ j_\mathcal{V}=j_\mathcal{U}$.

We can see that $k\circ k'\circ j_\mathcal{V}= j_\mathcal{V}$ and $k'\circ k\circ j_\mathcal{U}= j_\mathcal{U}$.

Now observe that every element of $\operatorname{Ult}(V,\mathcal{U})$ is of the form $j_\mathcal{U}(f)([\mathrm{Id}]_\mathcal{U})$. Moreover, $$k'\circ k(j_\mathcal{U}(f)([\mathrm{Id}]_\mathcal{U})) = j_\mathcal{U}(f)([s\circ t]_\mathcal{U}).$$

You can see from the proof by Hamkins that $[\mathrm{Id}]_\mathcal{U}=[s\circ t]_\mathcal{U}$, so $k'\circ k$ is the identity. Similarly, $k\circ k'$ is also the identity, hence $\operatorname{Ult}(V,\mathcal{U})= \operatorname{Ult}(V,\mathcal{V})$ and $j_\mathcal{U}=j_\mathcal{V}$. (You may also refer to another answer by Hamkins, which summarizes some properties of RK-orders.)