Isomorphism of $V_\lambda$ and the ultraproduct of $V_{\lambda_{\mathrm{otp}(x)}}, x \in \mathcal{P}_\kappa({\lambda})$ by a normal fine measure.

large-cardinalsset-theory

This question is actually exercise (20.5) from Set Theory by Thomas Jech. The original statement is:

Let $\lambda \ge \kappa$ and let $U$ be a normal measure on $\mathcal{P}_\kappa({\lambda})$. The ultraproduct $\mathrm{Ult}_U\{(V_{\lambda_x}, \in): x \in \mathcal{P}_\kappa({\lambda})\}$ is isomorphic to $(V_\lambda, \in)$.

Here, $\lambda_x$ is the order type of $x$.

I was able to shown (if I'm not mistaken) that if $M$ is the transitive collapse of $\mathrm{Ult}_U(V)$, and $N$ is the transitive collapse of $\mathrm{Ult}_U\{V_{\lambda_x}: \ x \in \mathcal{P}_\kappa({\lambda})\}$, then $N = V_\lambda\cap M$.

I have done this using a rank argument: Let $a \in M$ such that $\mathrm{rk}(a)<\lambda$, and let $f$ be the function on $\mathcal{P}_\kappa(\lambda)$ that represents $a$ in $\mathrm{Ult}_U(V)$ (i.e. $\pi ([f]_U) = a$ where $\pi$ is the collapsing function). Then as $x \mapsto \lambda_x$ represents $\lambda$, we have $\mathrm{rk}(f(x)) < \lambda_x$ for almost all $x$. So we may assume that this holds for all $x$, and then $[f]_U \in \mathrm{Ult}_U\{V_{\lambda_x}: \ x \in \mathcal{P}_\kappa({\lambda})\}$.

But why would $V_\lambda^M = V_\lambda$ hold in general? Moreover, what about the fact that $\left|N\right| \le \kappa^{\lambda^{<\kappa}}$, couldn't $V_\lambda$ be much larger?

Best Answer

As pointed out in a comment by spaceisdarkgreen, an almost identical question has been posted on MO in the past: https://mathoverflow.net/q/44289

It has recived an answer by Joel David Hamkins: https://mathoverflow.net/a/44291

For completeness, here is the answer given by Hamkins:

In general, $\lambda$-supercompactness, if consistent, does not imply $\lambda$-strongness. One can see this by observing that the smallest cardinal $\kappa$ that is $\kappa^+$-supercompact is never $\kappa^+$-strong, and in fact, cannot be even $(\kappa+3)$-strong. The reason is that $\kappa^+$-supercompactness is witnessed by a measure on $P_\kappa\kappa^+$, which amounts essentially to (is coded by) a subset of $P(\kappa^+)$, and hence is witnessed inside $V_{\kappa+3}$. Thus, if $j:V\to M$ were any embedding with critical point $\kappa$, by minimality it follows that $\kappa$ is not $\kappa^+$-supercompact in $M$, and hence $M$ cannot have the true $V_{\kappa+3}$. Thus, $\kappa$ is not $(\kappa+3)$-strong and thus definitely not $\kappa^+$-strong.

I haven't looked at the context of the exercise, but perhaps he is merely asking you to make the observation that you did in fact make, that the ultrapower will give you $V_\lambda^M$?

For some kinds of $\lambda$, it does follow that $\lambda$-supercompactness implies $\lambda$-strongness. For example, if $\lambda$ is a beth-fixed point, then every $\lambda$-supercompactness embedding is also $\lambda$-strong and even $(\lambda+1)$-strong, since in this case $|V_\lambda|=\lambda$. But if $\lambda$ is not a beth-fixed point, then a version of my argument above will still apply: if $\lambda$ is not a beth-fixed point and $j:V\to M$ is a Mitchell minimal $\lambda$-supercompactness embedding for $\kappa$, then $\kappa$ is not $\lambda$-supercompact in $M$, and this is witnessed inside $V_\lambda$ by the assumption on $\lambda$, and so $j$ cannot be a $\lambda$-strongness embedding.

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