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Notation: For each measurable cardinal $\kappa$ and a non-trivial $\kappa$-additive two-valued measure $\mu$ on it let $M_{\kappa,\mu}$ be the corresponding inner model.

Question: Assuming suitable consistency assumptions, are these consistent?

(a) $ZFC+\text{There exist class many measurable cardinals}+$ $\text{For each ordinal}~\alpha~\text{if}~\kappa_{\alpha}~\text{denotes}~\alpha~\text{th measurable cardinal then}$ $\text{there exists a non-trivial}~\kappa_{\alpha}~\text{-additive}~\text{two-valued measure on}~\kappa_{\alpha}~\text{like}~\mu_{\alpha}~\text{such that}$ $\forall \alpha,\beta \in Ord~~~~~\alpha<\beta\longrightarrow M_{\kappa_{\alpha},\mu_{\alpha}}\subseteq M_{\kappa_{\beta},\mu_{\beta}}$

(b) $ZFC+\text{There exist class many measurable cardinals}+$ $\text{For each ordinal}~\alpha~\text{if}~\kappa_{\alpha}~\text{denotes}~\alpha~\text{th measurable cardinal then}$ $\text{there exists a non-trivial}~\kappa_{\alpha}~\text{-additive}~\text{two-valued measure on}~\kappa_{\alpha}~\text{like}~\mu_{\alpha}~\text{such that}$ $\forall \alpha,\beta \in Ord~~~~~\alpha<\beta\longrightarrow M_{\kappa_{\alpha},\mu_{\alpha}}\supseteq M_{\kappa_{\beta},\mu_{\beta}}$

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  • $\begingroup$ Could someone explain why is the user posting this question not clickable? $\endgroup$ – Joel David Hamkins Jan 11 '14 at 2:36
  • $\begingroup$ @JoelDavidHamkins The user was deleted. You'd probably need to contact the moderators privately to see why, but most likely they cannot disclose details. $\endgroup$ – Andrés E. Caicedo Jan 11 '14 at 5:48
  • $\begingroup$ Ah, I see. Well, anyway, I think it is a very good question. $\endgroup$ – Joel David Hamkins Jan 11 '14 at 11:56
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I assume that by $M_{\kappa,\mu}$ you intend the ultrapower of $V$ by $\mu$.

Both of the statements are inconsistent, even in the case of only two measurable cardinals, let alone a proper class of them.

Statement (b) is inconsistent. If $\mu_\alpha$ is a measure on $\kappa_\alpha$ and $\kappa_\alpha\lt\kappa_\beta$, then $\mu_\alpha\in M_{\kappa_\beta,\mu_\beta}$, since the larger ultrapower includes all sets of rank below $\kappa_\beta$, and so under (b) we would have $\mu_\alpha\in M_{\kappa_\alpha,\mu_\alpha}$, but this is impossible, since the measure that you take an ultrapower by is never in the ultrapower itself.

Statement (a) is inconsistent. Suppose that $\kappa<\delta$ are measurable cardinals, witnessed by measures $\mu$ and $\nu$, respectively. Let $j_\mu:V\to M_\mu$ and $j_\nu:V\to M_\nu$ be the corresponding ultrapower maps. Suppose toward contradiction that $M_\mu\subset M_\nu$. It is not difficult to see that $j_\mu(\delta)=\delta$, since $j_\mu$ is continuous except at cardinals of cofinality $\kappa$. Consider the object $j_\mu(\nu)$, which is a measure on $\delta$ inside $M_\mu$. Since $\mu\in M_\nu$, as it has all objects in $V_\delta$, it follows that $M_\nu$ can construct the map $j_\mu\upharpoonright M_\nu$. Note that for $X\subset\delta$, we have $X\in\nu\iff j_\mu(X)\in j_\mu(\nu)$. Thus, if $j_\mu(\nu)$ were in $M_\nu$, then inside $M_\nu$ we could construct $\nu$. But as mentioned before, $\nu\notin M_\nu$, as the measure is never an element in the ultrapower. QED

Both arguments relied on the following:

Lemma. If $j:V\to M_\mu$ is the ultrapower embedding by a nonprincipal $\kappa$-complete measure $\mu$ on $\kappa$, then $\mu\notin M_\mu$.

Proof. If $\mu\in M_\mu$, then $M_\mu$ could reconstruct $j_\mu\upharpoonright P(\kappa)$, and thereby learn that $j_\mu(\kappa)$ is an ordinal of cardinality $2^\kappa$, which contradicts that it is supposed to be a measurable cardinal in $M_\mu$. QED


Perhaps a more focused summary of the argument is the following:

Theorem. If $\kappa<\delta$ are measurable cardinals, witnessed by measures $\mu$ and $\nu$, respectively, on $\kappa$ and $\delta$, with (transitive) ultrapowers $M_\mu$ and $M_\nu$, then $M_\mu\not\subset M_\nu$ and $M_\nu\not\subset M_\mu$.

Proof. Since $\mu$ is small relative to $\delta$, we have $\mu\in M_\nu$. But $\mu\notin M_\mu$ by the lemma, and so $M_\nu\not\subset M_\mu$. Conversely, since $M_\nu$ can construct $j_\mu\upharpoonright M_\nu$, as it has $\mu$ and also all functions from $\kappa$ to $M_\nu$, and since $X\in\nu\iff j_\mu(X)\in j_\mu(\nu)$ for $X\subset\delta$, it follows that $j_\mu(\nu)\notin M_\nu$, for otherwise we could construct $\nu$ inside $M_\nu$ contrary to the lemma, and so we have $M_\mu\not\subset M_\nu$. QED

So there can be not even a single instance of the phenomenon requested in the question. The phenomenon generalizes beyond measures to various extenders as well.

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