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Question: Is it consistent (relative to the existence of large cardinals) that there is an elementary embedding $j\colon V\to M$ (where $M$ is transitive model) that factors as $j = j_n \circ k_n$ for $n < \omega$, such that for every $n$, $\text{crit }j_n = \kappa$ (the same ordinal), but $j_{n} (\kappa) < j_{n + 1} (\kappa)$ for every $n$?

The motivation for this question is the following observation: It is possible to get from single measurable cardinal many different elementary embeddings $j_n\colon V \to M_n$ with the same critical point (by using iterated ultrapower) and a single model $M$ and elementary embedding $j \colon V \to M$ such that $j = k_n \circ j_n$ for $k_n \colon M_n \to M$. The question is whether the "dual" situation, which seems to be more problematic, is possible.

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Yes, this situation can occur. One should simply undertake the dual of the construction you had suggested with iterated ultrapowers.

Specifically, suppose that $\mu$ is a normal measure on a measurable cardinal $\kappa$ and let $j:V\to M_\omega$ be the embedding arising from iterating the ultrapower $\omega$ many times. Thus, $M_\omega$ is the direct limit of the system of embeddings $j_{n,k}:M_n\to M_k$, where $j_{n,n+1}$ is the ultrapower of $M_n$ by $\mu_n=j_{0,n}(\mu)$. The sequence $\langle\kappa_n\mid n<\omega\rangle$ is the critical sequence.

For any set $S\subset\{\kappa_n\mid n<\omega\}$, we may form the seed hull $$X_S=\{j(f)(\vec s)\mid f:\kappa^{<\omega}\to V, f\in V, \vec s\in S^{<\omega}\},$$ and this is an elementary substructure of $M$, which can be seen by verifying the Tarski-Vaught criterion, and it contains the range of $j$. Further, it is a basic fact of normal ultrapowers that no seed $\kappa_n$ can be generated from the others. That is, if $\kappa_n\notin S$, then $\kappa_n\notin X_S$. You can find this and related results in section 3 of my paper, Canonical seeds and Prikry trees, JSL 62, 1997 (adapted from a chapter of my dissertation). It uses the normality of $\mu$, and this particular fact is not necessarily true without that assumption, as shown in the paper.

Now, for each finite $n$, let $S_n=\{\kappa_m\mid m\geq n\}$ and let $\pi_n:X_{S_n}\cong N_n$ be the Mostowski collapse of $X_{S_n}$. Thus, rather than including all seeds up to $\kappa_n$, which is how you might have proved the situation you mentioned at the end of your question, we instead undertake the dual set, using only seeds from $\kappa_n$ and upward. Let $k_n=\pi_n\circ j:V\to N_n$, and let $j_n=\pi_n^{-1}:N_n\to M$, so that we have a commutative triangle of elementary embeddings $j=j_n\circ k_n$.

Since $S_n$ contains only $\kappa_m$ for $m\geq n$, it follows that $\kappa_i\notin X_{S_n}$ for $i<n$, and consequently $X_{S_n}\cap[\kappa,\kappa_n)=\emptyset$, leading to $\pi(\kappa_n)=\kappa$. Thus, $j_n(\kappa)=\kappa_n$. In particular, these all have the same critical point, and they reach higher as $n$ increases. So the situation is just what you requested.

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  • $\begingroup$ Thank you. This answers my question completely. It seems that in this situation $j_n$ are not definable in $N_n$ (since $\kappa$ is no longer measurable in $N_n$). Is it possible (maybe by starting with larger cardinal) that those embedding will be also definable in $N_n$? $\endgroup$ – Yair Hayut Nov 29 '14 at 15:58
  • $\begingroup$ Well, $\kappa$ could be measurable in $N_n$, if you use a normal measure $\mu$ that concentrates on measurables (that is, of nontrivial Mitchell rank), but your point instead should be that $\mu$ is not in $N_n$. You'll have a chance if you make an iteration not by a single measure $\mu$, but at step $n$ use a measure $\mu_n$ (actually its image in that model), such that they are increasing with respect to the Mitchell rank. This way, the later iterates will have the earlier measures, and so we'll get $j_n$ in $N_n$. But I haven't checked the details... $\endgroup$ – Joel David Hamkins Nov 29 '14 at 17:32
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This isn't really an answer, but here's a way you can get the maps as you want with the embeddings defined outside of the model. Assume $0^\sharp$ exists, and let $\langle \alpha_i : i < \omega^2 \rangle$ be the first $\omega^2$ indiscernibles for $L$. Let $k$ send $\alpha_i$ to $\alpha_{\omega+i}$, for $i < \omega^2$ and fix the rest. Let $j_n$ send $\alpha_i$ to $\alpha_{i+n}$ for $i < \omega$ and fix the rest. Then for all $i < \omega^2$, $j_n \circ k(\alpha_i) = k(\alpha_i)$. We can extend the $j_n$'s and $k$ to elementary embeddings from $L$ to $L$. Then putting $j = k = k_n$ for each $n$, we have a collection of maps satisfying the requirements.

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  • $\begingroup$ This is very nice. It shows that it should be possible in general and I think that similar constructions should give us also definable $j$ and $j_n$-s. $\endgroup$ – Yair Hayut Nov 28 '14 at 13:12

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