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Let $M=M_0$ be a ctm of ZF.

  1. If $(M_0, \ldots, M_n)$ is a sequence of ctms of ZF, where for all $i,$ either $M_{i+1}$ is an inner model of $M_i$ or a generic extension of $M_i,$ then call $M_n$ an $n$-model of $M.$

  2. Call $M_n$ an $n'$-model of $M$ if we can further demand that $M_1$ is an inner model of $M_0.$

  3. Call $M_n$ an $n''$-model if we can instead demand that $M_1$ is a generic extension of $M_0.$

  4. Let $C_n^M, C_n^{M'}, C_n^{M''}$ respectively denote the sets of $n$-models, $n'$-models, and $n''$-models of $M.$

  5. Let $C_{\infty}^M=\bigcup_{n<\omega} C_n^M.$

My intention is that $C_{\infty}^M$ represents the set of $o(M)$-height models which $M$ is capable of "reasoning" about.

My main question is whether there is some $n$ such that $C_{\infty}^M=C_n^M.$

It's not hard to cook up $M$ such that $C_2^{M'} \not \subset C_2^{M''}$ or $C_2^{M''} \not \subset C_2^{M'},$ so this hierarchy doesn't seem to collapse immediately (if at all), at least in general. It's a priori possible that such $n$ exists for some $M$ and not others; I'm not sure how find an example either way.

If such an $n$ cannot exist, what if we instead ask for something weaker, say $n$ such that $\{\varphi \in \mathcal{L}_{\{\in\}}: \exists N \in C_{\infty}^M(N \models \varphi)\}=\{\varphi \in \mathcal{L}_{\{\in\}}: \exists N \in C_n^M(N \models \varphi)\}?$ It seems like some strengthening of the Inner Model Hypothesis could justify the existence of such an $n.$

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  • $\begingroup$ @MonroeEskew Please explain how that makes the hierarchy collapse. $\endgroup$ – Elliot Glazer May 28 '18 at 5:53
  • $\begingroup$ @MonroeEskew I do, but even if we restrict to set forcing, I don't understand your argument that this collapses. $\endgroup$ – Elliot Glazer May 28 '18 at 6:08
  • $\begingroup$ @MonroeEskew You seem to have misread the definition I presented. We aren't taking ground models, we're taking generic extensions. The sequence will alternate between smaller and large models, it's not monotonic. $\endgroup$ – Elliot Glazer May 28 '18 at 6:11
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    $\begingroup$ Also, since you want to claim that $M$ can reason about all the models in this collection, I assume you mean your inner models to be definable? $\endgroup$ – Miha Habič May 28 '18 at 9:09
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    $\begingroup$ @MihaHabič, could you explain how this goes regarding Usuba's theorem? Since he only considers choice models in his proof, this sounds like a nontrivial modification. Maybe worth posting as an answer. $\endgroup$ – Monroe Eskew May 28 '18 at 13:35

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