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Let $\kappa$ be an inaccessible cardinal and let $M \subseteq V_{\kappa}$ be an inner model of $V_{\kappa}$, i.e., a transitive model of $\mathsf{ZF}$ containing all the ordinals up to $\kappa$.

My question is whether such a model is always a rank-truncation of an inner model of $\mathbf V$ (defined using $M$ as parameter). Equivalently this question can be phrased as follows:

If $M \subseteq V_{\kappa}$ is an inner model of $V_\kappa$ is then $\mathbf L(M) \cap V_\kappa = M$, where $\mathbf L(M)$ is the minimal inner model of $\mathbf V$ containing $M$ ?

If the answer to this question is negative, I am curious whether there can be a counterexample $M$ which is definable in $V_\kappa$ without parameters and/or which is a model of $\mathsf{ZFC}$. If the answer to this question is positive, I am curious whether it is still positive if we weaken the assumption that $\kappa$ is inaccessible to $\kappa$ being worldly.

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  • $\begingroup$ Great question! $\endgroup$ – Joel David Hamkins Jun 21 '17 at 15:11
  • $\begingroup$ It might also be sensible to ask about inner models of ZFC and not just ZF, which could conceivably give a different answer. $\endgroup$ – Joel David Hamkins Jun 21 '17 at 15:59
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    $\begingroup$ @AsafKaragila I had meant to ask about both cases (although I am interested primarily in the ZFC case). The strongest kind of counterexample would be where $M$ is an inner model of ZFC for $V_\kappa$, but admits no extension to an inner model of ZF for $V$. $\endgroup$ – Joel David Hamkins Jun 21 '17 at 16:17
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    $\begingroup$ @Joel: Yes, although the most interesting case would be M which satisfies ZFC, but has only ZF extensions (and no ZFC ones, that is). $\endgroup$ – Asaf Karagila Jun 21 '17 at 16:18
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    $\begingroup$ For a negative solution it would be enough to find some inner model $M \subseteq \mathbb{V}_\kappa$ with $Th(M, \in) \not \in M$. Is this possible? $\endgroup$ – Douglas Ulrich Jun 22 '17 at 2:14
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Theorem: Let $\kappa$ be strongly inaccessible in $V$, such that $V \models ZFC$. If $M\models ZF$, then $L(M) \cap V_\kappa = M$.

Proof: Let us prove by induction on $\alpha < \kappa$ that $L(M) \cap V_\alpha = M \cap V_\alpha$.

Let $x \in L(M) \cap V_{\alpha + 1}$, so $x \subseteq M \cap V_\alpha$. Let $\gamma$ be an ordinal such that $x\in L_{\gamma}(M)$.

Let us find $X \prec L_\gamma(M)$, such that:

  1. $x\in X$, $V_\alpha \subseteq X$,
  2. $X \cap \kappa \in \kappa$
  3. $X \cap M$ is transitive and equal to $M \cap V_\beta$ for some $\beta$

This is possible by the strong inaccessiblity of $\kappa$:

We define, by induction of $n < \omega$, $X_n$. Let $X_0 = M\cap V_{\alpha} \cup \{x\}$. Extend $X_0$ to $X_0'$ such that $X_0'\prec L_\gamma(M)$ and $|X_0'| = |X_0| < \kappa$. Now, take $X_1 \supseteq X_0'$ such that assumptions 2 and 3 hold ($M \cap X_1$ is transitive and $X \cap \kappa$ is ordinal). $|X_1| < \kappa$, by the inaccessiblity of $\kappa$. Let $X_1 \subseteq X_1' \prec L_\gamma(M)$, and so on. Let $X$ be $\bigcup X_n$.

Let $\bar{X}$ be the transitive collapse of $X$. $\bar{X} = L_{\bar{\gamma}}(\bar{M})$ where $\gamma < \kappa$ and $x\in \bar{X}$. Since $M \cap X$ is transitive, $\bar{M} = M \cap X$. Since $M\cap X = M \cap V_\zeta$ for some $\zeta$, $\bar{M} = M \cap V_\zeta \in M$ (as $\zeta \leq \bar{\gamma} \in \kappa$). In particular, $M$ can compute the model $L_{\bar{\gamma}}(\bar{M})$ and compute $x$. We conclude that $x \in M$.

Remark: Without the inaccessibility assumption, one can get a model $M$ of ZFC such that $L(M) \cap V_{\omega + 1} \neq M \cap V_{\omega + 1}$.

Let us start with $V = L$ and let $\kappa$ be a worldly cardinal of countable cofinality. Let $\langle \alpha_n \mid n < \omega\rangle$ be the first cofinal sequence of singular cardinals with limit $\kappa$, in the canonical well order of $L$. Let us consider the following class forcing in $L_\kappa$: for every singular cardinal $\mu\in L_{\kappa}$, let $\lambda = \mu^{++}$, and force with the lottery sum of $Add(\lambda, \lambda^{++})$ and the trivial forcing. Take the $<\kappa$ support product of those forcing notions. This forcing does not collapse cardinals. Let $c$ be a Cohen real. Let $G$ be an $L_\kappa$-generic filter such that $L_\kappa[G] \models 2^{\alpha_n^{++}} = \alpha_n^{+4}$ iff $c(n) = 1$. This is possible by splitting the forcing into class forcing that only decides for which cardinal $\lambda$ we force $2^\lambda = \lambda^{++}$ (this class forcing does not add sets) and the second step in which we change the value of the corresponding $2^\lambda$-s. In $L[c]$ one can find a generic for the first step that codes $c$ as above.

Let $M = L_\kappa[G]$. $c \notin M$ (as $M$ and $L$ share the same reals), while $c \in L(M)$.

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  • $\begingroup$ I am still curious whether an analogy of your theorem is provable in $\mathsf{ZF}$ alone (when we define $\kappa$ to be inaccessible iff $\langle V_\kappa, V_\kappa+1\rangle$ is a model of second order $\mathsf{ZF}$ or equivalently iff $\kappa$ is regular and there is no surjection from any $x\in V_\kappa$ onto $\kappa$, which seems to be the natural generalization). $\endgroup$ – Alexander Block Jun 25 '17 at 14:40
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    $\begingroup$ I suspect that it is true is ZF as well, and that a similar proof works. Essentially, we need to show that the ordinal $\gamma$ in the proof is below $\kappa^{+}$ and that the following weak version of Lowenheim-Skolem-Tarski theorem holds: for every model of the form $N = \langle V_\kappa, \in ,A\rangle$ there is $\alpha < \kappa$ such that $V_\alpha \prec N$. Then, we need to show that there is $A\subseteq V_\kappa$ such that $\langle V_\kappa, A\rangle$ codes the structure $\langle L_{\gamma}(M), \in, M, x, ... \rangle$ and conclude that $X'_n$ exist. $\endgroup$ – Yair Hayut Jun 25 '17 at 16:01
  • $\begingroup$ Thank you for your follow-up answer on the ZF context. I see how most of it works out; however, I currently do not see how to show $\gamma < \kappa^+$ without relying on a version of the Lowenheim-Skolem-Tarski theorem involving at least some fragment of Choice. Do you have a proof idea for this in mind that avoids Choice altogehter? If so, I would be interested to know more about it. $\endgroup$ – Alexander Block Jun 27 '17 at 14:45
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    $\begingroup$ Let's assume $V = L(M)$ and let $x$ be as above. $\kappa$ is inaccessible in $V$. Therefore, we can show that there is no surjection from $V_\kappa$ onto $\kappa^{+}$ [the image of each $V_\alpha$, $\alpha < \kappa$ under this surjection will be a set of ordinals of order type $<\kappa$]. Thus, in the generic extension by $Col(\omega, M)$ (that satisfies choice), $\kappa^{+}$ is not collapsed. In this model I can find $\gamma < \kappa^{+}$ such that $x \in L_\gamma(M)$. Since $L_\gamma(M)$ is absolute between transitive class models that contain $M$ we conclude that the same holds in $V$. $\endgroup$ – Yair Hayut Jun 27 '17 at 15:32

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