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Let $\theta$ be regular and uncountable. Fix a well-ordering $<$ of $H(\theta)$. Since the structure $(H(\theta),\in,<)$ has a definable well-ordering, every subset $A\subset H(\theta)$ admits a Skolem hull, namely the set of elements of $H(\theta)$ which are definable in $(H(\theta),\in,<)$ from parameters in $A$.

Let $M\prec (H(\theta),\in,<)$ be countable. Take $p\in M$ which is finite. Obviously, the Skolem hull is contained in $M$. But does it have to be a member of $M$? There's the obvious obstruction of the undefinability of truth, but I can't find a non-trvial example where the hull is definitely not in $M$. By non-trivial I mean the following: suppose $M$ is itself is the Skolem hull of $p$. Then obviously the hull of $p$ does not belong to $M$.

I'm specially interested in the case where $p$ is a finite $\in$-chain of countable elementary submodels of $(H(\theta),\in,<)$.

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1 Answer 1

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Claim 1: Assume ZFC + a measurable cardinal. Then we can find examples of such hulls $M$ where the hull $H$ of some $p\in M$ isn't equal to $M$ and isn't a member of $M$, and also such that the transitive collapse $\bar{H}$ of $H$ isn't in $M$.

Proof: Let $\kappa$ be measurable and $U$ be a measure on $\kappa$. Let $\theta$ be a sufficiently large regular cardinal and fix a wellorder $<$ of $\mathcal{H}_\theta$. Let $H\preceq (\mathcal{H}_\theta,<)$ be the definable hull of some $p\in\mathcal{H}_\theta$ such that $U\in H$. Let $A=\bigcap(U\cap H)$, so $A\in U$. Let $\alpha\in A$. Note $\alpha\notin H$. Let $M$ be the definable hull of $(p,\alpha)$. Then $p\in H\subsetneq M$. But $H\notin M$, because $\sup(\mathrm{OR}\cap M)=\sup(\mathrm{OR}\cap H)$. In fact, letting $\bar{H}$ be the transitive collapse of $H$ (where the wellorder $<$ collapses to a wellorder for $\bar{H}$) and $\sigma:\bar{H}\to H$ the uncollapse map and $\sigma(\bar{U})=U$, and $\bar{M}$ the transitive collapse of $M$ and $\pi:\bar{M}\to M$ the uncollapse map and $\pi(\bar{p})=p$, then we have $\bar{M}=\mathrm{Ult}(\bar{H},\bar{U})$ and $\bar{H}$ is the transitive collapse of the hull of $\bar{p}$ in $\bar{M}$ (where we can still refer to the relevant wellorder for definability), and the ultrapower map $\bar{H}\to\bar{M}$ is just the uncollapse map. So also, $\bar{H}\notin\bar{M}$ and $\bar{H}\notin M$.

EDIT:

Claim 2: Assume $V=L$ and let $\theta=\gamma^+$ where $\gamma$ is a cardinal. (Recall $L_\theta=\mathcal{H}_\theta$ in this context.) Let $<$ be the standard $L$-ordering $<_L$ restricted to $L_\theta$. Let $p,q\in L_\gamma$. Let $H_p$ be the hull of $\{p\}$ in $L_\gamma$, and $H_q$ that of $\{q\}$. (Note: I mean the uncollapsed hulls here, i.e. not their transitive collapses.) Then $H_p\notin H_q$ and $H_q\notin H_p$.

Proof: In fact, $\sup(H_p\cap\theta)=\sup(H_q\cap\theta)<\theta$, which immediately implies the claim. In fact, $$\sup(\mathrm{Hull}^{L_\theta}(\emptyset)\cap\theta)=\sup(\mathrm{Hull}^{L_\theta}(\gamma)\cap\theta),$$ where $\mathrm{Hull}^{L_\theta}(X)$ denotes the definable hull of parameters in $X$ computed over $L_\theta$, and the hull is uncollapsed. To see this, first note that $\gamma$ is definable over $L_\theta$ without parameters. Now let $n<\omega$. Then note that $$\sup(\mathrm{Hull}^{L_\theta}_{\Sigma_n}(\gamma))$$ (the supremum of the $\Sigma_n$-definable-hull of parameters in $\gamma$) is ${<\theta}$, and this supremum is definable over $L_\theta$ without parameters (is we can define a $\Sigma_n$ satisfaction predicate over $L_\theta$), and this supremum is therefore in $\mathrm{Hull}^{L_\theta}(\emptyset)$. Since $$\sup(\mathrm{Hull}^{L_\theta}(\gamma)\cap\theta)=\sup_{n<\omega}(\sup(\mathrm{Hull}_{\Sigma_n}^{L_\theta}(\gamma)\cap\theta)),$$ the claim easily follows.

However, if we consider the transitive collapses of the hulls, it is a very different picture:

Claim 3: Assume $V=L$. Let $\theta$ be any regular uncountable cardinal. Let $<$ be the usual $L$-ordering (over $L_\theta$). Let $p\in M\preccurlyeq L_\theta$ and suppose $H_p\neq M$ (where $H_p$ is as before). Let $\bar{H_p}$ be the transitive collapse of $H_p$. Then $\bar{H_p}\in M$.

Proof: This is just a simple variant of the solidity of the standard parameter in the fine structure theory of $L$. We may assume that there is no $q<_Lp$ such that $p$ is definable over $L_\theta$ from $q$ (otherwise, some such gets into $H_p$, and then we can replace $p$ with the least such). Let $\bar{M}$ be the transitive collapse of $M$ and $\pi:\bar{M}\to M$ be the uncollapse map. Let $\pi(\bar{p})=p$. Then $\bar{M}=L_\alpha$ for some $\alpha$, by condensation, and similarly, $\bar{H_p}=L_\beta$ for some $\beta\leq\alpha$. If $\beta<\alpha$ then $\bar{H_p}\in\bar{M}$, and it easily follows that $\pi(\bar{H}_p)=\bar{H}_p\in M$. So suppose $\beta=\alpha$, so $\bar{H_p}=\bar{M}=L_\alpha$. Note that $H_p,\bar{H}_p$ are isomorphic to $\mathrm{Hull}^{\bar{M}}(\{\bar{p}\})$. So letting $\bar{H}'$ be the transitive collapse of this hull and $\sigma:\bar{H}'\to\bar{M}$ the uncollapse map, we have $\bar{H}'=\bar{H_p}=L_\alpha=\bar{M}$, and $\sigma(\bar{q})=\bar{p}$ for some $\bar{q}$, and $$\text{(*) }\bar{H}'=\mathrm{Hull}^{\bar{H}'}(\{\bar{q}\});\text{ that is, } L_\alpha=\mathrm{Hull}^{L_\alpha}(\{\bar{q}\}).$$ Since $\sigma:L_\alpha\to L_\alpha$ is elementary, and $\sigma(\bar{q})=\bar{p}$, we must have $\bar{q}\leq_L\bar{p}$. Note (by (*)) $\bar{p}$ is definable over $L_\alpha$ from $\bar{q}$. But we chose $p$ so that it was not definable over $L_\theta$ from any $q<_Lp$, and it is then straightforward to see that this (together with $\bar{q}\leq_L\bar{p}$) implies $\bar{q}=\bar{p}$. But then $L_\alpha=\mathrm{Hull}^{L_\alpha}(\{\bar{p}\})$, but since $\bar{M}=L_\alpha$, this lifts to give that $M=\mathrm{Hull}^M(\{p\})$, so $H_p=M$, contradiction.

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    $\begingroup$ Thank you very much for your answer. Would you mind explaining why $\alpha\not\in H$ in your first claim? I'm taking my time to read the other two arguments. My knowledge of fine structure is quite rudimentary, and by that I mean it's almost non-existent but the temptation to make the "rudimentary" pun proved too strong to resist. Alas, modulo your motivating solidity comment, at a first glance it doesn't seem that you're explicitly using any fine structure, but I want to read it carefully. Thank you again. $\endgroup$
    – Reveillark
    Jun 21, 2021 at 15:22
  • $\begingroup$ No worries :) Well, actually it's more that it's similar to the proof of solidity at the $\Sigma_1$-level for $L_\alpha$, which itself doesn't involve any substantial fine structure. $\endgroup$
    – Farmer S
    Jun 21, 2021 at 15:56
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    $\begingroup$ In the proof of Claim 1, we get $\alpha\notin H$ because given $\beta\in H$, consider the set $X=\kappa\backslash\{\beta\}$: We have $X\in U\cap H$, so $\beta\notin A$. $\endgroup$
    – Farmer S
    Jun 21, 2021 at 15:58

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