# Skolem Hulls and countable elementary submodels of $H(\theta)$

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 $$(H(\theta),\in,<)$$.

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.

• 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. Jun 21, 2021 at 15:22
• 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. Jun 21, 2021 at 15:56
• 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 $\beta\notin A$. Jun 21, 2021 at 15:58