# Where do uncountable models collapse to?

Suppose $T$ is a complete first-order theory (in an finite, or at worst countable, language). Given any model $\mathcal{M}\models T$ of cardinality $\kappa$, we can ask whether $\mathcal{M}$ can be made isomorphic to any countable model $\mathcal{N}\models T$ in the ground model; that is, whether there is some $\mathcal{N}\models T$, which is countable, such that $V[G]\models \mathcal{M}\cong\mathcal{N}$ for $G$ $Col(\kappa,\omega)$-generic over $V$. For simplicity, we'll say "$\mathcal{M}$ collapses to $\mathcal{N}$".

In case $T$ has countably many countable models, the answer is yes: the set of models of $T$ is a countable Borel set, and so is absolute. The uncountable models of such a $T$ are thus partitioned into $\omega$-many classes. My question is about what the nonempty classes are. For example, suppose $T$ is $\aleph_1$-categorical. Then the countable models of $T$ form an elementary chain, with a "top" model $\mathcal{N}_\omega$, and it is easy to see that this is the only countable model which uncountable models can collapse to.

However, there are plenty of non-$\aleph_1$-categorical theories with only countably many countable models, and beyond the $\aleph_1$-categorical setting things are much less clear. Some examples:

• The language has predicates $P_i$ ($i\in\omega$); $T$ asserts that the $P_i$ name infinite disjoint sets. Then the countable models of $T$ are classified by how many elements they have not in $\bigcup_{i\in\omega} P_i$, and so there are countably many countable models. For each countable model of $T$, there is an uncountable model of $T$ which collapses to it.

• The language has a single binary relation $E$; $T$ asserts that $E$ is an equivalence relation and that there is exactly one class of cardinality $n$, for each $n$. Countable models of $T$ are determined by how many infinite classes they have; and every countable model of $T$ except the prime is collapsed to by some uncountable model of $T$.

I'm curious what we can say in general about the set of models which can be collapsed to. There's a lot of questions around here that one can ask; let me focus on:

Is there a theory $T$ with countably many countable models, including $\mathcal{M}_0\prec\mathcal{M}_1$, such that $\mathcal{M}_0$ can be collapsed to by some uncountable structure but $\mathcal{M}_1$ cannot?

That is, can there be "gaps" in the range of the collapse?

EDIT: As Paul Larson pointed out in the comments below, a countable model is collapsed to by some uncountable model iff it is extendible (see Definition 2.6 in http://shelah.logic.at/files/1003.pdf) - that is, iff it has an uncountable $L_{\omega_1\omega}$-elementary extension. One direction is trivial. To show that every extendible model is collapsed to, let $\mathcal{M}$ be extendible with $\mathcal{M}\prec_{\omega_1\omega}\mathcal{N}$ for $\mathcal{N}$ uncountable. Let $\varphi$ be the Scott sentence of $\mathcal{M}$, and let $V[G]$ be a forcing extension in which $\mathcal{N}$ is made countable. Then $\mathcal{N}\models\varphi$ in both $V$ and $V[G]$; moreover, the statement "$\varphi$ is the Scott sentence of $\mathcal{M}$" is $\Pi^1_2$ (actually, better, but this is enough) and so absolute between $V$ and $V[G]$. So $V[G]\models\mathcal{M}\cong\mathcal{N}$.

It's worth keeping in mind the broader result, due to Barwise and Karp if I recall correctly, that $\mathcal{M}\equiv_{\infty\omega}\mathcal{N}$ iff $\mathcal{M}\cong\mathcal{N}$ in some forcing extension.

• It seems that you could naturally rephrase this in terms of extendible models (i.e., countable models which have the same Scott sentence as some uncountable model) and then drop the mention of forcing. – Paul Larson Oct 22 '15 at 2:32
• John Baldwin used "extendible" in our paper with Saharon (<users.miamioh.edu/larsonpb/BlLrSh1003march2015_3.pdf>). I assume that it was used previously, but I don't really know. – Paul Larson Oct 22 '15 at 3:03
• You might also get some mileage out of Shelah's theory of model-theoretic forcing, from his publication #88r, depending on what you're interested in. – Paul Larson Oct 22 '15 at 3:08
• For instance, it seems that the model-theoretic forcing approach gives : if $T$ is a complete first order theory having an uncountable model and only countably many countable models, then there is a countable extendible model of $T$ having a copy of each countable model of $T$ as an elementary submodel. Again, I don't know how much insight this gives into your question. – Paul Larson Oct 22 '15 at 17:28
• I've been wondering about this too. I don't see that it matters. By the way, Su Gao's "On automorphism groups of countable structures" shows that a countable structure is extendible if and only if its automorphism group is not closed in the group of permutations of its domain. So one could rephrase the problem using this. Here's a variation I've wondered about : can you have countable $M \prec N$ such that $M$ is extendible and $N$ is rigid? – Paul Larson Feb 17 '17 at 21:20