A generalization of Vaught's two cardinal theorem

Question

I'm trying to determine whether or not the following generalization of Vaught's two cardinal theorem is true.

Let $T$ be a complete theory in a language with a unary predicate $U$ and a binary predicate $E$, where $T$ says that $E$ is the edge relation of an undirected graph. Let $a\underline{E} b$ mean $a=b\vee aEb$.

Suppose that there exists models $\mathfrak{M}\prec \mathfrak{N}$, with $\mathfrak{N}$ a proper elementary extension of $\mathfrak{M}$, such that for all $a\in U(\mathfrak{N})$ there exists $b\in \mathfrak M$ with $a\underline{E}^\mathfrak{N}b$.

Does it follow that there exists an $\omega_1$-sized model $\mathfrak{A}$ with a countable $X\subset A$ (not necessarily definable) such that for all $a\in U(\mathfrak{A})$ there exists $b\in X$ with $a\underline{E}^\mathfrak{A}b$? If it doesn't follow in general, does it follow assuming some stability?

Answer 1

In case anyone else is wondering about this very specific question, I found a relevant paper but also a counterexample (which I believe actually invalidates one of the results in that paper).

The counterexample is totally cateogrical, so it seems that there's no way to recover this with niceness assumptions: Let $X$ be an infinite set with no structure and let $\mathfrak{A}$ be a structure whose underlying set is $[X]^2$, the set of unordered pairs, and let $E$ be given by $xEy$ iff $|x\cap y|=1$. If we let $\mathfrak{A}$ be the starting model and then $\mathfrak{B}$ be an elementary extension whose underlying set is $[X\cup \{ a\}]^2$, where $a$ is some element not in $X$, then $\mathfrak{A}$ and $\mathfrak{B}$ satisfy the assumptions of the question, but there is no "$(\omega_1,\omega)$-model" of this theory.

I believe the error in that paper is in the proof of lemma 2.4 (which they entirely gloss over), basically all of the proof of Vaught's theorem goes through up until the point where you try to construct the $\omega_1$-chain. Normally the set you're trying to keep small is definable, so it is fixed setwise by the isomorphism between $\mathfrak{A}$ and $\mathfrak{B}$, but for this kind of thing the set is not definable, so there's no reason why it can't be moved by the isomorphism between $\mathfrak{A}$ and $\mathfrak{B}$, which indeed is exactly what's happening in the counterexample in the previous paragraph. The maximal (in $\mathfrak{B}$) homogeneous subset/covering set of $\mathfrak{A}$ is some set of disjoint elements whose union is all of $X$. $\mathfrak{A}$ and $\mathfrak{B}$ are isomorphic, but any isomorphism moves some element of any maximal (in $\mathfrak{B}$) homogeneous set outside of $\mathfrak{A}$.

There is a weaker analog that might be true, but I don't know how to show it. Maybe if you have an $\omega$-chain $\mathfrak{A}_0 \prec \mathfrak{A}_1 \prec \dots$ of proper extensions such that $\mathfrak{A}_0$ has a maximal homogeneous set/covering set for all of $\bigcup_{i<\omega}\mathfrak{A}_i$ then you can do it. None of the counterexamples I can construct for the original statement are counterexamples for this.

EDIT: Also for countable theories the two-cardinal version of this does hold. The stronger version of Vaught's two-cardinal theorem proved in Chang and Keisler implies that if you have a model $\mathfrak{A}$ with a maximal homogeneous set/covering set $X$ with $|X|<|A|$, then there is a model $\mathfrak{B}$ with a maximal homogeneous set/covering set $Y$ with $|Y|=\omega$ and $|B|=\omega_1$.