Robinson consistency theorem for $L_{\omega_1\omega}$-logic

Question

I'm reading Keisler's book Model Theory for Infinitary Logic. More specific, I'm interested in one of the exercises that shows that the Robinson consistency theorem does not hold in general for $L_{\omega_1\omega}$-logic (see also p.22). It is stated there in the following (weak) version:

Let $L',L''$ be expannsions of $L$ such that $L'\cap L''=L$. Let $T$ be a countable complete theory of $L_{\omega_1\omega}$. Let $\varphi$ be a sentence of $L'$ and $\psi$ one of $L''$. If $T\cup\{\varphi\}$ and $T\cup\{\psi\}$ each have a model, then $T\cup\{\varphi,\psi\}$ has a model.

I already found the solution for this exercise using Scott's isomorphism theorem (as indicated in the hints), but I can't think of a counterexample that shows that the above statement is wrong if you replace countable by uncountable.

I already know that for a counterexample one of the theories, say $T\cup\{\psi\}$ can't have a countable model, since otherwise the proof as in the "countable"-case would go through.

This leeds me to the intuition that this exercise is connected with another problem: show that there is a countable $L$ and an uncountable model (structure, if you like) $\mathcal{B}$ such that no countable model $\mathcal{A}$ is $L_{\omega_1\omega}$-elementarily equivalent to $\mathcal{B}$.

Thanks for any help or advice!

Martin

Answer 1

Here's an answer to your second question:

My language $L$ will consist of infinitely many unary predicates $U_n$ ($n\in\mathbb{N}$). My structure $\mathcal{B}$ will essentially be the powerset of the naturals - $\mathcal{B}$ will have domain $\mathcal{P}(\mathbb{N})$, and for $X\in\mathcal{B}$ we'll have $$U_n(X)\iff n\in X.$$ Note that for every $X\in\mathcal{P}(\mathbb{N})$, there's an infinitary sentence $\varphi_X\in\mathcal{L}_{\omega_1\omega}(L)$ saying roughly that $X$ is in $\mathcal{B}$: specifically, set $$\varphi_X=\exists z[(\bigwedge_{n\in X}U_n(z))\wedge(\bigwedge_{n\not\in X}\neg U_n(z))].$$ Clearly $\mathcal{B}$ satisfies each $\varphi_X$, but any countable $\mathcal{A}$ can only satisfy countably many of the $\varphi_X$s.