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

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

• By elementarily equivalent you mean with respect to $L_{\omega_1,\omega}$? – tomasz Nov 4 '16 at 12:48
• And what do you mean by "other" countable model? – tomasz Nov 4 '16 at 12:53
• To your 1st comment: Yes, I mean elementarily equivalent w.r.t. $L_{\omega_1\omega}$, i.e., satisfying the same $L_{\omega_1\omega}$-sentences. To your 2nd comment: "model" is not really the right word (that's why I wrote structure in parentheses) but some people also use it in this context. The exercise claims that there is a countable language $L$ and an uncountable ($L$-) structure $\mathcal{B}$ such that for any other countable ($L$-)structure $\mathcal{A}$, $\mathcal{A}$ and $\mathcal{B}$ are not $L_{\omega_1\omega}$ elementarily equivalent. – Martin Monath Nov 4 '16 at 16:42
• What I meant is that when you say "other countable $L$-structure", it sounds like there has been some countable $L$-structure in the context already. – tomasz Nov 5 '16 at 19:33
• Sorry for the misunderstanding: no, this is not what I meant. The precise statement is: there exists a countable language $L$ and an uncountable $L$-structure $\mathcal{B}$ such that for any $L$-structure $\mathcal{A}$: if $\mathcal{A}$ is countable, then it is not $L_{\omega_1\omega}$-elementarily equivalent to $\mathcal{B}$. Hope this clarifies the misunderstanding. – Martin Monath Nov 7 '16 at 9:38

## 1 Answer

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.

• Thank you very much, Noah for your answer. I already saw it some days ago. I still haven't solved the first question, but your solution helped me a lot in terms of the intuition for that problem. I'll write more when I have time for that. Thanks again! – Martin Monath Nov 14 '16 at 7:56