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!


  • $\begingroup$ By elementarily equivalent you mean with respect to $L_{\omega_1,\omega}$? $\endgroup$ – tomasz Nov 4 '16 at 12:48
  • $\begingroup$ And what do you mean by "other" countable model? $\endgroup$ – tomasz Nov 4 '16 at 12:53
  • $\begingroup$ 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. $\endgroup$ – Martin Monath Nov 4 '16 at 16:42
  • $\begingroup$ 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. $\endgroup$ – tomasz Nov 5 '16 at 19:33
  • $\begingroup$ 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. $\endgroup$ – Martin Monath Nov 7 '16 at 9:38

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.

  • $\begingroup$ 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! $\endgroup$ – Martin Monath Nov 14 '16 at 7:56

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