I'm trying to learn model theory, but I'm stuck trying to understand something. I know that $Th(\mathbb{Q}, <)$ is supposed to be $\omega$-categorical, by a back-and-forth argument. However I seem to be able to prove that it isn't. I'm hoping someone can help me understand where my proof is going wrong.

Let $L$ be the language of $(\mathbb{Q}, <)$ (No constant symbols, one 2-place relation symbol <). Consider the following set of $L_{(\mathbb{Q}, <)}$ sentences:

$\Sigma = \{ \exists x \ c_q < x \mid q \in \mathbb{Q} \}$.

By Compactness, we have a model $N = (A, <)$ of $\Sigma$ which also models the elementary diagram of $(\mathbb{Q}, <)$ (since $\mathbb{Q}$ has no upper bound). By Lowenheim-Skolem, we can assume that $N$ has a countable universe.

Since $(\mathbb{Q}, <)$ (with constants corresponding to the elements of $\mathbb{Q})$ does not model $\Sigma$, they are not elementary equivalent as $L_{(\mathbb{Q}, <)}$ structures.

But then this means that the $L$ reduct of $N$ is not isomorphic to $(\mathbb{Q}, <)$ either.

This is because for any $L$ structure $(X, <)$, if $f: \mathbb{Q} \to X$ is an isomorphism, then for all $L$ formula $\phi(x_1, \dots, x_n)$, and all $n$-tuples $(q_1, \dots, q_n) \in \mathbb{Q}^n$, we would have that $(\mathbb{Q}, <) \models \phi(q_1, \dots, q_n)$ if and only if $(X, <) \models \phi(f(q_1), \dots, f(q_n))$. So then interpreting $c_q$ as $f(q)$ turns $(X, <)$ into a $L_{(\mathbb{Q}, <)}$ structure which is elementary equivalent to $(\mathbb{Q}, <)$.

In fact, this seems to be able to show that the theory of any structure is not $\kappa$-categorical for any $\kappa$ at least as big as the size of its language. Where is my mistake?

  • $\begingroup$ $(0,1)$ is an elementary submodel of $\mathbb{Q}$. $\endgroup$ – Rene Schipperus Dec 13 '17 at 4:48
  • $\begingroup$ I'm sorry, I don't understand. $\endgroup$ – Grant Dec 13 '17 at 4:55
  • $\begingroup$ What is $L_{(\mathbb{Q},<)}$? Is it $L$ together with a constant symbol $c_q$ for each $q\in\mathbb{Q}$? In any case, it's not clear why you say $(\mathbb{Q},<)$ does not model $\Sigma$. For any $q\in \mathbb{Q}$, there does exist some $x\in\mathbb{Q}$ which is greater than $q$... $\endgroup$ – Eric Wofsey Dec 13 '17 at 4:57
  • $\begingroup$ Yeah, that's what I meant by $L_{(\mathbb{Q}, <)}$. And I goofed up a bit. I was thinking of writing the type $\Sigma(x) = \{ c_q < x \mid q \in \mathbb{Q} \}$. Then $(\mathbb{Q}, <)$ doesn't satisfy it, but some elementary extension of it does since it is finitely consistent. I think the rest still works. That is to say, I'm still not sure where it goes wrong. $\endgroup$ – Grant Dec 13 '17 at 5:03
  • $\begingroup$ Well then $x$ is an additional constant symbol in the language, and you have not said anything about formulas involving $x$ in your penultimate paragraph. $\endgroup$ – Eric Wofsey Dec 13 '17 at 5:13

I'm taking into account the comments, that is : you feel like an extension that realizes the type $\Sigma(x) = \{x<c_q : q\in \mathbb{Q}\}$ shouldn't be isomorphic to $\mathbb{Q}$.

Your mistake is that you conclude from "$N$ and $\mathbb{Q}$ are not isomorphic as $L_{(\mathbb{Q}, <)}$ structures" that $N$ and $\mathbb{Q}$ are not isomorphic as $L$-structures. The most basic example of where this fails is going from any language (say, the language of groups) to the empty language, where isomorphism means equinumerosity. It may seem as though "simply adding constants to the language" shouldn't change what isomorphism means, but it's quite clear that it does : in this new language, $\mathbb{Q}$ is rigid, it has no proper automorphism, whereas in $L$ it was homogeneous ! Obviously if you add a few points to it, you'll lose that property, and so even if you're order-isomorphic to $\mathbb{Q}$, you will not be isomorphic in the new language.

As for your last remark in the question : if $N$ is your extension realizing $\Sigma(x)$, and you find an order isomorphism with $\mathbb{Q}$ and then change the interpretations for $c_q$, obviously it won't realize the type anymore !

  • 1
    $\begingroup$ Thanks, I understand now! $\endgroup$ – Grant Dec 13 '17 at 7:39

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