It is a well-known theorem that well-orderings can not be characterized in $L_{\omega_1,\omega}$. In particular, if $\psi$ is an $L_{\omega_1,\omega}$-sentence in a vocabulary $\tau$ that contains a binary symbol $<$ and $<$ is a linear order in all models of $\psi$, then either $<$ is a well-ordering of order-type $\alpha<\omega_1$, or otherwise, $\psi$ has models where < is not well-ordered.

The proofs that I am aware of produce a model $N$ of size $\aleph_1$ such that $\mathbb{Q}$ embeds into $<^N$.

Is the following generalization of the above theorem true? Let $\psi$ be an $L_{\omega_1,\omega}$-sentence and $\psi$ has a model (of any size) where $<$ is not well-ordered. If $\psi$ as a model of size $\kappa$, then there is a model of size $\kappa$ where $<$ is not well-ordered.

My conjecture is "yes", but I could not find a proof in the literature. The main obstacle is that the methods used to produce the $\aleph_1$-sized model do not generalize to arbitrary $\kappa$.

  • $\begingroup$ I suppose that you want to assume that the domain of $<$ is uncountable in your model of size $\kappa$, since otherwise the answer is easily "no". Do you want to assume that the domain of $<$ is the entire model, or just of size $\kappa$? $\endgroup$ – Paul Larson Dec 19 '17 at 18:31
  • $\begingroup$ @PaulLarson: Let's assume that the domain of $<$ is the entire model. If we can deal with this case, then I think we can deal with the case where the domain of $<$ is a subset that may have different cardinality than the entire model. Also, I reworded my question to avoid your "easily no" possibility. $\endgroup$ – Ioannis Souldatos Dec 20 '17 at 10:11
  • $\begingroup$ Remind me, please, $L_{\omega_1,\omega}$ has Löwenheim-Skolem, or completeness? (or neither, or a partial property of some?) $\endgroup$ – Asaf Karagila Dec 20 '17 at 10:31
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    $\begingroup$ @Asaf It has downward Löwenheim-Skolem. I'm not sure what you mean by completeness. The logic is (for countable theories) complete wrt a nice infinitary calculus. There is no question of effectiveness, as already the number of sentences is continuum. $\endgroup$ – Emil Jeřábek Dec 20 '17 at 11:05
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    $\begingroup$ It's not compact. $\endgroup$ – Emil Jeřábek Dec 20 '17 at 11:13

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