Given a theory $T$ over a countable language with infinite models, and $\kappa$ an infinite cardinal, we can find a model of $T$ of size $\kappa$ whose infinite definable sets are all of size $\kappa$.

I have seen this claim in many lecture notes but have not been able to find one giving the actual proof. I only know of the compactness argument to find a model of size $\kappa$. I have read that the claim might follow from an application of the downward version of Löwenheim-Skolem's theorem, but I have no clue how to apply it here.

I could really use a hint. Thanks!


Outline: Let $M$ be a model of size $\kappa$. For every formula $\varphi(x)$ of the language that defines an infinite subset of $M$, add $\kappa$ constant symbols $c_{\alpha,\varphi}$, where $\alpha$ ranges over all ordinals $\lt \kappa$, together with axioms that say that if $\alpha\ne \beta$ then $c_{\alpha,\varphi}\ne c_{\beta,\varphi}$, and $\varphi(c_{\alpha,\varphi})$. The resulting theory $T'$ is consistent, and has a model of cardinality $\kappa$ with the required property.

  • $\begingroup$ Very neat. Thank you for the quick reply. $\endgroup$ – mdlt Apr 11 '15 at 3:31
  • $\begingroup$ You are welcome. A variant of Upward Lowenheim-Skolem, same argument. $\endgroup$ – André Nicolas Apr 11 '15 at 5:23
  • $\begingroup$ Of course, as written this only gives that all infinite sets definable without parameters have size $\kappa$. But if you start with a model with all elements named, and iterate the argument $\omega$ times, you can get it for sets definable with parameters too. $\endgroup$ – Alex Kruckman Apr 12 '15 at 9:17

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