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I've been researching model theoretic logics and all the related stuff (categoricity, Hanf numbers) lately, and have been confused on how the Łoś–Vaught test applied to infinitary logics.

The Łoś–Vaught test specifically states:

If a satisfiable theory is k-categorical for some k ≥ |σ| for σ is some signature and has no finite models, it is complete.

A large part of the proof of the Łoś–Vaught test is the upwards Löwenheim–Skolem theorem (to show models M≡N have an elementary extensions M' and N' of cardinality k such that M'≅N', this is also the reason why finite models are exempted in the definition).

My understanding of the Łoś–Vaught test is that it applies to any theory (higher order, ones based of infinitary logics) as long as it's categorical in some cardinality greater than it's signature. But I don't see how one can prove this in a theory with different Löwenheim–Skolem properties (An extreme case would to consider a theory based on the logic L(∞,ω) where it's lowenheim number and Hanf number is ∞. Essentially meaning that is has no upwards or downwards Löwenheim–Skolem property at all).

On a side note, it is possible to find a theory that is categorical in some infinite cardinality but isn't complete by the definition above (from Wikipedia)?

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    $\begingroup$ "My understanding of the Łoś–Vaught test is that it applies to any theory (higher order, ones based of infinitary logics) as long as it's categorical in some cardinality greater than it's signature." On what basis is this your understanding? $\endgroup$ Apr 15 at 3:48
  • $\begingroup$ I have found many different definitions including the one above and the one below my Noah Schweber. Just found out the definition above is very poor and general. $\endgroup$
    – SJe967
    Apr 15 at 4:21

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You write:

My understanding of the Łoś–Vaught test is that it applies to any theory (higher order, ones based of infinitary logics) as long as it's categorical in some cardinality greater than it's signature.

This is extremely false. For example, working in the empty (= just equality) language, there is a single second-order sentence $\sigma$ which says "The universe is infinite and does not have cardinality exactly $\aleph_1$." The theory $\{\sigma\}$ is categorical in every infinite cardinality (except $\aleph_1$ :P), but not complete since there are second-order sentences further restricting the cardinality of the universe (e.g. "The universe is countable" and "The universe has cardinality exactly $\aleph_{17}$" are each second-order expressible).

I suspect that you misheard/read/understood the statement

any first-order theory without finite models which is categorical in some infinite cardinality greater than that of its signature is complete,

which is true.

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  • $\begingroup$ Thank you,this mostly answers my question! Theories based on L(∞,ω are still first order and you can't use the Löwenheim–Skolem theorem to show elementary extensions. Is there another restriction that is implied? $\endgroup$
    – SJe967
    Apr 15 at 4:18
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    $\begingroup$ @SJe967 No, $\mathcal{L}_{\infty,\omega}$ is not first-order. Or rather, when we say "first-order logic" we almost always mean "finitary first-order logic." $\endgroup$ Apr 15 at 4:36
  • $\begingroup$ I see. Naturally, there would be much weaker results of the Łoś–Vaught test for infinitary logics but obviously not L∞,ω due to it's Hanf number and Löwenheim number being ∞. $\endgroup$
    – SJe967
    Apr 15 at 4:46
  • $\begingroup$ A quick question about uncountable signatures/languages. Working within a theory T based on a finitary first order logic L with an uncountable signature. Assume the Łoś–Vaught test applies to this theory. If we have a model M of cardinality ℵ_(0), less than that of the signature, we will get M≡N in order for T to be complete. For N is a model of cardinality greater than the signature. How does this occur when the Löwenheim-Skolem theorem only guarantees elementary extensions and substructures ≥ the size of the signature? $\endgroup$
    – SJe967
    Apr 16 at 12:13

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