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Below, $\alpha$ is a countable p.r.-closed ordinal $>\omega$.

Let $\mathcal{L}_\alpha=\mathcal{L}_{\infty,\omega}\cap L_\alpha$ (note that this is not the same as $\mathcal{L}_{\omega_1,\omega}\cap L_\alpha$ since elements of the latter are explicitly countable). By the Barwise completeness theorem if $\alpha$ is admissible or a limit of admissibles the set of $\mathcal{L}_\alpha$-validities is definable in $L_\alpha$ - indeed, it's $\Sigma_1$ over $L_\alpha$ - the point being that we have a good notion of proof.

However, if $\alpha$ is neither admissible nor a limit of admissible this argument breaks down. For some such $\alpha$ the result still holds, e.g. if $L_\theta$ is interpretable in $L_\alpha$ where $\theta$ is the next admissible above $\alpha$ (and in particular whenever the supremum of the $\alpha$-recursive ordinals is admissible), but that seems rather rare (e.g. club-many countable ordinals are non-Gandy).

My question is:

For which $\alpha$ is $\mathcal{L}_\alpha$-validity definable in $L_\alpha$?

(I'd also be interested in the finer-grained question of $\Sigma_1$-definability; for what it's worth, I'm not sure which is harder.) "Obviously" this should fail for most $\alpha$ ... but at present I don't even have an example of a single $\alpha$ for which it fails!

Closely related is the following question:

Supposing $\varphi\in\mathcal{L}_\alpha$ is a validity, how high up in the $L$-hierarchy do I have to look to find a proof of $\varphi$?

I suspect the answer to this question is not in general "unboundedly high below the next admissible."

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  • $\begingroup$ Note that unlike Barwise completeness, Barwise compactness does not apply to inadmissible limits of admissibles. For example, work in $\theta=\sup_{n\in\omega}\omega_n^{CK}$, for $i\in\omega$ let $\varphi_i$ be the sentence saying that $L_{\omega_i^{CK}}$ is an initial segment of the universe, and let $\psi$ be the sentence "$\omega$ is standard and there are only finitely many admissible ordinals." Then $\{\varphi_i:i\in\omega\}\cup\{\psi\}$ is $\mathcal{L}_\theta$-finitely satisfiable but clearly unsatisfiable. $\endgroup$
    – Noah Schweber
    Commented Mar 12, 2020 at 13:52

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