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I was reading Kanamori's The Higher Infinite, when I came across the fact that for any extendible cardinal $\kappa$ and any $\mathcal{L}_{\kappa,\kappa}^n$-theory $T$, $Sat(T)\Leftrightarrow \forall t\subseteq T(|t|<\kappa\rightarrow Sat(t))$

I thought this was interesting, as any $\mathcal{L}_{\omega,\omega}^n$-theory $T$ (that is, normal $n+1$-th order logic) is also a $\mathcal{L}_{\kappa,\kappa}^n$-theory. This implies that, assuming an extendible cardinal exists, the strong compactness cardinal of $n+1$-th order logic exists and is at most the least extendible cardinal. In fact, the strong compactness cardinal of normal logic, the union of all $n$-th order logics for $0<n<\omega$, is at most the least extendible cardinal.

This brings up the following questions: What is the consistency strength of asserting that $n$-th order logic has a strong compactness cardinal (for $n>1$)? If there is a strong compactness cardinal of $n$-th order logic, what large cardinal properties does it have?


EDIT: I have since done some thinking on this problem, and realized the strong compactness cardinal of "normal logic" (as I called it) is at least the supremum of all strong compactness cardinals of all $\mathcal{L}_{\kappa,\kappa}^n$ for natural $n$ and cardinals $\kappa$. This of course means that, letting the strong compactness cardinal of normal logic be $\lambda$, the strong compactness cardinal of $\mathcal{L}_{\lambda,\lambda}^n$ is at most $\lambda$. Because of this, $\lambda$ must be extendible.

This fact combined with the fact the the strong compactness cardinal of normal logic is at most the least extendible cardinal shows that the strong compactness cardinal of normal logic is in fact the least extendible cardinal.

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  • $\begingroup$ I wonder if they are basically very close, since to assert that an $\mathcal{L}_{\kappa,\kappa}^n$ theory $T$ has a model can be made in something like $\mathcal{L}_{\omega,\omega}^{n+1}$ as the assertion that there is a model and $\mathcal{L}_{\kappa,\kappa}$-truth predicate verifying every assertion of $T$ as true, and this now gets rid of the infinitary aspect of the theory. $\endgroup$ – Joel David Hamkins Nov 11 '17 at 12:37
  • $\begingroup$ @JoelDavidHamkins That may be. In any case, "normal logic" as I called it can express satisfiability of $\mathcal{L}_{\kappa,\kappa}^n$ theories, so the strong compactness cardinal of $\mathcal{L}_{\kappa,\kappa}^n$ is at most than that of normal logic for every $\kappa$ and $n$, and that is at most the least extendible cardinal. $\endgroup$ – Zetapology Nov 11 '17 at 16:29
  • $\begingroup$ @JoelDavidHamkins As you may see now, I have edited my question with the solution that the strong compactness cardinal of normal logic is the least extendible cardinal. $\endgroup$ – Zetapology Nov 11 '17 at 18:04
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Maybe this question was already resolved, but here is a solution in the literature.

Magidor showed (Theorem 4) that $\kappa$ is extendible iff $L^2_{\kappa\kappa}$ is $\kappa$-compact, and that $\kappa$ is the first extendible iff it is the minimal $\kappa$ such that $L^2_{\omega\omega}$ is $\kappa$-compact.

That is, although the usual statement is that $\kappa$ is extendible iff $L^n_{\kappa\kappa}$ is $\kappa$-compact for all $n\in \mathbb{N}$, it is equivalent to the same statement about just $L^2_{\kappa\kappa}$.

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  • $\begingroup$ By the way, this is also Theorem 23.4 in Kanamori. $\endgroup$ – Tim Campion Sep 11 '18 at 2:17

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