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I would like to get a better idea of how badly compactness fails in $\mathcal{L}_{\omega_1\omega}$.

Let $\Gamma$ be an arbitrary set of sentences from $\mathcal{L}_{\omega_1\omega}$. Let the underlying signature $\tau$ also have arbitrary cardinality. Is there some cardinal $\kappa $ such that if every $\Delta\subseteq\Gamma$ where $|\Delta|\leq\kappa$ is satisfiable, then $\Gamma$ is satisfiable?

It is relatively easy to show that any such $\kappa$ would need to be $\geq \beth_{\omega_1}$, but I am unsure of how to proceed beyond there. If there is no such $\kappa$, I am also interested in weakening the question.

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I suppose one could carefully read the proof that compactness does not fail in the weakly-compact case and tweak it to have the answer here. –  Asaf Karagila May 16 '12 at 9:02
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If you weaken your demands to just having the upward Lowenheim-Skolem property, then Morley proved that the Hanf number of $L_{\omega_1, \omega}$ is $\beth_{\omega_1}$. –  Haim May 16 '12 at 12:15

1 Answer 1

up vote 14 down vote accepted

I like the question very much.

First, let me mention briefly that the question has a flaw in the quantifier order, since you have first fixed the theory $\Gamma$ and then ask for a cardinal $\kappa$ such that if all subtheories $\Delta\subset\Gamma$ of size at most $\kappa$ are consistent, then $\Gamma$ is consistent. This is trivially affirmative, since we may simply let $\kappa=|\Gamma|$, in which case $\Delta=\Gamma$ is one of the allowed subtheories.

The actual question here is the following (and note that I replace your $\leq\kappa$ with $\lt\kappa$, since this is how one usually frames it with weakly and strongly compact cardinals):

Question. Is there are a cardinal $\kappa$ such that if $\Gamma$ is any $L_{\omega_1,\omega}$ theory in any signature, and every $\kappa$-small subtheory is consistent, then $\Gamma$ is consistent?

Let us call this property the $\kappa$-compactness property for $L_{\omega_1,\omega}$. Just to be clear, $L_{\omega_1,\omega}$ is the infinitary language in which one is allowed to form countable conjunctions and disjunctions, but still only finitely many quantifiers at a time. Meanwhile, $L_{\kappa,\lambda}$ allows conjunctions and disjunctions of size less than $\kappa$ and blocks of quantifiers of size less than $\lambda$. One instinctively thinks of the following large cardinals:

  • A cardinal $\kappa$ is weakly compact if and only if it is uncountable and $L_{\kappa,\kappa}$ has the $\kappa$-compactness property for theories in a language of size at most $\kappa$.

  • A cardinal $\kappa$ is strongly compact if and only if it is uncountable and $L_{\kappa,\kappa}$ has the $\kappa$-compactness property for any theory without any size restriction.

One crucial difference between $L_{\omega_1,\omega}$ and $L_{\kappa,\kappa}$ or even $L_{\omega_1,\omega_1}$ is that in $L_{\omega_1,\omega_1}$, one can express the assertion that a relation is well-founded, since you can say that it has no infinite descending sequence. This does not seem possible to express in $L_{\omega_1,\omega}$, because one can quantify only finitely many variables at a time.

Theorem. If $\kappa$ is strongly compact, then $L_{\omega_1,\omega}$ has the $\kappa$-compactness property.

Proof. This is immediate, since any $L_{\omega_1,\omega}$ theory is also a $L_{\kappa,\kappa}$ theory. QED

Theorem. If $L_{\omega_1,\omega}$ has the $\kappa$-compactness property, then there is a measurable cardinal.

Proof. Suppose that $L_{\omega_1,\omega}$ has the $\kappa$-compactness property. Let $\Gamma$ be the theory including the following assertions:

  • the full $L_{\omega_1,\omega}$ diagram of the structure $\langle \kappa,\in,\hat A\rangle_{A\subset \kappa}$, in the language with a predicate $\hat A$ for each $A\subset\kappa$ and constants $\hat\alpha$ for each $\alpha\in\kappa$.
  • the assertions $c\neq \hat\alpha$ for each $\alpha\in\kappa$.

Note that any $\kappa$-small subtheory of $\Gamma$ is consistent, since we may interpret $c$ inside $\kappa$ if only fewer than $\kappa$ many $\alpha$ are excluded. So by the $\kappa$-compactness property, $\Gamma$ has a model $\langle M,\hat\in,\hat A^M\rangle$. Let $U$ be the set of $A$ for which $M\models c\in\hat A$. This $U$ is an ultrafilter and it is countably complete, since the assertions $(\forall x. \bigwedge_n x\in A_n)\to x\in A$, whenever $A=\cap A_n$, are part of $\Gamma$. It is nonprincipal since $c\neq \hat\alpha$ for any $\alpha$. So there is a countably complete nonprincipal ultrafilter, and hence there is a measurable cardinal, since the degree of completeness of such an ultrafilter is always measurable. QED

In particular, the hypothesis is strictly stronger than a weakly compact cardinal.

I'm not yet sure of the exact strength, but I'm inclined to think it is equiconsistent with a strongly compact cardinal, in light of the following observation:

Theorem. If $\kappa$ is the least measurable cardinal, then $L_{\omega_1,\omega}$ has the $\kappa$-compactness property if and only if $\kappa$ is strongly compact.

Proof. Suppose $\kappa$ is the smallest measurable cardinal and $L_{\omega_1,\omega}$ has the $\kappa$-compactness property. Fix any regular cardinal $\theta\geq\kappa$, and let $\Gamma$ be the $L_{\omega_1,\omega}$ theory of $\langle\theta,\in,\hat\alpha,\hat A\rangle_{\alpha\in\theta,A\subset\theta}$, plus the assertion $\hat\alpha\lt c$ for each $\alpha\in\theta$. This theory is $\kappa$-satisfiable, and indeed, $\theta$-satisfiable. So it is satisfiable, and there is therefore a model $\langle M,\in^M,\hat\alpha^M,\hat A^M,c^M\rangle$. Let $U$ be the set of $A\subset \theta$ such that $M\models c\in \hat A$. This is a countably complete nonprincipal uniform ultrafilter on $\theta$. Since $\kappa$ is the least measurable cardinal, it follows that $U$ must be $\kappa$-complete. So we have a $\kappa$-complete nonprincipal uniform ultrafilter on every regular $\theta\geq\kappa$. By a theorem of Menas, this implies that $\kappa$ is strongly compact. QED

Note that results of Magidor show that it is indeed possible for the least measurable cardinal to be strongly compact.

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Thanks Joel. Also thanks for the question tweak. –  Toby Meadows May 16 '12 at 13:06
    
Correction, in the last paragraph, it should be Ketonen's theorem, not Menas'. –  Joel David Hamkins May 16 '12 at 13:46
    
Isn't the $\kappa$-compactness of $\mathcal L_{\kappa,\kappa}$ and $\mathcal L_{\kappa,\omega}$ equivalent? I recall it is so at east in the weakly compact case. –  Asaf Karagila May 16 '12 at 14:09
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I'm inclined to think that $L_{\omega_1,\omega}$ has the $\kappa$-compactness property if and only if there is a strongly compact cardinal $\leq\kappa$. The reverse direction is clear. For the forward direction, my argument above shows that for all regular $\theta\geq\kappa$, there is a countably complete uniform ultrafilter on $\theta$. But I'm not clear on whether this gives a strongly compact cardinal or not. But this property certainly fails in $L[\mu]$, and so a measurable cardinal is not sufficient. I think it will similarly fail in many other canonical inner models. –  Joel David Hamkins May 16 '12 at 15:38
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Sorry for a not quite on-topic comment, but in case someone reading this is wondering: that no $L_{\kappa,\omega}$ can define well-foundedness is known for fact, it was proved by Lopez-Escobar eudml.org/doc/213903 . For fascinating connections of $L_{\infty,\omega}$ extended with a means for expressing well-foundedness, see Barwise dx.doi.org/10.1016/0003-4843(72)90002-2 . –  Emil Jeřábek Dec 10 '13 at 23:50

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