Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
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
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.

share|improve this answer
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
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.