For a logic $\mathcal{L}$, let the compactness number of $\mathcal{L}$ (if it exists) be the least $\kappa$ such that every $(<\kappa)$-satisfiable $\mathcal{L}$-theory is satisfiable. Note that there is no restriction here on the cardinality of the language of the theory in question.

For example, an uncountable cardinal $\kappa$ is strongly compact iff it is the compactness number of its own infinitary logic $\mathcal{L}_{\kappa,\kappa}$; more interestingly, Magidor showed that $\mathsf{SOL}$ has a compactness number iff there is an extendible cardinal, in which case its compactness number is the least extendible cardinal.

My question is:

What is the strength of "For every $\kappa$, the compactness number of $\mathcal{L}_{\kappa,\kappa}$ exists?"

EDIT: Originally I said that I didn't know anything relevant, but I just noticed that one of the suggested related questions is very relevant, namely this one: there it is shown for example that the existence of a compactness number for $\mathcal{L}_{\omega_1,\omega_1}$ already implies the existence of a measurable cardinal, or more technically that the existence of a compactness number for $\mathcal{L}_{\omega_1,\omega_1}$ is equivalent to the existence of an $\omega_1$-strongly compact cardinal. A natural guess based on that is that the principle in question is equivalent to "For every $\kappa$ there is a $\kappa$-strongly compact cardinal," but I haven't had a chance to read through the argument in detail so I'm not too confident here.

  • $\begingroup$ Oh now I see you answered your question before I did. $\endgroup$ Oct 23, 2020 at 23:05
  • $\begingroup$ @GabeGoldberg I mean not really - I didn't see how to turn that into a genuine answer. $\endgroup$ Oct 23, 2020 at 23:06
  • $\begingroup$ Will Boney knows a lot about compactness characterizations of large cardinals (in various logics). I talked to him about this once but I forgot what he said, but he would be a good person to ask about questions like this. $\endgroup$ Oct 29, 2020 at 3:14

1 Answer 1


The compactness number for $\mathcal L_{\kappa,\kappa}$ is equal to the least $(\kappa,\infty)$-strongly compact cardinal. A cardinal is $(\kappa,\infty)$-strongly compact if for every set $X$, there is a $j : V\to M$ such that $\text{crit}(j)\geq \kappa$, and $j[X]$ can be covered by and element of $M$ of $M$-cardinality less than $j(\delta)$. I sketch a proof at the end because I don't know the reference.

But first: it follows easily that your hypothesis is equivalent to the existence of a proper class of almost strongly compact cardinals, which are (resp. should be) defined to be cardinals $\kappa$ such that for all $\gamma < \kappa$ every $\kappa$-complete filter can be extended to a $\gamma$-complete (resp. $\gamma^+$-complete) ultrafilter. Whether this is equivalent to the existence of a proper class of strongly compact cardinals is an open question. The true consistency strength is probably a proper class of supercompacts: all three of these hypotheses are equivalent under the Ultrapower Axiom. There is some evidence that the equivalence between a proper class of almost strong compacts and a proper class of strong compacts is a theorem of ZFC: the first almost strongly compact cardinal above an ordinal $\gamma$ is either strongly compact or else has countable cofinality (although the truth is I needed a little SCH to handle the case $\gamma = 0$). This is in Some combinatorial properties of Ultimate $L$ and $V$.

Now the proof. In one direction, you show that $\mathcal L_{\kappa,\kappa}$ is $\delta$-compact for any $\kappa$-strongly compact $\delta$. Fix a $\delta$-consistent theory $T$ in the signature $\tau$. Cover $j[T]$ by a theory $S\subseteq j(T)$ in $M$ of $M$-cardinality less than $j(\delta)$. You get a model $\mathfrak A$ of $S$ in $M$ by $j(\delta)$-consistency of $j(T)$. Take the reduct of $\mathfrak A$ to $j[\tau]$. This is essentially a model of $T$: more precisely, $j : T \to j[T]$ is an isomorphism of $\mathcal L_{\kappa,\kappa}$-theories because $\text{crit}(j)\geq \kappa$.

Conversely, if $\delta$ is the compactness number of $\mathcal L_{\kappa,\kappa}$, then for any set $X$ and any $\delta$-complete filter base $\mathcal B$ on $X$, you can build a $\delta$-consistent theory whose models are $\kappa$-complete ultrafilters on $X$ extending $\mathcal B$. (A $\delta$-complete filter base is a family of sets such that the intersection of any ${<}\delta$-sized subfamily is nonempty.) The signature has constants for all subsets of $X$ along with a predicate $W$. The theory contains the axiom "$W(A)$" for each $A\in \mathcal B$ and the axiom "If $W(\bigcup \mathcal P)$, then $\bigvee_{A\in \mathcal P}W(A)$" for every partition $\mathcal P$ of $X$ with $|\mathcal P| < \kappa$. The theory is $\delta$-consistent since if one takes a set $\mathcal A\subseteq P(X)$ of cardinality less than $\delta$, one obtains a model of the axioms in the signature restricted to constants from $\mathcal A$ by letting $W$ be the principal ultrafilter concentrated at $x\in \bigcap(\mathcal A\cap \mathcal B)$.

It follows that for any set $X$, there is a $\kappa$-complete ultrafilter on $P_{\delta}(X)$ extending the filter base $\langle A_x \rangle_{x\in X}$ where $A_x = \{\sigma \in P_\delta(X): x\in \sigma\}$. Such an ultrafilter is, by definition, fine. The associated ultrapower embedding $j : V\to M$ has critical point at least $\kappa$ and closure under $\kappa$-sequences by $\kappa$-completeness. Finally $\text{id}_\mathcal U$ is a cover of $j[X]$ by fineness, and $\text{id}_\mathcal U$ has $M$-cardinality less than $j(\delta)$ since it is an element of $j(P_{\delta}(X))$ by the definition of $M$-membership. So $\delta'$ is $(\kappa,\infty)$-strongly compact. But it is not too hard to show that the least $(\kappa,\infty)$-strongly compact cardinal is a limit cardinal, so $\delta$ must be $(\kappa,\infty)$-strongly compact.

  • $\begingroup$ Sorry, I'm having a silly moment; I don't immediately see the theory you describe in the second part of the last paragraph (getting from compactness numbers to large cardinals). At the very least it seems that we should have names for all subsets of $P_\delta(X)$, not just of $X$. Can you add some details? $\endgroup$ Oct 25, 2020 at 21:21
  • $\begingroup$ Yes you're right, I'll add the details. $\endgroup$ Oct 25, 2020 at 21:22
  • $\begingroup$ A couple years later, what happens with $\mathcal{L}_{\kappa,\omega}$ instead? (The inability to characterize well-foundedness seems like an issue for this argument, but I might be having a stupid moment.) $\endgroup$ Feb 2, 2022 at 4:15
  • $\begingroup$ The argument in the second-to-last paragraph above seems to yield that if $\delta$ is the compactness number for $\mathcal L_{\kappa,\omega}$, then every $\delta$-complete filter extends to a $\kappa$-complete ultrafilter; i.e., $\delta$ is $\kappa$-compact. Conversely, if $\delta$ is $\kappa$-compact, it is greater than or equal to the compactness number for $\mathcal L_{\kappa,\kappa}$, which in turn is at least the compactness number for $\mathcal L_{\kappa,\omega}$. Am I missing something? $\endgroup$ Feb 2, 2022 at 18:02
  • $\begingroup$ Nope, I'm just bad at math once it gets past 8pm or so. :P Thanks! $\endgroup$ Feb 2, 2022 at 18:16

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