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.



How badly does compactness fail in $\mathcal{L}_{\omega_1\omega}$?
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. 


