In Chang and Keisler's *Model Theory* I came across the following theorem (Theorem 7.2.13):


**Theorem** There exists a (first-order) sentence $\sigma$ such that for all infinite cardinals $\alpha$, $\sigma$ admits $(\alpha^{++},\alpha)$ iff there exists an $\alpha$-Kurepa tree.


**Definitions** 


(1) Fix a predicate $P$ in the language of $\sigma$. Then $\sigma$ admits $(\kappa,\lambda)$ if there is a model $M$ of $\sigma$ with $|M|=\kappa$ and $|P^M|=\lambda$. Obviously, $\kappa\ge\lambda$.


(2) An $\alpha$-Kurepa tree has height $\alpha^+$, each level has at most $\alpha$ elements and there are at least $\alpha^{++}$ elements of height $\alpha^+$.


Although $\sigma$ is not explicitly defined in Chang and Keisler, I am assuming it says something like:


There is a predicate $P$, there is a surjection from $P$ to each level, the levels of the tree are linearly ordered, and there is a surjection from $P$ to every initial segment of this linear order.


I see this sentence works under GCH, but I have a hard time when GCH fails. 
Consider for instance the case when $\alpha=\aleph\_0$ and, say $2^{\aleph\_0}=\aleph\_2$. Then we can cook up a model of the above sentence that resembles $(\omega^{<\omega},\subset)$ with $\omega^\omega$ many cofinal branches. This model will be have type $(\aleph\_2,\aleph\_0)$. 


By a theorem of 
[Devlin](http://matwbn.icm.edu.pl/ksiazki/fm/fm89/fm8914.pdf), we can find a model of ZFC where there is no Kurepa tree and the continuum is $\aleph\_2$. Thus, it is consistent that $\sigma$ will have models of type $(\aleph\_2,\aleph\_0)$, while there are no Kurepa trees.


*My question*: Do I miss something "obvious", or is the above theorem true only under GCH?



Is there a (first-order) sentence which admits $(\aleph_2,\aleph_0)$ iff a Kurepa tree exists?
The problem, as I see it, is to build into the sentence $\sigma$ something that makes a certain definable set $A$ (in this case a set indexing the levels of your tree) have cardinality $\aleph\_1$. You've already ensured that $A$ isn't any bigger than $\aleph\_1$ by having surjections from $P$ to all of the initial segments of $A$. I claim the same idea can be used to ensure that $A$ isn't any smaller than $\aleph\_1$, as follows. Since the universe of your model is guaranteed to have cardinality $\aleph\_2$, you can ensure that $A$ is big enough by having a linear ordering of the universe and requiring surjections from $A$ onto all the initial segments of this ordering. The point is that there is no linear ordering of cardinality $\aleph\_2$ in which all proper initial segments are countable. (Here, and also in the $\sigma$ that you described in the question, "initial segment" should mean a set of the form $\{x:x\leq a\}$ for some $a$, not just a downward-closed set.)


