TITLE: Is there a (first-order) sentence which admits $(\aleph_2,\aleph_0)$ iff a Kurepa tree exists?
QUESTION [10 upvotes]: 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, 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?

REPLY [5 votes]: 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.)