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.


(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?

  • $\begingroup$ What if you add an extra predicate $U$, and require it has size successor to $|P|$, and require the tree has height of size $U$. $\endgroup$ Nov 7, 2015 at 5:06
  • $\begingroup$ I think if we require a first order theory, it might be easier: say we have a model of enouph of $ZFC\setminus $power set, $P$ is an infinite cardinal, $P^+$ is the largest cardinal and require you have a Kurepa family coded by elements of the universe. $\endgroup$ Nov 7, 2015 at 5:09
  • $\begingroup$ @Mohammad Because of the Downward Lowenheim-Skolem theorem, we can not require that a certain predicate has size the successor to |P|. $\endgroup$ Nov 9, 2015 at 16:48
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    $\begingroup$ When I was writing the note, I had the following idea: We have an ordering of the universe, U is ordered, any large enough initial segment of it is in one-one correspondence with P, and any initial segment of the universe is the one-one image of a map from U. In general the idea is that if we can realize both a cardinal $\kappa$ and $\kappa^{++}$, then by adding a new predicate and extra axioms we can also realize $\kappa^+.$ $\endgroup$ Nov 10, 2015 at 8:18
  • $\begingroup$ @Mohammad Yes this will work. The same idea is behind Andreas Blass' answer. $\endgroup$ Nov 11, 2015 at 16:28

1 Answer 1


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.)

  • $\begingroup$ Sorry I was gone for the weekend. This will work. $\endgroup$ Nov 9, 2015 at 16:41

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