Slim Kurepa tree at a singular strong limit cardinal of uncountable cofinality

For a strong limit cardinal $\kappa$ the notion of $\kappa$-Kurepa tree is trivial: the full binary tree is a $\kappa$-Kurepa tree. Accordingly, we consider the following strengthening:

A slim $\kappa$-Kurepa tree is a tree $T$ of height $\kappa$ such that for every infinite $\alpha < \kappa$ the $\alpha$-th level of $T$ has cardinality $\left| \alpha \right|$, and $T$ has more than $\kappa$ many branches.

If $\kappa$ is a strong limit cardinal of countable cofinality, it's easy to construct a slim $\kappa$-Kurepa tree. On the other hand, if $\kappa$ is measurable (or just ineffable) then there is no slim $\kappa$-Kurepa tree. If $\kappa$ is inaccessible, then my understanding from comments here is that there is a $\mathord{<}\kappa$-closed forcing to create a slim $\kappa$-Kurepa tree (but this destroys measurability.) What about the uncountable cofinality singular case?

If $\kappa$ is a singular strong limit cardinal of uncountable cofinality, can there exist a slim $\kappa$-Kurepa tree?

EDIT: This turned out to be fairly easy; see my answer below. However, I would like to know where I can find this result proved (or at least mentioned) in print. So I will accept the first answer that tells me this.

• I am not sure about countable cofinality case. I know it is true if we assume square. Would you please provide a proof for this case in general. Apr 17, 2016 at 10:38
• In your definition of slimness, you should consider only nonzero $\alpha$. Apr 17, 2016 at 11:44
• @MohammadGolshani You are right; I don't see how to prove it in the general countable cofinality case. I don't know exactly what I had in mind, but maybe I was also assuming that $\kappa$ was strong limit, in which case we can take a cofinal sequence $(\kappa_i : i < \omega)$ and consider the tree corresponding to all subsets of the set $\bigcup_{i<\omega} [2^{\kappa_i}, 2^{\kappa_i} + \kappa_i)$. I will edit to add this assumption. Apr 21, 2016 at 1:50
• @Joel You are right, and in fact it seems to me that one should consider only infinite $\alpha$. I edited the question. Apr 21, 2016 at 1:54
• I think finite nonzero $\alpha$ are fine. If you insist on $|T(\alpha)|=|\alpha|$ for nonzero $\alpha$, then you can get get $\omega$ many nodes on level $\omega$ and then go to town. Apr 21, 2016 at 1:57

The following is proved by Erdos-Hajnal-Milner in On sets of almost disjoint subsets of a set. Acta Math. Acad. Sci. Hungar 19 1968 209–218'', from which the required result follows

Theorem. ssume $\aleph_0 < cf(\kappa) < \kappa$ and $\forall \theta< \kappa, \theta^{cf(\kappa)} < \kappa.$ Let $F \subseteq P(\kappa)$ be such that $\{\alpha < \kappa: |F \restriction \alpha| \leq \alpha \}$ is stationary. Then $|F| \leq \kappa.$

The answer is no. Martin Zeman showed me this proof. (Any mistakes were probably introduced by me.)

Let $\kappa$ be a singular strong limit cardinal of uncountable cofinality and let $T$ be a tree of height $\kappa$ such that for every $\alpha < \kappa$ the $\alpha$-th level of $T$ has cardinality $\left|\alpha \right|$. We will show that $T$ has at most $\kappa$ many cofinal branches.

Let $\gamma$ be the cofinality of $\kappa$ and let $(\kappa_\xi : \xi < \gamma)$ be a continuous increasing sequence of cardinals that is cofinal in $\kappa$. For every $\xi < \gamma$ let $(b^\xi_\alpha : \alpha < \kappa_\xi)$ enumerate the $\kappa_\xi$-th level of $T$.

For every branch $b$ of $T$, by a pressing-down argument there is a stationary subset $S \subset \gamma$ and an ordinal $\beta < \kappa$ such that for every ordinal $\xi \in S$ we have $b \restriction \kappa_\xi = b^\xi_\alpha$ for some $\alpha < \beta$.

So every branch is determined by a stationary subset $S \subset \gamma$ and a bounded function $S \to \kappa$, and there are only $\kappa$ many such functions.

• The fact that each branch is determined by a bounded function over a stationary set is how one proves that saturation implies that there are no Kurepa trees. You may want to investigate the history of the latter implication.
– saf
Aug 26, 2014 at 19:41