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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 $\alpha < \kappa$ the $\alpha$-th level of $T$ has cardinality $\left| \alpha \right|$, and $T$ has more than $\kappa$ many branches.

If $\kappa$ has 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.

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up vote 6 down vote accepted

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.

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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 '14 at 19:41

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