Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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.

share|improve this answer
1  
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 at 19:41

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.