MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
    
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. – Mohammad Golshani Apr 17 at 10:38
    
In your definition of slimness, you should consider only nonzero $\alpha$. – Joel David Hamkins Apr 17 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. – Trevor Wilson Apr 21 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. – Trevor Wilson Apr 21 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. – Joel David Hamkins Apr 21 at 1:57
up vote 5 down vote accepted

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 $|F| \leq \kappa.$

share|cite|improve this answer

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|cite|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 '14 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.