The question in the title is somewhat self contained but let me make some definitions and remarks to clarify.

Recall that $\mathsf{MA}^+(\sigma\text{-closed})$ is the statement that if $\mathbb P$ is a $\sigma$-closed poset, $\langle D_\alpha \; |\; \alpha < \omega_1\rangle$ is a sequence of dense subsets of $\mathbb P$ and $\dot{S}$ is a name for a stationary set then there is a filter $G \subseteq \mathbb P$ so that $G \cap D_\alpha \neq \emptyset$ for all $\alpha < \omega_1$ and $\{\alpha \; | \; \exists p \in G \; p \Vdash \check{\alpha} \in \dot{S}\}$ is stationary.

A Kurepa tree is a tree of height $\omega_1$, countable levels and $\geq \omega_2$ branches.

My question is simply if $\mathsf{MA}^+(\sigma\text{-closed})$ implies there are no Kurepa trees.

I have no intuition about this, as on the one hand I don't see a way that $\sigma$-closed forcing suffices to show there are no Kurepa trees and on the other hand I don't see a way to force $\mathsf{MA}^+(\sigma\text{-closed})$ without killing all Kurepa trees. Specifically:

It's well known that Silver collapsing an inaccessible to $\omega_2$ implies that there are no Kurepa trees so in natural models of $MA^+(\sigma\text{-closed})$ obtained by iterating all $\sigma$-closed forcing notions below some sufficiently large inaccessible (supercompact?) $\kappa$ there are no Kurepa Trees.

Meanwhile, the proof that $\mathsf{PFA}$ implies there are no Kurepa trees involves specializing an Aronszajn tree and such specialization forcings are not in general $\sigma$-closed.



1 Answer 1


Yes, it implies no Kurepa trees. First, note that the forcing axiom you consider implies the Weak Reflection Principle, which in turn implies (a strong form of ) Chang's Conjecture. Both of those facts are covered in the Foreman-Magidor-Shelah paper on Martin's Maximum. And Chang's Conjecture implies there are no Kurepa trees. I believe the proof of the latter appears in Foreman's chapter in the Handbook of Set Theory, but here is a sketch: suppose $T$ was a Kurepa tree on $\omega_1$; let $\langle b_i \ : \ i < \omega_2 \rangle$ be a 1-1 list of cofinal branches of $T$. By Chang's Conjecture, there is an $X \prec (H_{\omega_3},\in, \vec{b})$ such that $X \cap \omega_2$ has ordertype $\omega_1$, but $X \cap \omega_1 \in \omega_1$. Consider the collection $\{ b_i \ : \ i \in X \cap \omega_2 \}$. For any distinct $b_i$, $b_j$ in that collection, since $i \ne j$ and both indices are in $X$, it follows that $b_i$ and $b_j$ diverge at some level below $X \cap \omega_1$. This implies that level $X \cap \omega_1$ of $T$ has size $\omega_1$, contradicting that it is a thin tree.

  • $\begingroup$ This is exactly the type of answer I was hoping for. Thank you! $\endgroup$ Nov 26, 2018 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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