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

Recall that $\mathsf{MA}^+(\sigma-{\rm closed})$ 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 $\mathsf{MA}^+(\sigma - {\rm closed})$ 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 $\mathsf{MA}^+(\sigma-{\rm closed})$ 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 $MA^+(\sigma-{\rm closed})$ 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

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