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Question: Does every $\omega_1$-Aronszajn tree contains a Suslin sub-tree or a special Aronszajn sub-tree?

Recall that Suslin trees are $\omega_1$-trees (trees of height $\omega_1$, and countable width) that have only countable anti-chains. Special Aronszajn trees are $\omega_1$-trees that can be represented as a countable union of anti-chains. Aronszajn trees are $\omega_1$-trees without branches.

Every sub-tree of a special Aronszajn tree with height $\omega_1$ is special, and therefore - not Suslin. Similarly, every subtree of a Suslin tree is Suslin so it's not special.

So, my question is if those are the only possibilities or that (consistently) there are other kinds of Aronszajn trees in which every sub-tree is not a Suslin tree and not a special Aronszajn tree.

Note that under $MA$, for example, every Aronszajn tree is special, so it is consistent that the answer is (trivially) positive.

Edit: In this question, a sub-tree is a subset with the restricted order, namely if $\langle T,<_T\rangle$ is a tree, then for every $X\subset T$, $\langle X, <_T \cap (X\times X)\rangle$ is a subtree (usually not an $\omega_1$-tree by itself).

Sub Question: Is it consistent that there is an $\omega_1$-tree that has no special sub tree but it is not Suslin?

Mohammad Golshani asked this question in the comments. Maybe a solution for the sub-question will lead to an answer for the main question.

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    $\begingroup$ Could you clarify what you mean by 'subtree' here? $\endgroup$
    – user3462
    Commented Apr 7, 2014 at 12:45
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    $\begingroup$ I think more clarification of 'subtree' is needed here. If you count every subset of a tree as a subtree, then given any non-Suslin tree $T$, find some uncountable antichain $A$; isn't it special? $\endgroup$
    – Paul McKenney
    Commented Apr 7, 2014 at 17:54
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    $\begingroup$ You might also want to look at Chapter IX of Shelah's "Proper and Improper Forcing". $\endgroup$
    – Paul McKenney
    Commented Apr 7, 2014 at 18:06
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    $\begingroup$ @Paul: Funny you should say that. He carry a physical copy of that in his backpack. :-P $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 8, 2014 at 4:41
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    $\begingroup$ @PaulMcKenney : a subtree that contains a single uncountable antichain won't be a special Aronszajn tree since it won't be an $\omega_1$-tree (it has height $1$ and width $\omega_1$). In the definition of special Aronszajn tree I demanded that the tree is $\omega_1$-tree. $\endgroup$
    – Yair Hayut
    Commented Apr 8, 2014 at 4:46

1 Answer 1

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It is consistent that the answer is no. The following is proved in Beaodouin's thesis ``On uncountable trees and linear orders'', as Theorem 1.10:

Theorem. Assume $\kappa^{<\kappa}=\kappa$ and $\Diamond(E)$ holds, where $E \subseteq \{\alpha < \kappa^+: cf(\alpha)=\kappa \}$.Then there is a normal $\kappa^+$-Aronszajn tree which has no special or Souslin subtree of size $\kappa^+.$

I may mention that the theorem is not stated as above in the paper, but looking at the proof, it is clear that the above result is in fact proved. It is also mentioned in the paper that the case $\kappa=\omega$ has discovered by Hanazawa too.

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