8
$\begingroup$

Recall that a tree on $\omega_1$ is called Suslin if every chain and antichain are countable. If every level is countable and there are no cofinal branches, then it is called Aronszajn (in particular Suslin is Aronszajn). Moreover, an Aronszajn tree is called special if it is the countable union of antichains.

So special Aronszajn trees and Suslin trees represent some sort of endpoints on the spectrum of Aronszajn trees: either every antichain is countable, or the whole tree is a countable union of antichains.

Given an Aronszajn tree, we can always specialize it using a ccc forcing; and given a Suslin tree it is a ccc forcing by itself, and forcing with it will add a cofinal branch (which is an uncountable chain).

Therefore, given a Suslin tree, we have a choice of how we want to violate its Suslinity: either violate its Aronszajn-ness, or specialize it.

Can we make a Suslin tree into a non-special Aronszajn tree, while destroying its Suslinity?

If the answer is positive, what is the "best" way of doing so? (Either consistently or provably, from the existence of a Suslin tree.)

Can we do it with a ccc forcing, or with a $\sigma$-closed forcing, or with just a proper forcing? Do the properties of this forcing somehow depend on the tree (e.g. if the tree is rigid, or homogeneous, etc.)?

(As a minor bonus question, is it consistent that all Aronszajn trees are Suslin or special, but Suslin trees exist?)


Just a remark on triviality, when I say "tree" I always mean that any point has at least two successors, and that it has successors on every level of the tree. I am also going to assume the tree is Hausdorff, in the sense that at limit levels the points are always determined by their predecessors.

$\endgroup$
5
  • 2
    $\begingroup$ You can't achieve this by $\sigma$-closed forcing by Easton's lemma. For the bonus question, you can build a nonspecial non-Suslin tree as follows: start with an special A-tree with some uncountable mac and a Suslin tree, then replace the cone above each element in the mac by the Suslin tree. It is not Suslin since we start with an uncountable mac. It is nonspecial since forcing with this does not collapse $\omega_1$. $\endgroup$
    – Jing Zhang
    Commented Dec 1, 2018 at 22:50
  • $\begingroup$ I don't understand. If you added a Suslin tree, how are there no Suslin trees? $\endgroup$
    – Asaf Karagila
    Commented Dec 1, 2018 at 22:52
  • 2
    $\begingroup$ I'm saying you can't have universe where the only A-trees are either special or Suslin (and there is a Suslin tree). You can find other A-trees that are nonspecial and non-Suslin. $\endgroup$
    – Jing Zhang
    Commented Dec 1, 2018 at 22:55
  • $\begingroup$ Ohh, right. Sorry. $\endgroup$
    – Asaf Karagila
    Commented Dec 2, 2018 at 8:40
  • $\begingroup$ Here is a technical refinement to your remark, by the way: Is it consistent that every Aronszajn tree is special, or contains a Suslin subtree? $\endgroup$
    – Asaf Karagila
    Commented Jun 10, 2019 at 10:18

1 Answer 1

10
$\begingroup$

Chapter IX of Proper and Improper Forcing addresses this issue.

Shelah proves that Souslin's Hypothesis does not imply every Aronszajn tree is special, and he does this by investigating weak notions of specialness that are still incompatible with Souslinity. He shows that there are forcings that "specialize" Aronszajn trees in the weak sense, and that they can be iterated while still preserving at least one non-special Aronszajn tree.

Corollary 4.8 spells out more details:

He starts with a Souslin tree $T^*$ and preserves the fact that it is a non-special Aronszajn tree while also specializing all A-trees in a weak sense. He uses an $\aleph_1$-free iteration of proper forcing instead of countable support, but it is not clear this is necessary.

The concept of $(T^*, S)$-preserving (Definition 4.5) seems to be the property of forcing notions that ensures the tree $T^*$ never gets fully specialized.

The entire chapter is pretty technical, but it contains many interesting ideas that could and should be developed further.

$\endgroup$
3
  • $\begingroup$ Thanks Todd. What is an $\aleph_1$-free iteration, though? $\endgroup$
    – Asaf Karagila
    Commented Nov 30, 2018 at 18:03
  • 5
    $\begingroup$ That's a good question, because I don't think anyone other than Shelah really knows. He defines it in the first section of that chapter; it's an iteration that does something "bigger" than the inverse limit at ordinals of countable cofinality. He says that he can get away without it, but just wanted to present the material. $\endgroup$ Commented Nov 30, 2018 at 18:31
  • $\begingroup$ Hmm. That reminds me a bit of Woodin's extender algebra thing. But the whole definition just doesn't work when you insist on forcing "upwards". It's interesting, though. Thanks for pointing the finger to this definition! $\endgroup$
    – Asaf Karagila
    Commented Nov 30, 2018 at 19:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .