By an Aronszajn tree, I mean a tree of height $\omega_1$ with countable levels and no branch. Such a tree is Souslin if it has no uncountable antichains and special if it can be written as the countable union of antichains. It is well known that there is both a ccc forcing and a $\sigma$-closed forcing for adding Souslin trees and, given any Aronszajn tree $T$ (Souslin or otherwise) there is a ccc forcing for specializing $T$. Therefore we can add Souslin trees, and then specialize them, giving us new special Aronszajn trees. In this post I have a variety of questions about forcings which add Aronszajn trees that are neither special nor Souslin. The most basic of these is as follows.

Question 1: What are some of the most common ways of forcing to add an Aronszajn tree which is neither Souslin nor special? In this post Mohammad Golshani cites a theorem that immediately implies that such trees exist under certain $\diamondsuit$ hypotheses. Therefore forcings to add diamonds suffice but I'm curious whether there is a more explicit forcing where the generic is actually the tree.

My next question is about killing Souslin-ness (or whatever it's called) without specializing.

Question 2: Suppose $T$ is Souslin. Does there exist a forcing notion $\mathbb P$ such that in $V^\mathbb P$, $T$ is still Aronszajn but no longer Souslin and also not special?

I'm also curious about these questions in the case of fat Aronszajn trees: Aronszajn trees with the requirement of countable levels removed. The same ccc specializing forcing works to make such trees special so under MA all such trees are special. I'm curious whether non-special fat Aronszajn trees can be built by hand using forcing.

Question 3: What forcings will add a fat, non-special Aronszajn tree?

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    $\begingroup$ In these notes math.cmu.edu/~eschimme/Appalachian/EisworthMooreNotes.pdf a forcing to kill an A-tree is provided. It answers your second question. (Page 17.) $\endgroup$
    – Not Mike
    Jan 12, 2018 at 22:18
  • $\begingroup$ Regarding question 2, I suggest the following: suppose $T$ is the given Souslin tree and suppose all of its levels be countable. Let $P$ be the forcing notion for adding an antichain into $T$ using conditions which are finite antichains of $T$. The forcing is c.c.c and adds an uncountable antichain in $T$. I think $T$ is not special in the extension but I did not check it. $\endgroup$ Jan 14, 2018 at 4:52
  • $\begingroup$ I may mention that the work of Laver-shelah to kill all $\aleph_2$-Soulin trees uses similar forcing using countable antichains. $\endgroup$ Jan 14, 2018 at 4:52
  • $\begingroup$ @Not Mike thank you I'll check that out! $\endgroup$ Jan 15, 2018 at 13:31
  • $\begingroup$ @Mohammad Golshani Thanks as well. For both of the forcings you suggest do you know of a reference I can look at? $\endgroup$ Jan 15, 2018 at 13:31

1 Answer 1


Regarding your comment before Question 3, you mean all such trees with size $<\mathfrak{m}$ are special. In particular, it doesn't say anything about trees of size $\geq 2^{\omega}$. Indeed, $T(\mathbb{R})$ where elements are bounded well ordered subsets of $\mathbb{R}$ ordered by end-extension is not special. It definitely has no uncountable branches.

For question 1, you can first add a Suslin tree (there are various ways of doing this), of which the most straightfoward one is to use countable well-pruned trees with end-extension (due to Jech). For the second step (which also gives yet another answer to the second question), you can pick $S\subset \omega_1$ that is stationary and co-stationary, and use a Shelah style forcing $Q(T,S)$ to make $T$ $S$-st-special (this means there exists a regressive function $f$ whose domain is $T\restriction S$ such that $t<t'\in T\restriction S$ implies $f(t)\neq f(t')$), but not special (this is implied by the property of $Q(T,S)$ called $(T,S)$-preserving). Observe that if a tree is $S$-st-special, then there must not be any uncountable branch. You can find more details in Chapter IX of Shelah's proper forcing book.


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