# Can there be an almost-special not-fully-special Aronszajn tree?

Question. Can there be an Aronszajn tree $T$, such that no c.c.c. forcing extension adds a cofinal branch to $T$, but there is an $\omega_1$-preserving forcing extension adding a cofinal branch to $T$?

To give some background, an Aronszajn tree is an $\omega_1$-tree with no $\omega_1$-branches, and it is special, if it is the union of countably many antichains. Any forcing extension in which a special Aronszajn tree $T$ comes to have a cofinal branch must collapse $\omega_1$, since the branch has size $\omega_1^V$ and will contain at most one node from each of the antichains witnessing that $T$ is special. In other words, a special Aronszajn tree must remain Aronszajn in any $\omega_1$-preserving forcing extension, and this includes all c.c.c. extensions. Meanwhile, any Aronszajn tree can be forced to be special, and the specializing forcing is c.c.c. and absolute (consisting of finite partial specializing functions). Thus, this forcing remains c.c.c. in any extension in which $T$ remains Aronszajn.

Concerning the question,

• A special Aronszajn tree satisfies the first requirement of the question, since we cannot add a branch to it by c.c.c. forcing, but it doesn't satisfy the second requirement, since no $\omega_1$-preserving extension can add a cofinal branch. So special Aronszajn trees won't work.

• A Souslin tree satisfies the second requirement, but not the first, precisely because it is itself c.c.c. and adds a branch through itself. So Souslin trees won't work.

What is needed is a tree that is almost special, in the sense that it remains Aronszajn in every c.c.c. extension, but not fully special, in the sense that we can add a cofinal branch by some (non-c.c.c.) $\omega_1$-preserving forcing.

I am interested in this question in part because the existence of such a tree $T$ would provide an answer Arthur Fischer's question on two versions of absolutely ccc, because the forcing $\mathbb{P}$ to specialize $T$ is c.c.c. and would remain c.c.c. in any c.c.c. extensions, since the tree would still be Aronszajn there, but it would not be c.c.c. in the $\omega_1$-preserving extension in which $T$ gains a cofinal branch.

• Allow me to clarify that my question is intended to inquire about the consistency with ZFC of having such a tree. We already know, of course, that it is consistent with ZFC that every Aronszajn tree is special, and in such a model, there will be no such trees as I seek. But can there be another model of ZFC that does have such a strictly almost-special tree? – Joel David Hamkins Mar 26 '12 at 13:42
• I am thinking of this question. Is it the same as asking whether there is an Aroszajn tree which is not a Souslin tree, but which is not a special Aronzajn tree? So, it should have uncountable anti-chains, but not be the countable union of uncountable anti-chains. – user10290 Mar 26 '12 at 18:52
• Just a comment that may be relevant: Chapter IX of Proper and Improper Forcing contains a zoo of A-trees that are "not quite special". Not sure if anyone has looked at these with regard to your question, though. – Todd Eisworth Mar 26 '12 at 19:01
• Todd, thanks for the advice---that sounds quite promising! I hope someone will take a look at it. Erin, I'm not sure if being a non-special non-Souslin Aronszajn tree is sufficient for my requirement, but indeed these are necessary requirements. – Joel David Hamkins Mar 26 '12 at 21:09
• If I remember correctly, Shelah's trees are "nearly-special" in the sense that adding a branch will necessarily kill a stationary set, so there can't be a proper forcing that adds a branch. (But that doesn't mean that there is nothing relevant in Chapter IX.) – François G. Dorais Mar 27 '12 at 16:29

## 2 Answers

The answer is yes.

Force a generic $\square(\omega_1)$ sequence. This poset, $S$, is $\sigma$-closed, so it doesn't collapse $\omega_1$. By Todorcevic, in the $\omega_1$-tree obtained from the minimal walks through this sequence every branch of length $\omega_1$ codes a thread through the $\square(\omega_1)$ sequence, and vise verse (and this is absolute upward as long as $\omega_1$ is uncountable).

In particular, in the model obtained from forcing the generic $\square(\omega_1)$ sequence, this tree is Aronszajn and as long as we don't collapse $\omega_1$, we know that it's enough to check what kind of forcing can add a thread to a $\square(\omega_1)$ sequence in order to know which forcing posets can add a branch.

It is well known that the threading forcing $T$ of this generic $\square(\omega_1)$ sequence is $\omega_1$-distributive, and actually $S\ast T$ has a dense $\sigma$-closed subset.

On the other hand, c.c.c. forcing can't thread a $\square(\omega_1)$ sequence:

Let $\langle C_\alpha\mid\alpha < \omega_1\rangle$ be a $\square(\omega_1)$ sequence, and assume that there is some c.c.c. forcing threading it. Let $D$ be a name for this thread (so $D$ is forced to be a club in $\omega_1$). By the c.c.c. there is some $D^\prime \subset D$ a club in the "ground model" (meaning before the c.c.c. forcing). Now, if $\alpha < \beta$ in $\text{acc } D^\prime$, then $D^\prime \cap \beta \subset C_\beta$ and therefore $\alpha \in \text{acc }C_\beta$, and by the coherence $C_\alpha = C_\beta \cap \alpha$. This implies that $E=\bigcup_{\alpha \in \text{acc } D^\prime } C_\alpha$ is a thread in the ground model.

• Took you long enough! :-) – Asaf Karagila Oct 28 '13 at 10:52

To supplement this with another example, it is also possible to construct a tree that is $\omega$-distributive and $S$-st-special (in Shelah's terminology) from $\Diamond^*(S^c)$ for some $S$ bi-stationary (in fact I think $\Diamond(S^c)$ works). It appears in Shelah's proper forcing book Chapter IX Lemma 3.9. This tree satisfies your requirement since forcing with the tree itself adds no countable sequences of ordinals, and no forcing preserving stationary of $S$ (and preserving $\omega_1$) can add a branch by $S$-st-specialness.

$T$ is $S$-st-special if there exists $f: T\restriction S\to \omega_1$ such that $f(x)\in ht(x)$ and $x<_T y \rightarrow f(x)\neq f(y)$.