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I'd like to know more about forcing to add a cofinal branch to an $\omega_1$-tree.

Question 1:

What kinds of forcings add cofinal branches to $\omega_1$-trees? What kinds of forcings cannot?

The answer may depend on the type of tree. I don't expect a full characterization of such forcings, but would like to know what experience anyone has with forcing branches.

Question 2:

Is the following statement correct? If $\mathbb{P}$ is a notion of forcing which has the countable chain condition and $\mathbb{P} \times \mathbb{P}$ also has the countable chain condition, then $\mathbb{P}$ cannot force a cofinal branch in an $\omega_1$-tree.

If the statement is correct, what is the proof? A non-example is that a Souslin poset is ccc, but the product of two Souslin posets is $\it{not}$ ccc (since the set of pairs of immediate sucessors of each node in the tree gives an uncountable antichain) and a Souslin poset $\it{can}$ add a cofinal branch to a Souslin tree.

Question 3:

Sometimes (as in the case of a special Aronszajn tree) a notion of forcing which satisfies the countable chain condition cannot add a branch to an $\omega_1$-tree because any forcing which adds a cofinal branch to the tree will collapse $\omega_1$ and ccc posets preserve $\omega_1$.

Does anyone have any experience with an $\omega_1$-tree such that no ccc forcing can add a branch, but the reason is not because adding a cofinal branch will collapse $\omega_1$?

These questions are related to and arose because of Joel Hamkins' open question on mathoverflow: Can there be an almost-special not-fully-special Aronszajn tree?

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3 Answers 3

Tanmay Inamdar has pointed out to me that Lemma 2.8 of my paper with Stevo Todorcevic, ``Chain conditions in maximal models", is related to Question 3 here, and also to Joel and Arthur's questions. In this Lemma, we assume that $S$ is a stationary, co-stationary subset of $\omega_{1}$, and that $\Diamond$ holds on the compliment of $S$. By a relatively straightforward construction, we build an $\omega_{1}$-tree $T$ and a set $A \subseteq T$ such that if $P$ is any path through $T$ intersecting $A$ at uncountably many levels (as any generic branch will do), then this set of levels is a club disjoint from $S$. Furthermore, forcing with $T$ does not collapse $\omega_{1}$. This doesn't quite answer the questions above, as it is still possible that some c.c.c. forcing adds an uncountable branch to the tree formed by removing the nodes in $A$ from $T$. However, it seems that one can modify the construction to specialize $T\setminus A$, that is, to add a function $f \colon T \setminus A \to \omega$ such that whenever $p, q \in T \setminus A$ are compatible, and no member of $A$ lies in between $p$ and $q$, then $f$ is injective on the interval from $p$ to $q$. Then any outer model with an uncountable branch must either collapse $\omega_{1}$ or make $S$ nonstationary.

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An alternate argument is simply to let $B$ be a name for a new branch and choose for each $\alpha \in \omega_1$ conditions $p_\alpha$ and $q_\alpha$ and $t_\alpha$ in the $\alpha^\text{th}$ level of the tree such that $p_\alpha $ forces that $t_{p,\alpha}\in B$ and $q_\alpha $ forces that $t_{q,\alpha}\in B$ and $t\leq t_{q,\alpha}$ and $t\leq t_{p,\alpha}$. (This is possible because otherwise $B$ is not new.) Now it follows that $\{(p_\alpha, q_\alpha)\}_{\alpha\in\omega_1}$ must be an antichain in $\mathbb{P}\times\mathbb{P}$ (or at least can be thinned out to an uncountable antichain) --- this is the same argument as that the product of trees is never ccc.

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The answer to question 2 is yes, and I believe that this is due to Silver. (It may be in Jech's book.) The hypothesis that $\mathbb{P}\times\mathbb{P}$ is c.c.c. is equivalent to the assertion that $\mathbb{P}$ remains c.c.c. after forcing with $\mathbb{P}$.

Suppose now that $\mathbb{P}$ adds a new branch $t$ through a ground model tree $T$ of height $\omega_1$, and we don't even need it to have countable levels. Let $\dot t$ be a $\mathbb{P}$-name for this new branch, and let's assume $\Vdash\dot t$ is a new branch through $\check T$. Since $\mathbb{P}$ is c.c.c., $\omega_1$ is preserved. Consider the Boolean values $b_\alpha=\[\[\check t_\alpha\subset\dot t\]\]$, for $\alpha<\omega_1$, where $t_\alpha=t\upharpoonright\alpha$ is the new branch up to $\alpha$ (and note that for any fixed $\alpha$, this is an element of the ground model). If $\alpha<\beta$, then $b_\beta\leq b_\alpha$, and furthermore, these Boolean values must eventually descend, since otherwise we would be able to define $t$ in the ground model. So in the extension, we may refine this to a strictly descending $\omega_1$-sequence of Boolean values. The differences between these values makes an uncountable antichain in $\mathbb{P}$ in the extension, contrary to the assumption that $\mathbb{P}\times\mathbb{P}$ is c.c.c.

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Thank you. It is good to know. Could you explain the last sentence? It seems like you created an uncountable antichain in $\mathbb{P}$ in the extension with the descending $\omega_1$-sequence of Boolean values, but what do you mean by "the differences between these values"? And how does this contradict that $\mathbb{P} \times \mathbb{P}$ is c.c.c.? –  Erin Carmody Apr 10 '12 at 13:58
In a Boolean algebra, one has a notion of difference via $b-a=b\wedge\neg a$, and in any strictly descending sequences, the differences are incompatible. (Indeed, every well-ordered strictly descending chain from $1$ gives rise to a maximal antichain as its differences, and all maximal antichains arise this way. Thus, for Boolean algebras $V$-genericity is also characterized by $V$-completeness: any descending sequence in the ground model that is contained in the filter has a limit in the filter.) But to answer your question: my argument shows $\mathbb{P}$ is not c.c.c. in the extension. –  Joel David Hamkins Apr 10 '12 at 14:30
And this violates $\mathbb{P}\times\mathbb{P}$ is c.c.c., as I mention in my first paragraph. –  Joel David Hamkins Apr 10 '12 at 14:30
I see. Thanks. –  Erin Carmody Apr 10 '12 at 14:46

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