# Does Laver forcing add cofinal branches to $\omega_1$-trees?

Given an $\omega_1$-tree $T$ in the ground model, can Laver forcing add a cofinal branch to $T$? Assume GCH in the ground model.

Definitions:

An $\omega_1$-tree is a well-founded tree of height $\omega_1$ with all levels countable. A cofinal branch of $T$ is a maximal chain of type $\omega_1$. Laver forcing $\mathbb{L}$ consists of subtrees $T$ of $\omega^{<\omega}$ where, for some $x\in T$, every $y<_T x$ has only one child and every $y\geq_T x$ has $\aleph_0$ children. $q\leq_{\mathbb{L}} p$ iff $q\subset p$.

Related results:

Silver proved in 1971 that a countably closed forcing cannot add a cofinal branch to $T$: given $p_\varnothing$ forcing a new cofinal branch $B$, construct conditions $(p_s:s\in 2^{<\omega+1})$ where $p_t\leq p_s$ for $s\subset t$ but for $s\perp t$ nodes $b_s\perp b_t$ of $T$ are respectively forced into $B$ by $p_s$ and $p_t$. This contradicts countability of any level $T_\delta$ chosen high enough that $b_s\in T_{<\delta}$ for all finite $s$.

I can modify the above argument to show that Sacks forcing (perfect subtrees of $2^{<\omega}$) does not add a cofinal branch to $T$. Given $s\in 2^n$ and $p_s$, I find, for each $y\in p_s$ minimal with respect to having $n$ splitting nodes below it, conditions $$q_{sy0},q_{sy1}\leq p_s|y=\{x\in p_s: x\not\perp y\}$$ that respectively force nodes $c_{sy0},c_{sy1}$ into $B$ such that $c_{sy0}\perp c_{sz1}$ for all $y,z$. Then $p_{si}=\bigcup_y q_{syi}$ preserves the first $n$ 'levels' of splitting nodes of $p_s$, ensuring that $\{p_{f|n}:n<\omega\}$ has a common extension $q_f$ for each $f\in 2^{\omega}$. Finally, choose $\delta$ sufficiently high and then extend each $q_f$ to $p_f$ deciding $B$ at level $\delta$. If $f\not=g$, then $p_f$ and $p_g$ disagree about $B$ at level $\delta$.

The above fusion argument also applies to Miller forcing; just replace 'splitting' with 'infinitely splitting.' However, for Laver forcing, $p_{si}=\bigcup_y q_{syi}$ above may fail to be a condition because the 'trunks' of the $q_{syi}$ may be too long.

• Certainly not in the case where $T$ is a special Aronszajn tree, since Laver forcing is proper, and a branch would collapse $\omega_1$. This can probably be adapted to show that no fresh cofinal branches are added. – Asaf Karagila Mar 14 '17 at 17:53
• Also possibly related, mathoverflow.net/questions/244106/… and mathoverflow.net/questions/93658/… and maybe a few others too. – Asaf Karagila Mar 14 '17 at 17:56
• Branches through ground-model trees are always fresh, since all their initial segments, being in the tree, are in the ground model. – Joel David Hamkins Mar 14 '17 at 20:14
• Let $\tau$ be a Laver name for a cofinal branch through $T$, as forced by some condition $p$. In the generic extension $V[G]$, assuming $p \in G$, there will be some $n \in \omega$ such that cofinally many levels of the generic branch were decided by conditions with stem of length $n$. Some condition $p_{*} \leq p$ forces that some specific value $n_{*}$ has this property. Can you run your fusion argument below this $p_{*}$? – Paul Larson Mar 14 '17 at 21:30
• Asaf, we should restrict our ambitions from proper to Axiom A because Suslin trees are proper. Also, I think it more natural to focus on non-Aronszajn T here. The original motivation for Silver's argument was to show that a certain Levy collapse, thought of as a forcing iteration, never resurrects any of the Kurepa trees it kills. – David Milovich Mar 14 '17 at 23:21