countably closed forcing cannot add a branch to a $\aleph_2$-tree if $\neg\mathsf{CH}$

Question

I'm reading this survey. In it the author states the following result (fact 5.3) which is attributed to Silver:

If $2^{\aleph_0}>\aleph_1$, countably closed forcing cannot add a new branch to an $\aleph_2$-tree.

I cannot find any proof of this and I cannot come up proof. Could someone give an sketch of why this is true?

Thanks

Answer 1

Let $T$ be a tree of height $\omega_2$ and let $\mathbb{P}$ be a poset. Assume that the $\mathbb{P}$-name $\dot{f}$ is forced by the condition $p \in \mathbb{P}$ to be a branch through $T$ that is not in $V$. Then we can get a perfect binary tree $(p_s : s \in 2^{\mathord{<}\omega})$ of conditions in $\mathbb{P}$ below $p$ and a sequence of ordinals $(\alpha_n : n< \omega)$ less than $\omega_2$ such that for every $n < \omega$ and every pair of distinct $s,t \in 2^n$, the conditions $p_s$ and $p_t$ force $\dot{f} \restriction \alpha_n$ to be different nodes of $T$. Let $\alpha = \sup_{n< \omega} \alpha_n$.

If $\mathbb{P}$ is countably closed, then for each real $x \in 2^\omega$ there is a lower bound $p_x$ for $(p_{x \restriction n} : n<\omega)$. Then the various conditions $p_x$ force $\dot{f} \restriction \alpha$ to take continuum many different values from the $\alpha^{\text{th}}$ level of $T$. In particular if the continuum is at least $\omega_2$ then $T$ has a level of cardinality $\omega_2$, so it is not an $\omega_2$-tree.

I'm sure this argument appears in print somewhere, but I can't remember where.

Answer 2

Let $T$ be an $\aleph_2$-tree, and assume towards contradiction that $b$ is a branch in $V[G]\setminus V$.$\newcommand{\forces}{\Vdash}$

Then there is some $\dot b$ and $p\in G$ such that $p\forces\dot b\text{ is a branch in }\check T$. Because $b\notin V$, there are unbounded levels $C\subseteq\omega_2$, that for each such $\alpha\in C$ some $q_0,q_1\leq p$ decides incompatible data about where $b$ meets the $\alpha$-th level.

We construct an embedding of $2^\omega$ into $T$. This is impossible, because the width of the $\omega$-th level is $\aleph_2$, contradicting the fact that $T$ has no level of more than $\aleph_1$ elements, but the embedding must be bounded everywhere, so there is a level with $\aleph_2$ elements.

For each $f\in 2^{<\omega}$ we define $p_f\in P$ and $t_f\in T$. And since $P$ is countably closed each $f\in 2^\omega$ defines $p_f$ and $t_f$ as well.

1. Let $p_{\langle\rangle}=p$ and $t_{\langle\rangle}$ be any point $t\in T$ such that $p\forces t\in\dot b$.

2. Suppose that $p_f$ and $t_f$ were defined. There is some $q,q'\leq p_f$ which are incompatible and some $t\in T$ such that $q\forces t\in\dot b$ and $q'\forces t\notin\dot b$; let $p_{f^\frown 0}=q$ and $t_{f^\frown 0}=t$. Extend $q'$ a bit more to a condition $p_{f^\frown 1}$ such that for some $t_{f^\frown 1}\in T$, from the same level as $t$ above, and $p_{f^\frown 1}\forces \check t_{f^\frown 1}\in\dot b$.

3. Suppose that $f\in 2^\omega$ and every $n\in\omega$, $p_{f\upharpoonright n}$ and $t_{f\upharpoonright n}$ were defined. Let $q$ be some condition stronger than all the $p_{f\upharpoonright n}$'s, and $p_f\leq q$ is some condition which decides what is the element in $b$ from the level $\sup\{\alpha_n\mid n\in\omega\}$, where $\alpha_n$ is the level of $t_{f\upharpoonright n}$.

By each time choosing incompatible extensions we assure that this is in fact an embedding, and by limiting the levels we ensure that the levels of the $t_f$'s are not unbounded in the tree. And as remarked before, this is a contradiction to the fact that $T$ was an $\aleph_2$-tree.