$\omega_1$ trees in the ground model $M$ Let $M$ be a transitive and countable model of ZFC and
let $P \in M$ be an $\aleph_1$-closed notion of forcing. 
Suppose that $T$ is an $\omega_1$-tree in $M$
such that every level of $T$ is countable. What I would like to show is that $T$ has no new branches in $M[G]$ (generic extension.)
The way how I started doing it is the following:
1) Suppose for a contradiction that T has a branch $b \in M[G]$ which is not
in $M$. (note that then $b$ has to have length $\omega_1$)
Then there is a name $\overset{.}b$ for $b$ and a
condition $p_\emptyset \in G$ such that $p_\emptyset \Vdash \overset{.}b \neq \overset{ˇ}a$, for all $a \in M$.
2) I suppose that now, by induction, one should construct $p_s < p_\emptyset$ and nodes $x_s \in T$ for all finite sequences s of $0$’s and $1$’s, but I have problems in writing this construction.
 A: This result is due to Silver. 
$T$ may be assumed to be a tree on $\omega_1$. Let $R$ be the set of branches in $M$. Suppose in $M[G]$, there some branch $B$ through $T$ so that $B \notin R$. Let $\tau$ be some name for $B$ and $q$ be some condition so that $q$ forces that $\tau$ is a branch through $\check T$ and does not belong to $\check R$. In the following $\tau(\alpha)$ refers to the element of the branch $\tau$ on level $\alpha$.
Let $p_\emptyset = q$ and $\alpha_0 = 0$. 
Suppose for $n \in \omega$, that $\alpha_n$ has been defined and all $p_s$ have been defined where $s$ is a length $n$-string. 
Pick some length $n$ string $s$. Since $p_s \leq q$, $p_s \Vdash \tau \notin \check R$, i.e. $p_s$ forces that $\tau$ is not a ground model branch. Thus there must be some $\alpha^s > \alpha_n$ and conditions $p_0$ and $p_1$ below $p_s$ and ordinal $\beta_0,\beta_1 < \omega_1$ so that $p_0 \Vdash \tau(\check \alpha^s) = \check\beta_0$ and $p_1 \Vdash \tau(\check \alpha^s) = \check \beta_1$. Let $p_{s\hat{\ }i} = p_i$ for $i \in 2$. This define $p_t$ for all $t$ which are length $n + 1$ strings. 
Let $\alpha_\omega = \sup_{n \in \omega} \alpha_n$. 
Let $q_x$ be some element below $\langle p_{x\upharpoonright n} : n \in \omega\rangle$ using the $\aleph_1$-closedness.
Note that for every $\gamma < \alpha_\omega$, there is some ordinal $\epsilon_\gamma^x$ so that $q_x \Vdash \tau(\gamma) = \epsilon^x_\gamma$. So for each $x \in {}^\omega 2$, $\langle \epsilon_\gamma^x : \gamma < \alpha_\omega\rangle$ is a $\alpha_\omega$-length chain. By construction if $x \neq y$, then $\langle \epsilon_\gamma^x : \gamma < \alpha_\omega\rangle \neq \langle \epsilon_\gamma^y : \gamma < \alpha_\omega\rangle$. 
Choose $r_x \leq q_x$ and some ordinal $\gamma_x$ so that $r_x \Vdash \tau(\alpha_\omega) = \gamma_x$. By the above, if $x \neq y$, then $\gamma_x \neq \gamma_y$.
Then $\{\gamma_x : x \in {}^\omega 2\}$ is a uncountable set of ordinals in $T$ at level $\alpha_\omega$. Contradiction.
