Existence of branches on a tree. I would like to ask an easy detail about the proof of the following combinatorial question. Suppouse that $\kappa,\lambda$ are infinite regular cardinals and $\lambda<\kappa$. If $T$ is a tree of height $\kappa$ all whose levels have cardinality less than $\lambda$ then $T$ has a branch.
I'm following Todorcevic's Trees and linearly ordered sets survey and he says the following. Take $\delta<\kappa$ a cardinal of cofinality $\lambda$ and a node $t_\delta$ in the $\delta^{th}-$level of $T$. Now, there must exists a $s_\delta$ such that $$\{t\in T: s_\delta< t\}\cap L_\delta(T)=\{t_\delta\}$$
because if not $L_\delta(T)$ would have cardinality $\geq\lambda$. 
Is somebody willing justify me in detail this assertion?
 A: Heavily revised. (Actually, $\delta$ is any ordinal of cofinality $\lambda$ less than $\kappa$.) 
The result isn’t quite true as stated. Let $T$ be the set of all $\alpha$-sequences of zeroes for $\alpha<\omega_1$ together with the strings of the form $\sigma^\frown1$ such that $\sigma$ is a countable string of zeroes of limit length. $T$ is a subtree of the complete binary tree of height $\omega_1$ and is an $\omega$-tree of height $\omega_1$. Let $\delta<\omega_1$ be a limit ordinal, and let $t_\delta\in L_\delta(T)$; $t_\delta=\sigma^\frown i$ for the $\delta$-sequence of zeroes and some $i\in\{0,1\}$, and for each $s<_Tt_\delta$, $T^s\cap L_\delta(T)$ contains both $\sigma^\frown 0$ and $\sigma^\frown 1$.
Thus, we want to assume that if $s,t\in L_\alpha(T)$ for some limit ordinal $\alpha$, and 
$$\{u\in T:u<_Ts\}=\{u\in T:u<_Tt\}\;,$$
then $s=t$. (This can always be achieved by adding a new level immediately below each limit level of $T$; Todorčević discusses this shortly after the definition of normal tree.) 
Let $\delta<\kappa$ have cofinality $\lambda$, and fix $t_\delta\in L_\delta(T)$. Let $\langle\alpha_\xi:\xi<\lambda\rangle$ be an increasing $\lambda$-sequence cofinal in $\delta$, and for each $\xi<\lambda$ let $s_\xi$ be the predecessor of $t_\delta$ in $L_{\alpha_\xi}(T)$.
Suppose that 
$$|T^{s_\xi}\cap L_\delta(T)|>1$$
for each $\xi<\lambda$. Then for each $\xi<\lambda$ there is a $u_\xi\in T^{s_\xi}\cap(L_\delta(T)\setminus\{t_\delta\})$. For $u\in L_\delta(T)$ let $X(u)=\{\xi<\lambda:u_\xi=u\}$; clearly $s_\xi\le_T u$ for each $\xi\in X(u)$. Thus, if $|X(u)|=\lambda$, $s_\xi\le_T u$ for each $\xi<\lambda$, so $u$ and $t_\delta$ have the same predecessors in $T$, and therefore $u=t_\delta$. But for all $\xi<\lambda$ we have $u_\xi\ne t_\delta$, so $|X(u)|<\lambda$ for each $u\in L_\delta(T)$. And $|L_\delta(T)|<\lambda$, so 
$$\left|\,\bigcup_{u\in L_\delta(T)}X(u)\,\right|<\lambda\;,$$
which is absurd, since $\bigcup_{u\in L_\delta(T)}X(u)=\lambda$. Thus, $T^{s_\xi}\cap L_\delta(T)=\{t_\delta\}$ for some $\xi<\lambda$.
