$\kappa$-tree and cofinal branch Definitions: Let $\left<T, \leq_T\right>$ be a tree,

*

*(level of an element) The level of $x\in T$ is defined to be $Lev_T(x)= \mathrm{otp}\{y\in T\ |\ y<_Tx\}$.


*Let $T_\alpha = \{x\in T\ |\ Lev_T(x)=\alpha\}$, the height of $T$ is $\min\{\alpha\ |\ T_\alpha =\emptyset\}$


*($\kappa$-tree) Let $\kappa$ be a regular cardinal $T$ is called a $\kappa$-tree if the height of $T$ is $\kappa$ and $\forall\alpha<\kappa,\ |T_\alpha|<\kappa$.


*(cofinal branch) A set $b\subset T$ is a cofinal branch if for every $x,y\in b,\ x\leq_T y\mathrm{\ or\ }y\leq_T x$ and for all $\alpha<\kappa,\ b\cap T_\alpha\neq\emptyset$
Question: Let $T$ be a $\kappa$-tree and assume that for every $A\subseteq T$, $|A|=\kappa$ there exists some $x,y\in A$ s.t $x<_Ty$. Prove that either $\exists z\in T$ s.t
$$ b=\left\{y\in T\ \big|\ z\leq_Ty\vee y \leq_T z\right\}$$
is a cofinal branch, or there is no cofinal branch.
My Proof: Assume that there is some cofinal branch $b\subset T$ and that there is no $z\in T$ s.t
$$ b_z =\left\{y\in T\ \big|\ z\leq_Ty\vee y \leq_T z\right\}$$
is a cofinal branch. We define a set in the following way, first consider $x_\emptyset = \min b$ then $b_{x_\emptyset}$ is not a cofinal branch thus there exists $\emptyset < \alpha_\emptyset < \kappa$ s.t $\exists y\in b_{x_\emptyset}\cap T_{\alpha_\emptyset}$ and $y\notin b$ choose one and call it $y_{\alpha_\emptyset}$.
Assume we have defined $y_{\alpha_\beta}$ let $z\in b\cap T_{\alpha_\beta}$ then $b_z$ is not a cofinal branch and by the same reasoning there exists $\alpha_\beta<\alpha_{\beta+1}<\kappa$ s.t $\exists y \in b_z\cap T_{\alpha_{\beta+1}}$ s.t $y\notin b$ choose one and call it $y_{\alpha_{\beta+1}}$.
Assume we have defined $y_{\alpha_\gamma}$ for $\gamma<\beta$ define $\alpha = \sup_{\gamma<\beta}\alpha_\gamma$ then from regularity $\alpha <\kappa$ and we can define using the above procedure when taking $z\in T_\alpha\cap b$ the $\alpha_\beta$.
Consider the set $A=\{\alpha_\beta\}$, $\sup\alpha_\beta = \kappa$ then again from regularity $|A|=\kappa$ and $A$ contradicts the assumption because each element is on a different branch hence not in relation to any other element.
My Problem: The definition of the set $A$ does not seem right to me I think I'm doing here something I cant (maybe the use of regularity of $\kappa$?)
Any help would be appreciated. Thanks in advance.
 A: It’s written up a bit clumsily, but your basic idea is fine. The only thing missing is a clearer explanation of why $A$ is an antichain (a subset of $T$ in which no two elements are related by the tree order). Here is a slightly neater way to carry out a very similar idea.
Say that a node of $T$ splits if it has more then one immediate successor in $T$.
Suppose that $b$ is a cofinal branch, let $S=\{x\in b:x\text{ splits}\}$, and suppose that $|S|=\kappa$. The definition of $S$ ensures that each $x\in S$ has an immediate successor $y_x\in T\setminus b$; let $A=\{y_x:x\in S\}$. $|A|=\kappa$, so there are $x,z\in S$ such that $y_x<_Ty_z$. Let $\alpha=\operatorname{Lev}_T(x)$, and let $u$ be the unique member of $b$ in $T_{\alpha+1}$. Then $u\le_Tz<_Ty_z$, so $u$ and $y_x$ are distinct predecessors of $y_z$ in $T_{\alpha+1}$, which is impossible. Thus $|S|<\kappa$. The regularity of $\kappa$ now ensures that there is an $x\in b$ such that $y<_Tx$ for each $y\in S$ and hence that $b=b_x$, as desired.
