For the definitions used in this question I refer you to this previous question I asked.
Let $\kappa$ be an infinite regular cardinal and let $T$ be a $\kappa$ tree.
Prove that if there is some cardinal $\mu$ s.t $\mu^+ < \kappa$ and $|T_\alpha|\leq\mu$ for all $\alpha$ then $T$ has a cofinal branch.
I don't understand how to use the fact that $\mu^+<\kappa$ (I know that $\mu^+$ is regular as it is a successor cardinal).
We previously proved that we can look at a subtree $T'\subset T$ s.t $T'$ is a $\kappa$-tree and $\forall \alpha<\beta<\kappa$ and $$\forall x\in T'_\alpha,\ \ \exists y\in T'_\beta,\ x \leq_T y$$
Moving to this $T'$ I tried building the branch inductively but I get stuck at the limit steps. My thought is that there is where I use the assumption.
Best Answer
Let $L=\{\alpha<\kappa:\operatorname{cf}\alpha=\mu^+\}$, and for each $\alpha\in L$ let $x_\alpha\in T_\alpha$. For each $x\in T_\alpha\setminus\{x_\alpha\}$ there is a $\varphi_\alpha(x)<\alpha$ such that $x_\alpha$ and $x$ have different predecessors in $T_{\varphi_\alpha(x)}$. $|T_\alpha|\le\mu$, and $\operatorname{cf}\alpha=\mu^+>\mu$, so there is an $\eta(\alpha)<\alpha$ such that $\varphi_\alpha(x)<\eta(\alpha)$ for each $x\in T_\alpha\setminus\{x_\alpha\}$. Thus, the predecessor of $x_\alpha$ in $T_{\eta(\alpha)}$ is different from the predecessors in $T_{\eta(\alpha)}$ of all of the other members of $T_\alpha$.
The function $\eta:L\to\kappa$ is pressing-down (regressive) on the stationary set $L$, so the pressing-down lemma ensures that $\eta$ is constant on a cofinal subset of $L$. That is, there are a cofinal $C\subseteq L$ and a $\beta<\kappa$ such that $\eta(\alpha)=\beta$ for every $\alpha\in C$. $|T_\beta|\le\mu$, so there are a $y\in T_\beta$ and a cofinal $C_0\subseteq C$ such that $y<_Tx_\alpha$ for each $\alpha\in C_0$.
To finish the argument, use the definition of $\eta$ to show that $\{x_\alpha:\alpha\in C_0\}$ is a chain of length $\kappa$ in $T$ and therefore determines a cofinal branch through $T$.