# existence of a branch of length $\kappa$ in a tree of height $\kappa$ whose levels are finite

Let $$(T,\leq)$$ be a tree of height $$\kappa$$ ($$\kappa$$ infinite cardinal), all levels of which are finite. Prove that $$T$$ has a branch of length $$\kappa$$. A hint is given: Let $$U_\alpha=\{x\in T_\alpha\mid |\{y\in T\mid x\leq y\}|=\kappa\}$$. Prove that $$\{|U_\alpha|\mid\alpha<\kappa\}\subseteq\mathbb{N}$$ is bounded. Let $$m$$ be its maximum ($$m\geq 1$$). Show $$T$$ has exactly $$m$$ branches of length $$\kappa$$. (Exercise 3.3, Ch 12 from Introduction to Set Theory by Hrbacek and Jech.)

First of all, can we think of $$T$$ as some subtree of $$\kappa^{<\omega}$$ or relate it to some $$\theta^{<\lambda}$$? I'm not really sure how to form the correspondence.

I don't see what goes wrong if $$\{|U_\alpha|\mid\alpha<\kappa\}\subseteq\mathbb{N}$$ is unbounded. We get a countable set of distinct nodes, with each node having $$\kappa$$-many nodes above it; does that give us a subset of $$T$$ with size greater than $$\kappa$$?

I'm new to this and I'm quite confused. Thanks in advance for any help.

The hint actually does make a misleading oversight. In fact, it not generally true that $$|U_\alpha|$$ is bounded. First off, there is clearly a counterexample with height $$\omega$$: simply take a binary tree of that height. But it fails for some larger heights as well. For instance, we can make a tree of height $$\aleph_\omega$$ that starts with one branch, splits into two branches at level $$\aleph_1,$$ then the two branches each split into two at $$\aleph_2,$$ and so on. Then, it's clear that $$|U_\alpha|$$ is unbounded, but the levels are all finite. A similar counterexample will work whenever the height $$\kappa$$ has cofinality $$\omega.$$
But notice these counterexamples still have a branch of length $$\kappa.$$ In fact, we can see for $$cf(\kappa) =\omega$$ that we will always have a branch of length $$\kappa,$$ by just repeating the argument for Konig's lemma on a cofinal sequence of levels. So the theorem is still true for these $$\kappa,$$ even if the hint is not.
On the other hand, if $$cf(\kappa) > \omega,$$ then $$|U_\alpha|$$ is bounded like the hint suggests. Otherwise let $$\alpha_n$$ be the smallest level such that $$|U_\alpha|\ge n.$$ Then $$\alpha_n$$ will be a cofinal sequence in $$\kappa$$, contradicting $$cf(\kappa) > \omega.$$