König's Tree Lemma/Proof 2

Proof
We will show that we can choose an infinite sequence of nodes $t_0, t_1, t_2, \ldots$ of $T$ such that:
 * $t_0$ is the root node


 * $t_{n+1}$ is a child of $t_n$

Then the sequence $t_0, t_1, t_2, \ldots$ is such a branch of infinite length.

Let $I$ be the set of all nodes in $T$ that have infinitely many descendants.

Define a relation $\mathcal R$ on $I$ by letting $x \mathrel{\mathcal R} y$ $y$ is a child of $x$.

Since $T$ has infinitely many nodes, its root node is in $I$.

Let $t \in I$.

By the premise, $t$ has a finite number of children.

Suppose for the sake of contradiction that all of these childen had a finite number of descendants.

Then by Finite Union of Finite Sets is Finite, that would mean that $t$ had a finite number of descendants, contradicting the fact that $t$ is in $I$.

So $t$ has at least one child with infinitely many descendants.

That is, $\mathcal R \left({t}\right)$ is non-empty for each $t \in I$.

That is, $\mathcal R$ is a left-total relation.

Thus by Dependent Choice for Finite Sets, there is an infinite sequence $\left\langle{t_n}\right\rangle$ such that:
 * $t_0$ is the root of $T$.
 * For all $n \in \N$, $t_n \mathrel{\mathcal R} t_{n+1}$. That is, $t_{n+1}$ is a child of $t_n$.