König's Tree Lemma/Proof 1

Theorem
Let $T$ be a rooted tree with an infinite number of nodes, each with a finite number of children.

Then $T$ has a branch of infinite length.

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 $t_n$;


 * Each $t_n$ has infinitely many descendants.

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

Take the root node $t_0$.

By definition, it has a finite number of children.

Suppose that all of these childen had a finite number of descendants.

Then that would mean that $t_0$ had a finite number of descendants, and that would mean $T$ was finite.

So $t_0$ has at least one child with infinitely many descendants.

Thus, we may pick $t_1$ as any one of those children.

Now, suppose node $t_k$ has infinitely many descendants.

As $t_k$ has a finite number of children, by the same argument as above, $t_k$ has at least one child with infinitely many descendants.

Thus we may pick $t_{k+1}$ which has infinitely many descendants.

The assertion follows by the Axiom of Dependent Choice.