König's Lemma/Proof 2

Proof
Let $G$ be an infinite graph which is both connected and locally finite.

From Vertices in Locally Finite Graph, $G$ has an infinite number of vertices each of finite degree.

Let $v$ be a vertex of $G$.

Let $T$ be the set of all finite open paths that start at $v$.

Define the ordering $\preccurlyeq$ on $T$ as follows:

Let $\left({v_0, e_0, v_1, e_1, \ldots, v_m}\right) \preccurlyeq \left({v'_0, e'_0, v'_1, e'_1, \ldots, v'_n}\right)$ :
 * $m \le n$

and:
 * $\forall i < m: \left({v'_i,e'_i,v'_{i+1}}\right) = \left({v_i,e_i,v_{i+1}}\right)$

Observe that $T$ is a tree.

For each vertex $x$, let $E \left({x}\right)$ be the set of all edges incident on $x$.

$E \left({x}\right)$ is finite because $G$ is locally finite by definition.

For every $\left({v_0, e_0, v_1, e_1, \ldots, v_{n+1}}\right)\in T$ we have that $e_n \in E \left({v_n}\right)$.

It follows that $T$ is finitely branching.

By definition of connectedness, to every vertex $x$ there is a walk $w$ from $v$.

Suppose $w = \left({v_0, e_0, v_1, e_1, \ldots, v_n}\right)$ contains a cycle $\left({v_i, e_i, v_{i+1}, \ldots, v_j}\right)$ where $v_i = v_j$.

Then we can delete $\left({e_i, v_{i+1}, \ldots, v_j}\right)$ from $w$ to get a shorter walk from $v$ to $x$.

Repeatedly deleting cycles until none remain, we obtain an open path from $v$ to $x$.

There exist an infinite number of points in $G$, as, by definition of locally finite, $G$ is itself infinite.

We have just demonstrated that for each $x \in G$ there exists an open path from $v$ to $x$.

Hence $T$, the set of finite paths that start at $v$, is infinite.

By Kőnig's tree lemma, $T$ has an infinite branch $B$.

The union $\bigcup B$ is then an infinite path in $G$ that contains $v$.