König's Lemma

Lemma

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

Then every vertex lies on an open path of infinite length.

Proof 1

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

From Vertices in Locally Finite Graph, $G$ has an infinite number of vertices $v_1, v_2, \ldots, v_k, \ldots$, each of finite degree.

Let $\mathcal V_k$ be the set of all vertices adjacent to $v_k$.

As $G$ is a connected graph, between $v_k$ and every other vertex of $G$ there exists at least one open path from $v_k$ to every other vertex of $G$.

Take any vertex of $G$ and call it $v_1$.

Let $\mathcal P_1$ be the set of all paths from $v_1$.

Each element of $\mathcal P_1$ must start with an edge joining $v_1$ to some element of $\mathcal V_1$.

There must be some $v_r \in \mathcal V_1$ such that there is an infinite path from $v_r$ in $G$ which does not pass through $v_1$. Otherwise, every path from $v_1$ would be finite, and since there is a path from $v_1$ to each other vertex of the graph, all vertices are contained within one of these finite paths. There are a finite number of paths from $v_1$, so all vertices of $G$ are contained within a finite set of finite sets, contradicting the assumption that $G$ is infinite.

By the axiom of dependent choice, we may pick one of the vertices of $V_1$ such that there exists an infinite path through it that does not include $v_1$, and call this $v_2$.

Each such infinite path must start with one of the elements of $\mathcal V_2$.

Repeating the above argument shows that there is some $v_s \in \mathcal V_2$ such that there is an infinite path from $v_s$ in $G$ which does not pass through $v_2$.

Thus we can construct by induction an infinite path.

The induction hypothesis states that there are infinitely many vertices which can be reached by a path from a particular vertex $v_i$ that does not go through one of a finite set of vertices.

The induction argument is that one of the vertices adjacent to $v_i$ satisfies the induction hypothesis, even when $v_i$ is added to the finite set.

The result of this induction argument is that for all $n$ we can choose a vertex $v_n$ as per the construction.

The set of vertices chosen in the construction is then a path, because each one was chosen to be adjacent to the previous one, and the construction guarantees that the same vertex is never chosen twice.

Proof 2

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 paths that start at $v$.

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

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

$m \le n$

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, $v$ has an infinite branch $B$.

The union $\bigcup B$ is an infinite path.

$\blacksquare$

Note

If the graph $G$ is countably infinite, then the axiom of dependent choice is not required in order to be able to choose an infinite path from a given vertex that does not pass through a given finite set of vertices.

Source of Name

This entry was named for Dénes Kőnig.