# König's Lemma/Countable

## Theorem

Let $G = \struct {V, E}$ be a graph with countably infinitely many vertices which is connected and is locally finite.

Then every vertex of $G$ lies on a path of infinite length.

## Proof

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

Recursively define a sequence $\sequence {S_n}$:

Let $S_0 = \set r$.

Let $S_{n + 1}$ be the set of all vertices that are adjacent to some element of $S_n$ but not adjacent to any element of $S_k$ for $k < n$.

That is, $S_n$ is the set of vertices whose shortest path(s) to $r$ have $n$ edges.

Since $G$ is connected:

$V = \displaystyle \bigcup_{n \mathop \in \N} S_n$

Note that $\set {S_n: n \in \N}$ is pairwise disjoint.

Define a relation $\RR$ on $V$ by letting $p \mathrel \RR q$ if and only if:

$p$ is adjacent to $q$

and:

there exists an $n \in \N$ such that $p \in S_n$ and $q \in S_{n+1}$.

Let $\RR^+$ be the transitive closure of $\RR$.

Let $V'$ be the set of all $v \in V$ such that $\map {\RR^+} v$ is infinite.

It will be demonstrated that $r \in V'$:

Let $v \in V$.

Since $G$ is connected, there is a path from $r$ to $v$.

By the Well-Ordering Principle, there is a path $P$ from $r$ to $v$ of minimal length.

Then the vertices along $P$ will lie in successive sets $S_k$.

Thus $v \in \map {\RR^+} r$.

As this holds for all $v \in V$, and $V$ is infinite, $\map {\RR^+} r$ is infinite.

Therefore $r \in V'$.

Let $\RR'$ be the restriction of $\RR$ to $V'$.

It is to be shown that $\RR'$ is a left-total relation.

Let $v \in V'$.

Then $\map {\RR^+} v$ is infinite.

Since $\map \RR v$ is finite, Finite Union of Finite Sets is Finite shows that $\map \RR v$ must have an element $u$ such that $\map {\RR^+} u$ is infinite.

So $u \in V'$ and $v \mathrel {\RR'} u$.

We have that $V$ is assumed to be countably infinite.

Without loss of generality, let $V = \N$.

Recursively define a sequence $\sequence {v_k}$ in $V'$ thus:

$v_0 = r$
$v_{n + 1} = \map \min {\map {\RR'} {v_n} }$

Then $\sequence {v_k}$ is the sequence of vertices along an infinite path starting at $r$.

$\blacksquare$