König's Lemma/Proof 1

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Lemma

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


Then every vertex lies on a path of infinite length.


Proof

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.

$\blacksquare$


Axiom:Axiom of Countable Choice for Finite Sets

This theorem depends on Axiom:Axiom of Countable Choice for Finite Sets.

Although not as strong as the Axiom of Choice, Axiom:Axiom of Countable Choice for Finite Sets is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.


Source of Name

This entry was named for Dénes Kőnig.