König's Lemma

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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 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.

$\blacksquare$


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 open paths that start at $v$.


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

Let $\tuple {v_0, e_0, v_1, e_1, \ldots, v_m} \preccurlyeq \tuple {v'_0, e'_0, v'_1, e'_1, \ldots, v'_n}$ if and only if:

$m \le n$

and:

$\forall i < m: \tuple {v'_i, e'_i, v'_{i + 1} } = \tuple {v_i, e_i, v_{i + 1} }$


Observe that $T$ is a tree.



For each vertex $x$, let $\map E x$ be the set of all edges incident on $x$.

$\map E x$ is finite because $G$ is locally finite by definition.

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

It follows that $T$ is finitely branching.

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


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

Then we can delete $\tuple {e_i, v_{i + 1}, \ldots, v_j}$ 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 $v$.


$\blacksquare$


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.


Note


If the graph $G$ is assumed to be countably infinite, then the result will hold in pure Zermelo-Fraenkel set theory without any choice.


Source of Name

This entry was named for Dénes Kőnig.


Linguistic Note

In the conventional naming of König's Lemma, the diacritics on the ö in the König of König's Lemma and on the ő of Dénes Kőnig do not match.

While this is irritatingly confusing, unfortunately, it is how it is.