Talk:König's Lemma

Cardinality
The note at the bottom says that if the graph is countable, then DC is not needed. If I'm not mistaken, however, every locally finite, connected graph must be countable, and that DC is never needed:

Choose an arbitrary vertex $q$.

Let $S_0 = \{ q \}$.

Let $S_{n+1}$ be the set of all vertices that are adjacent to an 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 $q$ have $n$ edges.

Since the graph is connected, every vertex of the graph is in $\displaystyle \bigcup_{n \in \N} S_n$.

But by the Axiom of Countable Choice for Finite Sets, $\displaystyle \bigcup_{n \in \N} S_n$ is countable, so there are countably many vertices in the graph. No further choice is required to prove that there are countably many edges.

I believe it should be possible to prove this theorem using Dependent Choice for Finite Sets. --Dfeuer (talk) 18:14, 27 May 2013 (UTC)


 * Indeed, the set of minimal-length finite paths that start at $q$ (that is, paths that start at $q$ and end at some vertex $v$ with the property that there is no shorter path from $q$ to $v$) can be arranged into a tree in a natural way, and the tree lemma should then apply. --Dfeuer (talk) 18:23, 27 May 2013 (UTC)


 * More intuitively still, finite sequences $(x_n)$ such that $x_n \in S_n$ and $x_{n+1}$ is adjacent to $x_n$ for each $n$. I guess I'll have to write this up. --Dfeuer (talk) 18:28, 27 May 2013 (UTC)