# Eccentricity in infinite tournaments

Definitions. A tournament is an oriented complete graph, that is, it's what you get by taking a (finite or infinite) complete graph and assigning a unique direction to each edge. If $$T$$ is a tournament, $$V(T)$$ is its set of vertices and $$E(T)$$ its set of (directed) edges. If $$u,v$$ are vertices of a tournament (or digraph for that matter), the distance $$d(u,v)$$ is the minimum length of a (directed) path from $$u$$ to $$v$$, so $$d(u,u)=0$$; we put $$d(u,v)=\infty$$ if there is no path from $$u$$ to $$v$$. The eccentricity of a vertex $$u$$ of a tournament $$T$$ is $$e(u)=\sup\{d(u,v):v\in V(T)\}$$.

Motivation. It is a common exercise in graph theory to show that every finite tournament has a vertex of eccentricity at most two. (In fact, an arbitrary tournament $$T$$ contains either a vertex $$u$$ with $$e(u)\le2$$ or else an infinite sequence of vertices $$v_1,v_2,\dots,v_n,\dots$$ such that $$v_jv_i\in E(T)$$ whenever $$i\lt j$$.) Thus the following problem is only of interest for infinite tournaments.

Problem 1. Show that, if a tournament has a vertex of finite eccentricity, then it has a vertex of eccentricity at most three.

Problem 2. Find a tournament in which the eccentricity of every vertex is exactly three.

Context. These problems occurred to me after considering this question about finite tournaments. They are rather easy but I found them amusing, maybe you will too.

• Looks like no one is interested in answering these silly questions, so I have to answer them myself. :-( – bof Mar 20 at 8:09

Solution. Let $$u$$ be a vertex with $$e(u)=n\lt\infty$$, and suppose $$n\gt3$$. Choose a vertex $$w$$ with $$d(u,w)=n$$. Now consider any vertex $$v$$. If $$d(u,v)\le n-2$$ then $$d(w,v)=1$$; if $$d(u,v)=n-1$$ then $$d(w,v)\le2$$; and if $$d(u,v)=n$$ then $$d(w,v)\le2$$. Hence $$e(u)\le3$$.
By the way, for a simple example of a strongly connected tournament with no vertex of finite eccentricity, take $$V(T)=\mathbb N$$ and $$E(T)=\{(u,v):v=u+1\text{ or }v\lt u-1\}$$.
Solution. Let $$V(T)=\mathbb N$$ and $$E(T)=\{(u,v):v\equiv u+1\pmod3\}\cup\{(u,v): v\equiv u\pmod3,\ v\lt u\}$$. Then $$e(u)=d(u,u+3)=3$$ for every vertex $$u$$.