Prove that every nontrivial tournament has at least one serf. > **Serf definition** : A vertex $z$ in a nontrivial tournament is called a **serf** if for every vertex $x$ distinct from $z$, either $x$ adjacent to $z$ or $x$ is adjacent to a vertex that is adjacent to $z$. Prove that every nontrivial tournament has at least one serf. I'm not sure if I understand this correctly, but a tournament is an oriented complete graph, and in a complete graph of order $n$ every vertex un-oriented adjacent to $n-1$ other vertex, so the only way for a vertex to not be a serf is that vertex has to be a source. Assume the contrary that there exists a tournament that doesn't have any serf, then that tournament has every vertex is a source, which is impossible, thus every tournament must have at least one serf. is my argument acceptable?

HINT: I’ll phrase this in terms of a real round-robin tournament rather than a graph. Let $p$ be a player with the smallest possible score, and let $B$ be the set of players beaten by $p$. Use the fact that every $b\in B$ has a score that is at least as big as $p$’s to show that every $b\in B$ beats someone who beats $p$. Thus, a player with minimal score must be a serf.