Two vertices which don't lie in a common cycle in a tournament How to show that when two vertices don't lie in a common cycle in a tournament then they differ in their out-degrees ? 
I'm considering the adjacent vertices of one the two vertices to start with . 
 A: If two vertices do not lie on a common cycle, then they must be in different strongly connected components of the tournament. We can order the strongly connected components of a tournament as $S_1, \dots, S_k$ in such a way that, whenever $v \in S_i$ and $w \in S_j$ for $i < j$, the edge $vw$ is oriented $v \to w$.
(This is standard, and follows from the fact that there cannot be cycles containing more than one strongly connected component. If that were the case, then the union of those components would itself be strongly connected.)
Once we have that, then under the same assumptions on $v$ and $w$, the out-degree of $v$ is at least $|S_{i+1}| + |S_{i+2}| + \dots + |S_k|$, whereas the out-degree of $w$ is at most $(|S_j|-1) + |S_{j+1}| + \dots + |S_k|$, so $\deg^+ v > \deg^+ w$.
A: Ok, let $u$ and $v$ not belong to a cycle and have equal outdegree. Since the graph $G$ is tournament then without loss of generality there is an arc $(u, v)$. Let's consider vertex $w$ that is a head of arc $(v, w)$. Since there is no cycle containing $u$ and $v$ and $G$ is tournament then there is also arc $(u, w)$. Therefore
$$N^+(v) \subseteq N^+(u) \setminus \{\,v\,\}$$
and
$$\deg^+ v = |N^+(v)| \le |N^+(u)| - 1 = \deg^+ u - 1$$
that implies
$$\deg^+ v < \deg^+ u,$$
where $\deg^+ x = |N^+(x)|$ is outdegree of vertex $x$, $N^+(x) = \{\,y \mid (x, y) \in A(G)\,\}$ and $A(G)$ is set of arcs of $G$.
