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Thm 3.2 (Landau).  If vertex $u$ has maximum score in a tournament $(V, A)$, then for all $v \in V$, 
either $(u, v) \in A$ or there exists $w \in V$ such that $(u, w) \in A$ and $(w, v) \in A$. 
\v
Or in other works if $u$ has maximum score than for every other player
 $v$, either $u$ 
beats $v$ or $u$ beats another player, $w$, who beats $v$.


Proof:  Take $v \in V$.  

Case 1:  If $(u, v) \in A$, then the conclusion holds.

Case 2:  Suppose  $(u, v) \not\in A$.  Then we need to find $w \in V$ such that $(u, w) \in A$ and 
$(w, v) \in A$.

Suppose $s(u) = k$ and $\{w_1, ..., w_k\}$ is the set of all vertices such that $(u, w_i) \in A$.

We need a $j$ such that $(w_j, v) \in A$.


Proof by contradiction.  Assume there does not exist a $j$ such that $(w_j, v) \in A$.

Hence for all $j$, $(w_j, v) \not\in A$.  

Since $(V, A)$ is a tournament, $(w_j, v) \not\in A$ implies $(v, w_j) \in A$.
[Thus $s(v) \geq k$].

Since $(V, A)$ is a tournament, $(u, v) \not\in A$ implies  $(v, u) \in A$.  

Thus $s(v) \geq k+1 > s(u)$.

But this contradicts the hypothesis that  $u$ has maximum score.  
\v\v
\vfill
Hence our assumption is 
wrong and there does exist a $j$ 
such that $(w_j, v) \in A$.

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