I have written proofs for 3 problems and I am hoping someone can point out any flaws in my logic. Let me start with some definitions.
In a n-tournament (complete digraph) with n number of players, there is a "x beats y" or $x \rightarrow y$ relationship which is antisymmetric. Also we will say that the score of A is the number of players A beats.
We say that a player k is an alpha player if, for every player r different from k, either k beats r or else there is some player s such that k beats s and s beats r.
We will define the set P as the set of all players in a tournament and we will assume that every tournament has an alpha.
The problems
Prove that in every tournament if a player is beaten then it is beaten by a alpha player.
Prove that no tournament can have exactly 2 alphas.
Prove that there is no 4-player tournament in which every player is an alpha.
Proofs:
Suppose player $c\in P$ is beaten then the set of players who have beaten c, lets call it D and $D \subseteq P$, is not empty. If a player z has not beaten c then c must have beaten this player,so z belongs to the set of players beaten by c, call it L and $L \subseteq P$. Since we assumed that every tournament has an alpha then the set D must have an alpha named A. if A beats c directly then A must beat all the players in L through c so A beats every player in P. Therefore c is beaten by an alpha A.
We will prove by contradiction. Suppose a tournament has two alphas, w and z. Then w beats z directly or through another player so z is beaten by an alpha. Now z must beat w directly or indirectly so w is beaten by an alpha. However, since the relationship is anti symmetric they cannot beat each other directly. And if w beats z through a player k and z beats w then k beats w through z then k is also an alpha. This contradicts our assumption.
Say we have four players A,B,C and D. In each of the possible pairings a player can score a point so there are 6 points available. The minimum score of a player is 1 otherwise a player with a score of 0 cant be an alpha. If player score is 3 then the rest of the players will have 1 score point each but this also means that none of the other players can beat A so A is the only alpha. A player with 1 point, leaves 5 points left which can be distributed so that another player has 1 point and the remaining two get 2 points each. Now suppose A and D have 2 points each, one of them, say A, must beat a player with 1 point, say C, and a player with 2 points, and the other, say D, must beat both players who have 1 point. Since A only beats C and D then B beats A. In any way, C cant beat D so C is not an alpha. This can be shown if the variables names are interchanged.
Any hints or feedback will be appreciated. Thank you all!
Your proofs for 1 and 3 look fine to me. In your argument for 2, you have not justified the claim "then k is also an alpha." In fact, here is a counterexample showing that it can't be justified.
Consider a tournament with 5 players: w,z,u,v,k. Suppose that:
w beats u,v,k.
z beats w,u.
u beats v,k.
v beats z,k.
k beats z.
Now w and z are alphas; z beats w directly, and w beats z through k, but k is not an alpha.
This is a counterexample to your argument. It's not a counterexample to 2, because z is also an alpha.
(I could give you hints for proving 2, but you didn't ask for any, so I won't spoil your fun.)