I am reading a book title "A walk through Combinatorics" by Miklos Bona (fourth edition). The exercise I am talking is on page 3. Here is the cited passage.
Example 1.3. A chess tournament has $n$ participants, and any two players play one game against each other. Then it is true that in any given point of time, there are two players who have finished the same number of games.
His solution:
Solution. First we could think that the Pigeon-hole Principle will not be applicable here as the number of players ("balls") is $n$, and the number of possibilities for the number of games finished by any one of them ("boxes") is also $n$.
Indeed, a player could finish either no games, or one game, or two games, and so on, up to and including $n-1$ games.
The fact, however, that two players play their games against each other, provides the missing piece of our proof. If there is a player $A$ who has completed all his $n-1$ games, then there cannot be any player who completed zero games because at the very least, everyone has played with $A$. Therefore, the values $0$ and $n-1$ cannot both occur among the numbers of games finished by the players at any one time. So the number of possibilities for these numbers ("boxes") is at most $n-1$ at any given point of time, and the proof follows.
Question
It seems to me, the assumption
If there is a player $A$ who has completed all his $n-1$ games, then there cannot be any player who completed zero games because at the very least, everyone has played with $A$.
makes the proof lose its generality because at any given point of time, there may be none has played with all other players. Could you elaborate his proof? (I am not a native English speaker so I might have misunderstood his sentences.)
Best Answer
To make a comment to answer.
The argument of the author shows that two events: "somebody has played all $n-1$ games" and "somebody played no game" are not consistent because the former player has played with all others. Therefore at any moment only a proper subset of the numbers of played games: $\{0,1,\dots,n-1\}$ (pigeonholes) is available for $n$ players (pigeons).