Consider a tournament involving $2n$ teams, with the following properties: every day of the tournament involves $n$ matches (with no teams repeated in a given day); each team plays every other team exactly once during the tournament, for a total of $2n − 1$ days; and finally, there are no ties. Show that it is possible to choose one winning team from each day of the tournament, with no team chosen more than once.
As a first step I said let $X$ be the days and $Y$ be the winner, I need to use Hall Marriage theorem but I don't know how, can anyone help?
Okay, I figured it out. I won't give you the proof I'm turning in but I'll tell you the general idea.
If we can show that the conditions for Hall's Marriage Theorem are met, then we are done. For the conditions to not be met, there has to be a set of days $X'$ such that there are more days than the number of teams that won on those days. $|X'| > |\Gamma (X')|$. Since there are no ties in this tournament, each element of $X'$ is connected to $n$ winners in $Y$. This means we need to see if it's possible for only $n$ teams to win over the course of $n+1$ days. ($n+1$ is the minimum requirement in order satisfy the above inequality and break Hall's marriage theorem)
However, since the total number of teams is $2n$ and the tournament is a round robin this is impossible. It would only take one of the $n$ losers $n$ days to be defeated by all of the $n$ winners. On the $n+1$ day, this loser will have to fight another loser, and at least one of them has to win since there are no ties. This means it's impossible for there to be only $n$ winners over the course of $n+1$ days.