Prove that T is transitive if and only if the score of the $k^{th}$ vertex ($s_k$) = '$k-1$' for $k =1,2,ldots. n $

We want to show that a transitive tournament with $n$ vertices has the score sequence of $(0,1,2,…,n−1)$. Now suppose that it holds for all the transitive tournaments.

The proof is by induction on the number of vertices of $T$, the result being trivial for transitive tournaments with at most two vertices.

Let T be a transitive tournament with $n+1$ vertices. We know that a tournament is an orientation of a complete graph. In addition, transitivity implies that $T$ is acyclic. So $T$ has a vertex $V$ with out-degree of $n$. If we delete $V$ and it's incident edges, then the graph that remains is a transitive tournament with $n$ vertices and it has a vertex with out-degree of $n-1$. By our induction hypothesis, this graph has the score sequence of $(0,1,2,…,n−1)$. Appending the score of V to it gives the score sequence of $T$ which is $(0,1,2,...,n)$.

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Now we want to show that if score sequence $(S_1,S_2,S_3,...,S_n)$ where $S_k=k-1$ for all $k=1,2,3,…,n$ then $T$ is transitive.

Remove leaf vertex $V$ and it's incident edges. Removing $V$ will decrease all of the remaining vertices in-degrees by one so there will be another leaf vertex. Keep removing leaf vertices, we will eventually remove all vertices: The tournament is acyclic therefore transitive.