Let $(V,A)$ be a tournament. A subset of vertices $V'\subseteq V$ is stable if there exists no $v\in V\setminus V'$ such that $V'\cup${$v$} contains an inclusion-maximal transitive subtournament with source $v$, i.e., for every transitive subtournament $T\subseteq V'\cup${$v$} with $v\in T$ and $(v,x)\in A$ for all $x\in T\setminus${$v$} there is a $w\in V'$ such that $(w, x)\in A$ for all $x\in T$.

Is it true that no tournament contains two disjoint stable sets?

The statement would imply that every tournament contains a unique minimal stable set, which would have several appealing consequences in the social sciences. The statement is a weak version of a conjecture by Schwartz (see arxiv.org/pdf/0803.2138v3 and the references therein). Computer experiments have shown that there exists no counter-example with less than 13 vertices.

Let $(V,A)$ be a tournament. A subset of vertices $V'\subseteq V$ is stable if there exists no $v\in V\setminus V'$ such that $V'\cup${$v$} contains an inclusion-maximal transitive subtournament with source $v$.

(In other words, $V'$ is stable if for every transitive subtournament $T\subseteq V'\cup${$v$} with $v\in T$ and $(v,x)\in A$ for all $x\in T\setminus${$v$}, there is a $w\in V'$ such that $(w, x)\in A$ for all $x\in T$.)

Is it true that no tournament contains two disjoint stable sets?

The statement would imply that every tournament contains a unique minimal stable set, which would have several appealing consequences in the social sciences. The statement is a weak version of a conjecture by Schwartz (see this paper and the references therein). Computer analysis has shown that there exists no counter-example with less than 13 vertices.