Let $(V,A)$ be a [tournament](http://en.wikipedia.org/wiki/Tournament_%28graph_theory%29). 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](http://arxiv.org/pdf/0803.2138v3) and the references therein). Computer analysis has shown that there exists no counter-example with less than 13 vertices. 



Disjoint stable sets in tournaments
The statement is false. The existence of a counter-example using a probabilistic argument was shown by Chudnovsky, Kim, Liu, Norin, Scott, Seymour, and Thomasse.


