On a Theorem of Richardson

Given a set 5 with an irreflexive relation >, define the operators dom and max as follows. For any subset T of S, dom T is the set of all x in 5 such that for some t\nT,t>x; max T= T — dom T. A solution of (S, >) is a subset T of 5 satisfying T = S — dom T. A solution of an w-person game in the sense of [3] is precisely a solution, as just defined, of a certain system (S, >) associated with the game. Von Neumann and Morgenstern investigated various combinatorial criteria for solutions in [3, pp. 587-602]. Richardson has given further criteria; in particular, in [4] he established that if every unoriented cycle in S is even then S has a solution.2 An unoriented cycle is a subset (xi, • • • , x„) of 5 such that for i = 1, • ■ • , n — 1, either Xj>x,-+i or x,-+i>x,-, and also xi>x„ or x„>Xi. This note establishes a reduction for general (S, >) which yields a simplified proof of Richardson's theorem. The reduction is an elaboration of one which has been studied by Gillies [l]. Obviously any solution T must contain max S, since every point is in Tor in dom T. Hence no solution can contain any point of dom max S, since no point of T is in dom T. Inductively let Si = S, Ma = max Sa, Da = dom Ma, Sa+i*=Sa — Ma — Da; and for limit ordinals a, let 5a = H [Sp] B) then TC\R is a solution of (R, >), and (2) if TC^R is a solution of (R, >) then (Tr\R)\JM is a solution of (S, >). This establishes a one-to-one correspondence between solutions of R and solutions of S. By construction of M, max R is empty. Therefore solving 5 reduces to solving R, where max R is empty. In Richardson's case this is done by a theorem of Konig [2]: if all unoriented cycles in R are even, then R can be "colored in two colors," i.e. partitioned into two sets, P, Q, such that x>y never holds if x and y are both in P or both in Q.