= {x E Rd:S,[x -( a v o ) ] > 0, i = 1, ..., dk ; 1 f i [ x -( a v o ) ] i= 1 which gives at once the result stated above. 3. PROPERTIES OF d-CHOQUET SIMPLICES WITHOUT LINEALITY RESTRICTION PROPoSITlON 3.1. If S is a d-Choquet simplex and i f S n (as + x) = BS + y (a, B > 0), one can write S n (aS + x ) = yS + z or SO n (US' + x ) = ySo + z only i j z E y + T(S).Since [S n (crS + x)]' = So n (aso + x ) = BSo + y = ySo + z, it suffices to prove the property for an open Choquet simplex and this is plain from THEOREM 2.1. common point. Let F,, F, be two proper faces of S, x 1 E iFl and x , E i F 2 . Consider S n (S + x 2 -x , ) . Since x , belongs to both S and S + x2 -x , , one can find a 2 0 and x E Rd such that S n (S + x , -x l ) = US + x . According to THEOREM 2.1, the point x , belongs to the closure of an open set K, included in S, which is the intersection of an open ball with center x1 with m open half spaces [ m 5 dk + 1, where k = dim r(S)] whose bounding hyperplanes contain x and are independent. Therefore K + X , -X I c S + ~2 -X I and So n ( K + x2 -x , ) c US + x.The faces F, and F, have a common point or the set on the left-hand side is not empty; hence, a > 0. Let us first give the proof for the case in which S can be translated in such a way that So is of type 2.1.a; assume that the indexing is such that K = B(x,) n { x : J ( x ) > 0, i = 1, ..., m}.
One hasSo n ( K + x, -x l ) = So n B(x2) n { x : f i x ) >f;(x,), i = 1, ..., m) = B(x,) n { x : J ( x ) >J(x,), i = 1, ..., m;f;(x) > 0, i = 1, ..., dk ; d -k and the second set in this last intersection is nonvoid (and open) if in this case, indeed, a segment ] x 2 : a [ is included in this set and B(x,) n ]x2: a[ # (21, hence the conclusion; if 1 J(x2) = 1, d -k i = 1 S n ( S + xtx l ) c S n (S+ x ,xi) c i = 1 and, since this last set is a hyperplane, a = 0 and S n ( S + x ,xl) = { x , } .