Summary: Betweenness preserving permutations.
We let
be the identity map of the set of points.
Denition. We say a permutation of the set of points is betweenness preserving if
[s(a; b)] = s((a); (b)) whenever a and b are distinct points.
We let
B
be the set of betweenness preserving permutations of the set of points.
Theorem. B is a subgroup of the group of permutations of the set of points.
Proof. Exercise.
Remark. We will use repeatedly the fact that if 2 B and X is a set of points then
1 [ [X ]] = ( 1 Æ )[X ] = [X ] = X:
Proposition. Suppose A is a set of points and 2 B. Then
[b(A)] = b( [A]):
Proof. Suppose y 2 [b(A)]. Then there is x in b(A) such that y = (x). Since x 2 b(A) there is a in A such
that s(a; x) A. Let b = (a) and note that b 2 [A]. Since 2 B we have [s(a; x)] = s((a); (x)) = s(b; y).
Were it the case that y 2 [A] we would have x = 1 (y) 2 1 [ [A]] = A which is impossible since x 2 b(A)
implies x 62 A. Thus y 2 b( [A]).
Suppose y 2 b( [A]). Then y 62 [A] and there is b in [A] such that s(b; y) [A]. Let x = 1 (y) and
