
theorem Th50:
  for S,T being complete
  Scott non empty reflexive transitive antisymmetric TopRelStr
  st the RelStr of S = the RelStr of T
  for F being Subset of S, G being Subset of T st F = G
  holds F is open implies G is open
proof
  let S,T be complete
  Scott non empty reflexive transitive antisymmetric TopRelStr such that
A1: the RelStr of S = the RelStr of T;
  let F be Subset of S, G be Subset of T;
  assume that
A2: F = G and
A3: F is open;
  F is upper inaccessible by A3,WAYBEL11:def 4;
  then G is upper inaccessible by A1,A2,Th47,WAYBEL_0:25;
  hence thesis by WAYBEL11:def 4;
end;
