reserve e,u for set;
reserve X, Y for non empty TopSpace;

theorem Th11:
  for X, Y being TopSpace for A,B being Subset of [:X,Y:] st A c=
  B holds Base-Appr A c= Base-Appr B
proof
  let X, Y be TopSpace, A,B be Subset of [:X,Y:] such that
A1: A c= B;
  let e be object;
  assume e in Base-Appr A;
  then consider X1 being Subset of X, Y1 being Subset of Y such that
A2: e = [:X1,Y1:] and
A3: [:X1,Y1:] c= A and
A4: X1 is open & Y1 is open;
  [:X1,Y1:] c= B by A1,A3;
  hence thesis by A2,A4;
end;
