reserve A, B for non empty set,
  A1, A2, A3 for non empty Subset of A;
reserve X for TopSpace;
reserve X for non empty TopSpace;
reserve X1, X2 for non empty SubSpace of X;
reserve X0, X1, X2, Y1, Y2 for non empty SubSpace of X;
reserve X, Y for non empty TopSpace;
reserve f for Function of X,Y;
reserve X,Y,Z for non empty TopSpace;
reserve f for Function of X,Y,
  g for Function of Y,Z;
reserve X, Y for non empty TopSpace,
  X0 for non empty SubSpace of X;
reserve f for Function of X,Y;
reserve f for Function of X,Y,
  X0 for non empty SubSpace of X;
reserve X, Y for non empty TopSpace,
  X0, X1 for non empty SubSpace of X;
reserve f for Function of X,Y,
  g for Function of X0,Y;
reserve X0, X1, X2 for non empty SubSpace of X;
reserve f for Function of X,Y,
  g for Function of X0,Y;
reserve X for non empty TopSpace,
  H, G for Subset of X;
reserve A for Subset of X;
reserve X0 for non empty SubSpace of X;

theorem
  for X0 being SubSpace of X holds X0 is open SubSpace of X iff modid(X,
  X0) is continuous Function of X,X modified_with_respect_to X0
proof
  let X0 be SubSpace of X;
  reconsider A = the carrier of X0 as Subset of X by TSEP_1:1;
  thus X0 is open SubSpace of X implies modid(X,X0) is continuous Function of
  X,X modified_with_respect_to X0
  proof
    assume X0 is open SubSpace of X;
    then
A1: A is open by TSEP_1:16;
    X modified_with_respect_to X0 = X modified_with_respect_to A & modid(X
    ,X0) = modid(X,A) by Def10,Def11;
    hence thesis by A1,Th101;
  end;
  thus modid(X,X0) is continuous Function of X,X modified_with_respect_to X0
  implies X0 is open SubSpace of X
  proof
    assume
A2: modid(X,X0) is continuous Function of X,X modified_with_respect_to X0;
    X modified_with_respect_to X0 = X modified_with_respect_to A & modid(X
    ,X0) = modid(X,A) by Def10,Def11;
    then A is open by A2,Th101;
    hence thesis by TSEP_1:16;
  end;
end;
