theorem
  for X1, X2 being non empty SubSpace of X st X = X1 union X2 holds X1,
  X2 are_separated iff X1 misses X2 & for Y being non empty TopSpace, f being
  Function of X,Y st f|X1 is continuous Function of X1,Y & f|X2 is continuous
  Function of X2,Y holds f is continuous Function of X,Y
proof
  let X1, X2 be non empty SubSpace of X such that
A1: X = X1 union X2;
  thus X1,X2 are_separated implies X1 misses X2 & for Y being non empty
TopSpace, f being Function of X,Y st f|X1 is continuous Function of X1,Y & f|X2
  is continuous Function of X2,Y holds f is continuous Function of X,Y
  proof
    assume
A2: X1,X2 are_separated;
    hence X1 misses X2 by TSEP_1:63;
    X1,X2 are_weakly_separated by A2,TSEP_1:78;
    hence thesis by A1,Th120;
  end;
  thus X1 misses X2 & (for Y being non empty TopSpace, f being Function of X,Y
  st f|X1 is continuous Function of X1,Y & f|X2 is continuous Function of X2,Y
  holds f is continuous Function of X,Y) implies X1,X2 are_separated
  proof
    assume
A3: X1 misses X2;
    assume
A4: for Y being non empty TopSpace, f being Function of X,Y st f|X1 is
continuous Function of X1,Y & f|X2 is continuous Function of X2,Y holds f is
    continuous Function of X,Y;
    for Y being non empty TopSpace, f being Function of X,Y st f|X1 is
continuous Function of X1,Y & f|X2 is continuous Function of X2,Y holds f|(X1
    union X2) is continuous Function of X1 union X2,Y
    proof
      let Y be non empty TopSpace, f be Function of X,Y;
      assume f|X1 is continuous Function of X1,Y & f|X2 is continuous
      Function of X2,Y;
      then f is continuous Function of X,Y by A4;
      hence thesis;
    end;
    hence thesis by A3,Th124;
  end;
end;
