theorem Th15:
  for A be non empty Subset of X holds rng chi(A,A) = {1}
proof
  let A be non empty Subset of X;
A1: chi(A,A)|A is constant by Th14;
  dom chi(A,A) = A by FUNCT_3:def 3; then
A2: A = A /\ dom chi(A,A);
A3: dom chi(A,A) = A by FUNCT_3:def 3;
  ex x being Element of X st x in dom chi(A,A) & chi(A,A).x=1
  proof
    consider x being Element of X such that
A4: x in dom chi(A,A) by A3,SUBSET_1:4;
    take x;
    thus thesis by A4,FUNCT_3:def 3;
  end;
  then
A5: 1 in rng chi(A,A) by FUNCT_1:def 3;
  A meets dom chi(A,A) by A2;
  then ex y being Element of REAL st rng (chi(A,A)|A)={y} by A1,PARTFUN2:37;
  hence thesis by A5,TARSKI:def 1;
end;
