reserve X for non empty TopSpace,
  D for Subset of X;

theorem Th19:
  for D being non empty set holds ADTS(D) = 1TopSp(D) iff ex d0
  being Element of D st D = {d0}
proof
  let D be non empty set;
  thus ADTS(D) = 1TopSp(D) implies ex d0 being Element of D st D = {d0}
  proof
    set d0 = the Element of D;
    assume
A1: ADTS(D) = 1TopSp(D);
    take d0;
    thus thesis by A1,TARSKI:def 2;
  end;
  given d0 being Element of D such that
A2: D = {d0};
  thus thesis by A2,ZFMISC_1:24;
end;
