reserve X for non empty TopSpace,
  D for Subset of X;
reserve D for non empty set,
  d0 for Element of D;

theorem
  STS(D,d0) = 1TopSp(D) iff D = {d0}
proof
  set G = {P where P is Subset of D : d0 in P & P <> D};
  thus STS(D,d0) = 1TopSp(D) implies D = {d0}
  proof
    now
      let x be object;
      assume x in G;
      then ex Q being Subset of D st x = Q & d0 in Q & Q <> D;
      hence x in bool D;
    end;
    then
A1: G c= bool D by TARSKI:def 3;
    reconsider P = {d0} as Subset of D;
    assume
A2: STS(D,d0) = 1TopSp(D);
A3: d0 in P by TARSKI:def 1;
    assume D <> {d0};
    then P in G by A3;
    hence contradiction by A2,A1,XBOOLE_1:38;
  end;
  assume D = {d0};
  then G = {} by Lm2;
  hence thesis;
end;
