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

theorem
  D \ {d0} is non empty implies for A being Subset of STS(D,d0) holds (A
  = D \ {d0} implies A is open & A is dense) & (A <> D & A is open & A is dense
  implies A = D \ {d0})
proof
  assume
A1: D \ {d0} is non empty;
  reconsider P = {d0} as Subset of STS(D,d0);
  let A be Subset of STS(D,d0);
  set Z = A`;
  thus A = D \ {d0} implies A is open & A is dense
  proof
    assume
A2: A = D \ {d0};
    hence A is open by Th22;
    [#]STS(D,d0) \ ([#]STS(D,d0) \ P) = Z by A2;
    then P = Z by PRE_TOPC:3;
    then Z is boundary by A1,Th21;
    hence thesis by TOPS_3:18;
  end;
  thus A <> D & A is open & A is dense implies A = D \ {d0}
  proof
    assume A <> D;
    then
A3: Z <> {}STS(D,d0) by TOPS_3:2;
    assume
A4: A is open;
    assume A is dense;
    then Z is boundary by TOPS_3:18;
    then Z = {d0} by A1,A3,A4,Th21;
    then A = P`;
    hence thesis;
  end;
end;
