
theorem Th20:
  for T being non empty TopSpace st T is normal holds for A,B
being open Subset of T st A <> {} & Cl(A) c=B holds ex C being Subset of T st C
  <> {} & C is open & Cl(A) c= C & Cl(C) c= B
proof
  let T be non empty TopSpace;
  assume
A1: T is normal;
  let A,B be open Subset of T;
  assume that
A2: A <> {} and
A3: Cl(A) c=B;
  now
    per cases;
    case
A4:   B <> [#](T);
      reconsider W = Cl(A), V = [#](T) \ B as Subset of T;
A5:   W <> {} & V <> {}
      proof
A6:     [#](T) \ B <> {}
        proof
          assume [#](T) \ B = {};
          then B = [#](T) \ {} by PRE_TOPC:3;
          hence thesis by A4;
        end;
        A c= Cl(A) by PRE_TOPC:18;
        hence thesis by A2,A6;
      end;
A7:   W misses V
      proof
        assume W meets V;
        then consider x being object such that
A8:     x in W /\ V by XBOOLE_0:4;
        x in Cl(A) & x in [#](T) \ B by A8,XBOOLE_0:def 4;
        hence thesis by A3,XBOOLE_0:def 5;
      end;
      B = [#](T) \ V by PRE_TOPC:3;
      then V is closed;
      then consider C,Q being Subset of T such that
A9:   C is open and
A10:  Q is open and
A11:  W c= C and
A12:  V c= Q and
A13:  C misses Q by A1,A7;
      take C;
      C <> {} & Cl(A) c= C & Cl(C) c= B
      proof
        consider Q0,C0 being Subset of [#](T) such that
A14:    Q0 = Q & C0 = C;
        C0 c= Q0` by A13,A14,SUBSET_1:23;
        then Cl(C) c= Q` by A10,A14,TOPS_1:5;
        then Cl(C) misses Q by SUBSET_1:23;
        then
A15:    V misses Cl(C) by A12,XBOOLE_1:63;
        B`` = B;
        hence thesis by A5,A11,A15,SUBSET_1:23;
      end;
      hence thesis by A9;
    end;
    case
A16:  B = [#](T);
      consider C being Subset of T such that
A17:  C = [#](T);
      take C;
      Cl(C) c= B by A16;
      hence thesis by A17;
    end;
  end;
  hence thesis;
end;
