reserve GX for TopSpace;
reserve A, B, C for Subset of GX;
reserve TS for TopStruct;
reserve K, K1, L, L1 for Subset of TS;

theorem
  GX is connected iff for A being Subset of GX st A is open closed holds
  A = {}GX or A = [#]GX
proof
A1: now
    assume not GX is connected;
    then consider P,Q being Subset of GX such that
A2: [#]GX = P \/ Q and
A3: P,Q are_separated and
A4: P <> {}GX and
A5: Q <> {}GX;
    reconsider P, Q as Subset of GX;
    Q = [#]GX \ P by A2,A3,Th1,PRE_TOPC:5;
    then
A6: P <> [#]GX by A5,PRE_TOPC:4;
    P is open closed by A2,A3,Th4;
    hence ex P being Subset of GX st P is open closed & P <> {}GX & P <> [#]GX
    by A4,A6;
  end;
  now
    given A1 being Subset of GX such that
A7: A1 is open closed and
A8: A1 <> {}GX and
A9: A1 <> [#]GX;
A10: Cl ([#]GX \ A1) = [#]GX \ A1 by A7,PRE_TOPC:23;
A11: A1 misses A1` by XBOOLE_1:79;
    Cl A1 = A1 by A7,PRE_TOPC:22;
    hence not GX is connected by A8,A9,A10,A11,Th12;
  end;
  hence thesis by A1;
end;
