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
  A is closed & B is closed & A \/ B is connected & A /\ B is connected
  implies A is connected & B is connected
proof
  assume that
A1: A is closed and
A2: B is closed;
  set AB = A \/ B;
A3: A \/ B = [#](GX|(A \/ B)) by PRE_TOPC:def 5;
  then reconsider B1 = B as Subset of GX|AB by XBOOLE_1:7;
  reconsider A1 = A as Subset of GX|AB by A3,XBOOLE_1:7;
A4: [#](GX|(A \/ B)) \ (A1 /\ B1) = (A1 \ B1) \/ (B1 \ A1) by A3,XBOOLE_1:55;
  B /\ ([#](GX|(A \/ B))) = B by A3,XBOOLE_1:7,28;
  then
A5: B1 is closed by A2,PRE_TOPC:13;
  A /\ ([#](GX|(A \/ B))) = A by A3,XBOOLE_1:7,28;
  then A1 is closed by A1,PRE_TOPC:13;
  then
A6: A1 \ B1,B1 \ A1 are_separated by A5,Th9;
  assume that
A7: A \/ B is connected and
A8: A /\ B is connected;
A9: GX|(A \/ B) is connected by A7;
A10: A1 /\ B1 is connected by A8,Th23;
  (A1 /\ B1) \/ (B1 \ A1) = B1 by XBOOLE_1:51;
  then
A11: B1 is connected by A9,A4,A6,A10,Th20;
  (A1 /\ B1) \/ (A1 \ B1) = A1 by XBOOLE_1:51;
  then A1 is connected by A9,A4,A6,A10,Th20;
  hence thesis by A11,Th23;
end;
