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,B being Subset of GX st [#]GX = A \/ B &
  A <> {}GX & B <> {}GX & A is open & B is open holds A meets B
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;
A6: Q is open by A2,A3,Th4;
    P is open by A2,A3,Th4;
    hence
    ex A,B being Subset of GX st [#]GX = A \/ B & A <> {}GX & B <> {}GX &
    A is open & B is open & A misses B by A2,A3,A4,A5,A6,Th1;
  end;
   (ex A,B being Subset of GX st [#]GX = A \/ B & A <> {}GX &
      B <> {}GX &  A is open & B is open & A misses B)
    implies not GX is connected by Th3;
  hence thesis by A1;
end;
