theorem
  X = X1 union X2 & X1,X2 are_separated implies X1 is closed SubSpace of X
proof
  reconsider A2 = the carrier of X2 as Subset of X by Th1;
  reconsider A1 = the carrier of X1 as Subset of X by Th1;
  assume X = X1 union X2;
  then
A1: A1 \/ A2 = [#]X by Def2;
  assume X1,X2 are_separated;
  then A1,A2 are_separated;
  then A1 is closed by A1,CONNSP_1:4;
  hence thesis by Th11;
end;
