let X0 be non empty proper SubSpace of X; :: thesis: ( not X0 is open & not X0 is closed )
assume A1: ( X0 is open or X0 is closed ) ; :: thesis: contradiction
reconsider A = the carrier of X0 as non empty Subset of X by Lm1;
set B = A ` ;
A2: A ` <> the carrier of X by TOPS_3:1;
A3: A is proper by Def1;
then A4: A <> the carrier of X ;
now :: thesis: contradictionend;
hence contradiction ; :: thesis: verum