theorem Th22:
  X is open bounded_below implies not lower_bound X in X
proof
  assume that
A1: X is open and
A2: X is bounded_below;
  assume lower_bound X in X;
  then consider N being Neighbourhood of lower_bound X such that
A3: N c= X by A1,Th18;
  consider t such that
A4: t>0 and
A5: N = ].lower_bound X-t,lower_bound X+t.[ by Def6;
A6: lower_bound X - t/2 < lower_bound X by A4,XREAL_1:44,215;
A7: lower_bound X - t/2 - t/2 < lower_bound X - t/2 by A4,XREAL_1:44,215;
  lower_bound X < lower_bound X + t by A4,XREAL_1:29;
  then lower_bound X - t/2 < lower_bound X + t by A6,XXREAL_0:2;
  then lower_bound X - t/2 in {s: lower_bound X-t<s & s<
  lower_bound X+t} by A7;
  hence contradiction by A2,A3,A5,A6,SEQ_4:def 2;
end;
