
theorem Th11:
  for X being open Subset of REAL, a, b being Real st X c= [.a,b.]
  holds not a in X & not b in X
proof
  let X be open Subset of REAL, a,b be Real;
  assume
A1: X c= [.a,b.];
  assume
A2: a in X or b in X;
  per cases by A2;
  suppose
    a in X;
    then consider g be Real such that
A3: 0<g and
A4: ].a-g,a+g.[ c= X by RCOMP_1:19;
    g/2 <> 0 by A3;
    then
A5: a - g/2 < a-0 by A3,XREAL_1:15;
    g > g/2 by A3,XREAL_1:216;
    then
A6: a-g < a-g/2 by XREAL_1:15;
    a+0 < a+g by A3,XREAL_1:8;
    then a-g/2 < a + g by A5,XXREAL_0:2;
    then a-g/2 in {l where l is Real : a-g<l & l<a+g } by A6;
    then
A7: a-g/2 in ].a-g,a+g.[ by RCOMP_1:def 2;
    ].a-g,a+g.[ c= [.a,b.] by A1,A4;
    then a-g/2 in [.a,b.] by A7;
    then a-g/2 in { l where l is Real: a <= l & l <= b }
           by RCOMP_1:def 1;
    then ex l be Real st l = a-g/2 & a <= l & l <= b;
    hence thesis by A5;
  end;
  suppose
    b in X;
    then consider g be Real such that
A8: 0 < g and
A9: ].b-g,b+g.[ c= X by RCOMP_1:19;
    g/2 <> 0 by A8;
    then
A10: b+g/2 > b+0 by A8,XREAL_1:6;
    g > g/2 by A8,XREAL_1:216;
    then
A11: b+g/2 < b+g by XREAL_1:8;
    b-g < b-0 by A8,XREAL_1:15;
    then b-g < b+g/2 by A10,XXREAL_0:2;
    then b+g/2 in {l where l is Real: b-g<l & l<b+g } by A11;
    then
A12: b+g/2 in ].b-g,b+g.[ by RCOMP_1:def 2;
    ].b-g,b+g.[ c= [.a,b.] by A1,A9;
    then b+g/2 in [.a,b.] by A12;
    then b+g/2 in {l where l is Real: a <= l & l <= b}
           by RCOMP_1:def 1;
    then ex l be Real st l = b+g/2 & a<=l & l <=b;
    hence thesis by A10;
  end;
end;
