let p, q be real number ; ].p,q.[ is open
defpred S1[ Real] means ( $1 <= p or q <= $1 );
consider X being Subset of REAL such that
A1:
for r being Real holds
( r in X iff S1[r] )
from SUBSET_1:sch 3();
now let s1 be
Real_Sequence;
( rng s1 c= X & s1 is convergent implies lim s1 in X )assume that A2:
rng s1 c= X
and A3:
s1 is
convergent
;
lim s1 in X
(
lim s1 <= p or
q <= lim s1 )
proof
assume A4:
( not
lim s1 <= p & not
q <= lim s1 )
;
contradiction
then
0 < (lim s1) - p
by XREAL_1:50;
then consider n1 being
Element of
NAT such that A5:
for
m being
Element of
NAT st
n1 <= m holds
abs ((s1 . m) - (lim s1)) < (lim s1) - p
by A3, SEQ_2:def 7;
0 < q - (lim s1)
by A4, XREAL_1:50;
then consider n being
Element of
NAT such that A6:
for
m being
Element of
NAT st
n <= m holds
abs ((s1 . m) - (lim s1)) < q - (lim s1)
by A3, SEQ_2:def 7;
consider k being
Element of
NAT such that A7:
max (
n,
n1)
< k
by SEQ_4:3;
n1 <= max (
n,
n1)
by XXREAL_0:25;
then
n1 <= k
by A7, XXREAL_0:2;
then A8:
abs ((s1 . k) - (lim s1)) < (lim s1) - p
by A5;
- (abs ((s1 . k) - (lim s1))) <= (s1 . k) - (lim s1)
by ABSVALUE:4;
then
- ((s1 . k) - (lim s1)) <= - (- (abs ((s1 . k) - (lim s1))))
by XREAL_1:24;
then
- ((s1 . k) - (lim s1)) < (lim s1) - p
by A8, XXREAL_0:2;
then
- ((lim s1) - p) < - (- ((s1 . k) - (lim s1)))
by XREAL_1:24;
then
p - (lim s1) < (s1 . k) - (lim s1)
;
then A9:
p < s1 . k
by XREAL_1:9;
n <= max (
n,
n1)
by XXREAL_0:25;
then
n <= k
by A7, XXREAL_0:2;
then
(
(s1 . k) - (lim s1) <= abs ((s1 . k) - (lim s1)) &
abs ((s1 . k) - (lim s1)) < q - (lim s1) )
by A6, ABSVALUE:4;
then
(s1 . k) - (lim s1) < q - (lim s1)
by XXREAL_0:2;
then A10:
s1 . k < q
by XREAL_1:9;
dom s1 = NAT
by FUNCT_2:def 1;
then
s1 . k in rng s1
by FUNCT_1:def 3;
hence
contradiction
by A1, A2, A10, A9;
verum
end; hence
lim s1 in X
by A1;
verum end;
then A11:
X is closed
by Def4;
then A13:
].p,q.[ ` c= X
by MEMBERED:def 9;
then
X c= ].p,q.[ `
by MEMBERED:def 9;
then
].p,q.[ ` = X
by A13, XBOOLE_0:def 10;
hence
].p,q.[ is open
by A11, Def5; verum