deffunc H1( set ) -> Element of REAL = 1 - (h . $1);
defpred S1[ set ] means $1 in dom h;
consider IT being PartFunc of C,REAL such that
A1:
( ( for x being Element of C holds
( x in dom IT iff S1[x] ) ) & ( for x being Element of C st x in dom IT holds
IT . x = H1(x) ) )
from SEQ_1:sch 3();
for x being set st x in C holds
x in dom IT
then
C c= dom IT
by TARSKI:def 3;
then A3:
dom IT = C
by XBOOLE_0:def 10;
then A4:
for c being Element of C holds IT . c = 1 - (h . c)
by A1;
A5:
rng h c= [.0 ,1.]
by RELAT_1:def 19;
for y being set st y in rng IT holds
y in [.0 ,1.]
proof
reconsider A =
[.0 ,1.] as
closed-interval Subset of
REAL by INTEGRA1:def 1;
let y be
set ;
( y in rng IT implies y in [.0 ,1.] )
assume
y in rng IT
;
y in [.0 ,1.]
then consider x being
Element of
C such that A6:
x in dom IT
and A7:
y = IT . x
by PARTFUN1:26;
A8:
A = [.(lower_bound A),(upper_bound A).]
by INTEGRA1:5;
then A9:
0 = lower_bound A
by INTEGRA1:6;
x in C
;
then
x in dom h
by FUNCT_2:def 1;
then A10:
h . x in rng h
by FUNCT_1:def 5;
then
0 <= h . x
by A5, A9, INTEGRA2:1;
then
0 + 1
<= 1
+ (h . x)
by XREAL_1:8;
then
1
- (h . x) <= 1
by XREAL_1:22;
then A11:
IT . x <= 1
by A1, A6;
A12:
1
= upper_bound A
by A8, INTEGRA1:6;
then
h . x <= 1
by A5, A10, INTEGRA2:1;
then
0 <= 1
- (h . x)
by XREAL_1:50;
then
0 <= IT . x
by A1, A6;
hence
y in [.0 ,1.]
by A7, A9, A12, A11, INTEGRA2:1;
verum
end;
then A13:
rng IT c= [.0 ,1.]
by TARSKI:def 3;
IT is total
by A3, PARTFUN1:def 4;
then
IT is Membership_Func of C
by A13, RELAT_1:def 19;
hence
ex b1 being Membership_Func of C st
for c being Element of C holds b1 . c = 1 - (h . c)
by A4; verum