let X be RealLinearSpace; for x being sequence of X
for A being finite Subset of X st A c= rng x holds
ex n being Nat st A c= rng (x | (Segm n))
let x be sequence of X; for A being finite Subset of X st A c= rng x holds
ex n being Nat st A c= rng (x | (Segm n))
defpred S1[ Nat] means for A being finite Subset of X st card A = $1 & A c= rng x holds
ex n being Nat st A c= rng (x | (Segm n));
A1:
S1[ 0 ]
A3:
for k being Nat st S1[k] holds
S1[k + 1]
proof
let k be
Nat;
( S1[k] implies S1[k + 1] )
assume A4:
S1[
k]
;
S1[k + 1]
let A be
finite Subset of the
carrier of
X;
( card A = k + 1 & A c= rng x implies ex n being Nat st A c= rng (x | (Segm n)) )
assume A5:
(
card A = k + 1 &
A c= rng x )
;
ex n being Nat st A c= rng (x | (Segm n))
then
A <> {}
;
then consider w being
object such that A6:
w in A
by XBOOLE_0:def 1;
reconsider w =
w as
Element of the
carrier of
X by A6;
A7:
card (A \ {w}) =
(card A) - (card {w})
by A6, CARD_2:44, ZFMISC_1:31
.=
(k + 1) - 1
by A5, CARD_2:42
;
reconsider A0 =
A \ {w} as
finite Subset of
X ;
A0 c= A
by XBOOLE_1:36;
then
A0 c= rng x
by A5;
then consider n0 being
Nat such that A8:
A0 c= rng (x | (Segm n0))
by A4, A7;
consider n1 being
object such that A9:
(
n1 in NAT &
w = x . n1 )
by A5, A6, FUNCT_2:11;
reconsider n1 =
n1 as
Nat by A9;
set n =
(n0 + n1) + 1;
take
(n0 + n1) + 1
;
A c= rng (x | (Segm ((n0 + n1) + 1)))
A10:
A = A0 \/ {w}
by A6, XBOOLE_1:45, ZFMISC_1:31;
n0 < (n0 + n1) + 1
by NAT_1:11, NAT_1:13;
then
x | (Segm n0) c= x | (Segm ((n0 + n1) + 1))
by NAT_1:39, RELAT_1:75;
then
rng (x | (Segm n0)) c= rng (x | (Segm ((n0 + n1) + 1)))
by RELAT_1:11;
then A11:
A0 c= rng (x | (Segm ((n0 + n1) + 1)))
by A8;
n1 < (n0 + n1) + 1
by NAT_1:11, NAT_1:13;
then
(
n1 in Segm ((n0 + n1) + 1) &
dom x = NAT )
by FUNCT_2:def 1, NAT_1:44;
then
{w} c= rng (x | (Segm ((n0 + n1) + 1)))
by A9, FUNCT_1:50, ZFMISC_1:31;
hence
A c= rng (x | (Segm ((n0 + n1) + 1)))
by A10, A11, XBOOLE_1:8;
verum
end;
A12:
for k being Nat holds S1[k]
from NAT_1:sch 2(A1, A3);
let A be finite Subset of the carrier of X; ( A c= rng x implies ex n being Nat st A c= rng (x | (Segm n)) )
assume A13:
A c= rng x
; ex n being Nat st A c= rng (x | (Segm n))
card A is Nat
;
hence
ex n being Nat st A c= rng (x | (Segm n))
by A12, A13; verum