let X be RealHilbertSpace; :: thesis: ( the addF of X is commutative & the addF of X is associative & the addF of X is having_a_unity implies for Y being Subset of X holds
( Y is summable_set iff for e being Real st e > 0 holds
ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < e ) ) ) )

assume A1: ( the addF of X is commutative & the addF of X is associative & the addF of X is having_a_unity ) ; :: thesis: for Y being Subset of X holds
( Y is summable_set iff for e being Real st e > 0 holds
ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < e ) ) )

let Y be Subset of X; :: thesis: ( Y is summable_set iff for e being Real st e > 0 holds
ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < e ) ) )

A2: now :: thesis: ( ( for e being Real st e > 0 holds
ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < e ) ) ) implies Y is summable_set )
defpred S1[ object , object ] means ex D2 being set st
( D2 = \$2 & \$2 is finite Subset of X & not D2 is empty & D2 c= Y & ( for z being Real st z = \$1 holds
for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & D2 misses Y1 holds
||.(setsum Y1).|| < 1 / (z + 1) ) );
assume A3: for e being Real st e > 0 holds
ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < e ) ) ; :: thesis: Y is summable_set
A4: for x being object st x in NAT holds
ex y being object st
( y in bool Y & S1[x,y] )
proof
let x be object ; :: thesis: ( x in NAT implies ex y being object st
( y in bool Y & S1[x,y] ) )

assume x in NAT ; :: thesis: ex y being object st
( y in bool Y & S1[x,y] )

then reconsider xx = x as Nat ;
reconsider e = 1 / (xx + 1) as Real ;
0 / (xx + 1) < 1 / (xx + 1) by XREAL_1:74;
then consider Y0 being finite Subset of X such that
A5: ( not Y0 is empty & Y0 c= Y ) and
A6: for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < e by A3;
reconsider Y0 = Y0 as object ;
take Y0 ; :: thesis: ( Y0 in bool Y & S1[x,Y0] )
thus Y0 in bool Y by A5; :: thesis: S1[x,Y0]
take Y0 ; :: thesis: ( Y0 is set & Y0 = Y0 & Y0 is finite Subset of X & not Y0 is empty & Y0 c= Y & ( for z being Real st z = x holds
for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < 1 / (z + 1) ) )

thus ( Y0 is set & Y0 = Y0 & Y0 is finite Subset of X & not Y0 is empty & Y0 c= Y & ( for z being Real st z = x holds
for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < 1 / (z + 1) ) ) by A5, A6; :: thesis: verum
end;
consider B being sequence of (bool Y) such that
A7: for x being object st x in NAT holds
S1[x,B . x] from ex A being sequence of (bool Y) st
for i being Nat holds
( A . i is finite Subset of X & not A . i is empty & A . i c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & A . i misses Y1 holds
||.(setsum Y1).|| < 1 / (i + 1) ) & ( for j being Nat st i <= j holds
A . i c= A . j ) )
proof
A8: for x being object st x in NAT holds
( B . x is finite Subset of X & not B . x is empty & B . x c= Y & ( for z being Real st z = x holds
for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & B . x misses Y1 holds
||.(setsum Y1).|| < 1 / (z + 1) ) )
proof
let x be object ; :: thesis: ( x in NAT implies ( B . x is finite Subset of X & not B . x is empty & B . x c= Y & ( for z being Real st z = x holds
for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & B . x misses Y1 holds
||.(setsum Y1).|| < 1 / (z + 1) ) ) )

assume x in NAT ; :: thesis: ( B . x is finite Subset of X & not B . x is empty & B . x c= Y & ( for z being Real st z = x holds
for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & B . x misses Y1 holds
||.(setsum Y1).|| < 1 / (z + 1) ) )

then S1[x,B . x] by A7;
hence ( B . x is finite Subset of X & not B . x is empty & B . x c= Y & ( for z being Real st z = x holds
for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & B . x misses Y1 holds
||.(setsum Y1).|| < 1 / (z + 1) ) ) ; :: thesis: verum
end;
deffunc H1( Nat, set ) -> set = (B . (\$1 + 1)) \/ \$2;
ex A being Function st
( dom A = NAT & A . 0 = B . 0 & ( for n being Nat holds A . (n + 1) = H1(n,A . n) ) ) from NAT_1:sch 11();
then consider A being Function such that
A9: dom A = NAT and
A10: A . 0 = B . 0 and
A11: for n being Nat holds A . (n + 1) = (B . (n + 1)) \/ (A . n) ;
defpred S2[ Nat] means ( A . \$1 is finite Subset of X & not A . \$1 is empty & A . \$1 c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & A . \$1 misses Y1 holds
||.(setsum Y1).|| < 1 / (\$1 + 1) ) & ( for j being Nat st \$1 <= j holds
A . \$1 c= A . j ) );
A12: now :: thesis: for n being Nat st S2[n] holds
S2[n + 1]
let n be Nat; :: thesis: ( S2[n] implies S2[n + 1] )
assume A13: S2[n] ; :: thesis: S2[n + 1]
A14: for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & A . (n + 1) misses Y1 holds
||.(setsum Y1).|| < 1 / ((n + 1) + 1)
proof
let Y1 be finite Subset of X; :: thesis: ( not Y1 is empty & Y1 c= Y & A . (n + 1) misses Y1 implies ||.(setsum Y1).|| < 1 / ((n + 1) + 1) )
assume that
A15: ( not Y1 is empty & Y1 c= Y ) and
A16: A . (n + 1) misses Y1 ; :: thesis: ||.(setsum Y1).|| < 1 / ((n + 1) + 1)
A . (n + 1) = (B . (n + 1)) \/ (A . n) by A11;
then B . (n + 1) misses Y1 by ;
hence ||.(setsum Y1).|| < 1 / ((n + 1) + 1) by ; :: thesis: verum
end;
defpred S3[ Nat] means ( n + 1 <= \$1 implies A . (n + 1) c= A . \$1 );
A17: for j being Nat st S3[j] holds
S3[j + 1]
proof
let j be Nat; :: thesis: ( S3[j] implies S3[j + 1] )
assume that
A18: ( n + 1 <= j implies A . (n + 1) c= A . j ) and
A19: n + 1 <= j + 1 ; :: thesis: A . (n + 1) c= A . (j + 1)
per cases ( n = j or n <> j ) ;
suppose n = j ; :: thesis: A . (n + 1) c= A . (j + 1)
hence A . (n + 1) c= A . (j + 1) ; :: thesis: verum
end;
suppose A20: n <> j ; :: thesis: A . (n + 1) c= A . (j + 1)
A . (j + 1) = (B . (j + 1)) \/ (A . j) by A11;
then A21: A . j c= A . (j + 1) by XBOOLE_1:7;
n <= j by ;
then n < j by ;
hence A . (n + 1) c= A . (j + 1) by ; :: thesis: verum
end;
end;
end;
A22: S3[ 0 ] ;
A23: for j being Nat holds S3[j] from A24: ( A . (n + 1) = (B . (n + 1)) \/ (A . n) & B . (n + 1) is finite Subset of X ) by ;
thus S2[n + 1] by ; :: thesis: verum
end;
for j0 being Nat st j0 = 0 holds
for j being Nat st j0 <= j holds
A . j0 c= A . j
proof
let j0 be Nat; :: thesis: ( j0 = 0 implies for j being Nat st j0 <= j holds
A . j0 c= A . j )

assume A25: j0 = 0 ; :: thesis: for j being Nat st j0 <= j holds
A . j0 c= A . j

defpred S3[ Nat] means ( j0 <= \$1 implies A . j0 c= A . \$1 );
A26: now :: thesis: for j being Nat st S3[j] holds
S3[j + 1]
let j be Nat; :: thesis: ( S3[j] implies S3[j + 1] )
assume A27: S3[j] ; :: thesis: S3[j + 1]
A . (j + 1) = (B . (j + 1)) \/ (A . j) by A11;
then A . j c= A . (j + 1) by XBOOLE_1:7;
hence S3[j + 1] by ; :: thesis: verum
end;
A28: S3[ 0 ] ;
thus for j being Nat holds S3[j] from :: thesis: verum
end;
then A29: S2[ 0 ] by ;
A30: for i being Nat holds S2[i] from
now :: thesis: for y being object st y in rng A holds
y in bool Y
let y be object ; :: thesis: ( y in rng A implies y in bool Y )
assume y in rng A ; :: thesis: y in bool Y
then consider x being object such that
A31: x in dom A and
A32: y = A . x by FUNCT_1:def 3;
reconsider i = x as Nat by ;
A . i c= Y by A30;
hence y in bool Y by A32; :: thesis: verum
end;
then rng A c= bool Y by TARSKI:def 3;
then A is sequence of (bool Y) by ;
hence ex A being sequence of (bool Y) st
for i being Nat holds
( A . i is finite Subset of X & not A . i is empty & A . i c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & A . i misses Y1 holds
||.(setsum Y1).|| < 1 / (i + 1) ) & ( for j being Nat st i <= j holds
A . i c= A . j ) ) by A30; :: thesis: verum
end;
then consider A being sequence of (bool Y) such that
A33: for i being Nat holds
( A . i is finite Subset of X & not A . i is empty & A . i c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & A . i misses Y1 holds
||.(setsum Y1).|| < 1 / (i + 1) ) & ( for j being Nat st i <= j holds
A . i c= A . j ) ) ;
defpred S2[ object , object ] means ex Y1 being finite Subset of X st
( not Y1 is empty & A . \$1 = Y1 & \$2 = setsum Y1 );
A34: for x being object st x in NAT holds
ex y being object st
( y in the carrier of X & S2[x,y] )
proof
let x be object ; :: thesis: ( x in NAT implies ex y being object st
( y in the carrier of X & S2[x,y] ) )

assume x in NAT ; :: thesis: ex y being object st
( y in the carrier of X & S2[x,y] )

then reconsider i = x as Nat ;
A . i is finite Subset of X by A33;
then reconsider Y1 = A . x as finite Subset of X ;
reconsider y = setsum Y1 as set ;
not A . i is empty by A33;
then ex Y1 being finite Subset of X st
( Y1 is non empty set & A . x = Y1 & y = setsum Y1 ) ;
hence ex y being object st
( y in the carrier of X & S2[x,y] ) ; :: thesis: verum
end;
ex F being sequence of the carrier of X st
for x being object st x in NAT holds
S2[x,F . x] from then consider F being sequence of the carrier of X such that
A35: for x being object st x in NAT holds
ex Y1 being finite Subset of X st
( not Y1 is empty & A . x = Y1 & F . x = setsum Y1 ) ;
reconsider seq = F as sequence of X ;
now :: thesis: for e being Real st e > 0 holds
ex k being Nat st
for n, m being Nat st n >= k & m >= k holds
||.((seq . n) - (seq . m)).|| < e
let e be Real; :: thesis: ( e > 0 implies ex k being Nat st
for n, m being Nat st n >= k & m >= k holds
||.((seq . n) - (seq . m)).|| < e )

assume A36: e > 0 ; :: thesis: ex k being Nat st
for n, m being Nat st n >= k & m >= k holds
||.((seq . n) - (seq . m)).|| < e

A37: e / 2 > 0 / 2 by ;
ex k being Nat st 1 / (k + 1) < e / 2
proof
set p = e / 2;
consider k1 being Nat such that
A38: (e / 2) " < k1 by SEQ_4:3;
take k = k1; :: thesis: 1 / (k + 1) < e / 2
((e / 2) ") + 0 < k1 + 1 by ;
then 1 / (k1 + 1) < 1 / ((e / 2) ") by ;
then 1 / (k + 1) < 1 * (((e / 2) ") ") by XCMPLX_0:def 9;
hence 1 / (k + 1) < e / 2 ; :: thesis: verum
end;
then consider k being Nat such that
A39: 1 / (k + 1) < e / 2 ;
now :: thesis: for nn, mm being Nat st nn >= k & mm >= k holds
||.((seq . nn) - (seq . mm)).|| < e
let nn, mm be Nat; :: thesis: ( nn >= k & mm >= k implies ||.((seq . nn) - (seq . mm)).|| < e )
assume that
A40: nn >= k and
A41: mm >= k ; :: thesis: ||.((seq . nn) - (seq . mm)).|| < e
( nn in NAT & mm in NAT & k in NAT ) by ORDINAL1:def 12;
then reconsider k = k, m = mm, n = nn as Element of NAT ;
consider Y2 being finite Subset of X such that
not Y2 is empty and
A42: A . m = Y2 and
A43: seq . m = setsum Y2 by A35;
consider Y0 being finite Subset of X such that
not Y0 is empty and
A44: A . k = Y0 and
A45: seq . k = setsum Y0 by A35;
A46: Y0 c= Y2 by A33, A41, A44, A42;
consider Y1 being finite Subset of X such that
not Y1 is empty and
A47: A . n = Y1 and
A48: seq . n = setsum Y1 by A35;
A49: Y0 c= Y1 by A33, A40, A44, A47;
now :: thesis: ( ( Y0 = Y1 & ||.((seq . n) - (seq . m)).|| < e ) or ( Y0 <> Y1 & ||.((seq . n) - (seq . m)).|| < e ) )
per cases ( Y0 = Y1 or Y0 <> Y1 ) ;
case A50: Y0 = Y1 ; :: thesis: ||.((seq . n) - (seq . m)).|| < e
now :: thesis: ( ( Y0 = Y2 & ||.((seq . n) - (seq . m)).|| < e ) or ( Y0 <> Y2 & ||.((seq . n) - (seq . m)).|| < e ) )
per cases ( Y0 = Y2 or Y0 <> Y2 ) ;
case Y0 = Y2 ; :: thesis: ||.((seq . n) - (seq . m)).|| < e
then (seq . n) - (seq . m) = 0. X by ;
hence ||.((seq . n) - (seq . m)).|| < e by ; :: thesis: verum
end;
case A51: Y0 <> Y2 ; :: thesis: ||.((seq . n) - (seq . m)).|| < e
ex Y02 being finite Subset of X st
( not Y02 is empty & Y02 c= Y & Y02 misses Y0 & Y0 \/ Y02 = Y2 )
proof
take Y02 = Y2 \ Y0; :: thesis: ( not Y02 is empty & Y02 c= Y & Y02 misses Y0 & Y0 \/ Y02 = Y2 )
A52: Y2 \ Y0 c= Y2 by XBOOLE_1:36;
A53: Y0 \/ Y02 = Y0 \/ Y2 by XBOOLE_1:39
.= Y2 by ;
hence not Y02 is empty by A51; :: thesis: ( Y02 c= Y & Y02 misses Y0 & Y0 \/ Y02 = Y2 )
thus Y02 c= Y by ; :: thesis: ( Y02 misses Y0 & Y0 \/ Y02 = Y2 )
thus ( Y02 misses Y0 & Y0 \/ Y02 = Y2 ) by ; :: thesis: verum
end;
then consider Y02 being finite Subset of X such that
A54: ( not Y02 is empty & Y02 c= Y ) and
A55: Y02 misses Y0 and
A56: Y0 \/ Y02 = Y2 ;
||.(setsum Y02).|| < 1 / (k + 1) by A33, A44, A54, A55;
then A57: ||.(setsum Y02).|| < e / 2 by ;
setsum Y2 = (setsum Y0) + (setsum Y02) by A1, A55, A56, Th2;
then A58: ||.((seq . n) - (seq . m)).|| = ||.(((setsum Y0) - (setsum Y0)) - (setsum Y02)).|| by
.= ||.((0. X) - (setsum Y02)).|| by RLVECT_1:15
.= ||.(- (setsum Y02)).|| by RLVECT_1:14
.= ||.(setsum Y02).|| by BHSP_1:31 ;
e * (1 / 2) < e * 1 by ;
hence ||.((seq . n) - (seq . m)).|| < e by ; :: thesis: verum
end;
end;
end;
hence ||.((seq . n) - (seq . m)).|| < e ; :: thesis: verum
end;
case A59: Y0 <> Y1 ; :: thesis: ||.((seq . n) - (seq . m)).|| < e
now :: thesis: ( ( Y0 = Y2 & ||.((seq . n) - (seq . m)).|| < e ) or ( Y0 <> Y2 & ||.((seq . n) - (seq . m)).|| < e ) )
per cases ( Y0 = Y2 or Y0 <> Y2 ) ;
case Y0 = Y2 ; :: thesis: ||.((seq . n) - (seq . m)).|| < e
then A60: (seq . k) - (seq . m) = 0. X by ;
ex Y01 being finite Subset of X st
( not Y01 is empty & Y01 c= Y & Y01 misses Y0 & Y0 \/ Y01 = Y1 )
proof
take Y01 = Y1 \ Y0; :: thesis: ( not Y01 is empty & Y01 c= Y & Y01 misses Y0 & Y0 \/ Y01 = Y1 )
A61: Y1 \ Y0 c= Y1 by XBOOLE_1:36;
Y0 \/ Y01 = Y0 \/ Y1 by XBOOLE_1:39
.= Y1 by ;
hence ( not Y01 is empty & Y01 c= Y & Y01 misses Y0 & Y0 \/ Y01 = Y1 ) by ; :: thesis: verum
end;
then consider Y01 being finite Subset of X such that
A62: ( not Y01 is empty & Y01 c= Y ) and
A63: Y01 misses Y0 and
A64: Y0 \/ Y01 = Y1 ;
seq . n = (seq . k) + (setsum Y01) by A1, A45, A48, A63, A64, Th2;
then A65: (seq . n) - (seq . k) = (seq . k) + ((setsum Y01) + (- (seq . k))) by RLVECT_1:def 3
.= (seq . k) - ((seq . k) - (setsum Y01)) by RLVECT_1:33
.= ((setsum Y0) - (setsum Y0)) + (setsum Y01) by
.= (0. X) + (setsum Y01) by RLVECT_1:5
.= setsum Y01 by RLVECT_1:4 ;
(seq . n) - (seq . m) = ((seq . n) - (seq . m)) + (0. X) by RLVECT_1:4
.= ((seq . n) - (seq . m)) + ((seq . k) - (seq . k)) by RLVECT_1:5
.= (seq . n) - ((seq . m) - ((seq . k) - (seq . k))) by RLVECT_1:29
.= (seq . n) - (((seq . m) - (seq . k)) + (seq . k)) by RLVECT_1:29
.= (seq . n) - ((seq . k) - ((seq . k) - (seq . m))) by RLVECT_1:33
.= (setsum Y01) + (0. X) by ;
then ||.((seq . n) - (seq . m)).|| <= ||.(setsum Y01).|| + ||.(0. X).|| by BHSP_1:30;
then A66: ||.((seq . n) - (seq . m)).|| <= ||.(setsum Y01).|| + 0 by BHSP_1:26;
||.(setsum Y01).|| < 1 / (k + 1) by A33, A44, A62, A63;
then ||.(setsum Y01).|| < e / 2 by ;
then ||.((seq . n) - (seq . m)).|| < e / 2 by ;
then ||.((seq . n) - (seq . m)).|| + 0 < (e / 2) + (e / 2) by ;
hence ||.((seq . n) - (seq . m)).|| < e ; :: thesis: verum
end;
case A67: Y0 <> Y2 ; :: thesis: ||.((seq . n) - (seq . m)).|| < e
ex Y02 being finite Subset of X st
( not Y02 is empty & Y02 c= Y & Y02 misses Y0 & Y0 \/ Y02 = Y2 )
proof
take Y02 = Y2 \ Y0; :: thesis: ( not Y02 is empty & Y02 c= Y & Y02 misses Y0 & Y0 \/ Y02 = Y2 )
A68: Y2 \ Y0 c= Y2 by XBOOLE_1:36;
Y0 \/ Y02 = Y0 \/ Y2 by XBOOLE_1:39
.= Y2 by ;
hence ( not Y02 is empty & Y02 c= Y & Y02 misses Y0 & Y0 \/ Y02 = Y2 ) by ; :: thesis: verum
end;
then consider Y02 being finite Subset of X such that
A69: ( not Y02 is empty & Y02 c= Y ) and
A70: Y02 misses Y0 and
A71: Y0 \/ Y02 = Y2 ;
||.(setsum Y02).|| < 1 / (k + 1) by A33, A44, A69, A70;
then A72: ||.(setsum Y02).|| < e / 2 by ;
ex Y01 being finite Subset of X st
( not Y01 is empty & Y01 c= Y & Y01 misses Y0 & Y0 \/ Y01 = Y1 )
proof
take Y01 = Y1 \ Y0; :: thesis: ( not Y01 is empty & Y01 c= Y & Y01 misses Y0 & Y0 \/ Y01 = Y1 )
A73: Y1 \ Y0 c= Y1 by XBOOLE_1:36;
Y0 \/ Y01 = Y0 \/ Y1 by XBOOLE_1:39
.= Y1 by ;
hence ( not Y01 is empty & Y01 c= Y & Y01 misses Y0 & Y0 \/ Y01 = Y1 ) by ; :: thesis: verum
end;
then consider Y01 being finite Subset of X such that
A74: ( not Y01 is empty & Y01 c= Y ) and
A75: Y01 misses Y0 and
A76: Y0 \/ Y01 = Y1 ;
setsum Y1 = (setsum Y0) + (setsum Y01) by A1, A75, A76, Th2;
then A77: (seq . n) - (seq . k) = (setsum Y01) + ((seq . k) + (- (seq . k))) by
.= (setsum Y01) + (0. X) by RLVECT_1:5
.= setsum Y01 by RLVECT_1:4 ;
setsum Y2 = (setsum Y0) + (setsum Y02) by A1, A70, A71, Th2;
then A78: (seq . m) - (seq . k) = (setsum Y02) + ((seq . k) + (- (seq . k))) by
.= (setsum Y02) + (0. X) by RLVECT_1:5
.= setsum Y02 by RLVECT_1:4 ;
(seq . n) - (seq . m) = ((seq . n) - (seq . m)) + (0. X) by RLVECT_1:4
.= ((seq . n) - (seq . m)) + ((seq . k) - (seq . k)) by RLVECT_1:5
.= (seq . n) - ((seq . m) - ((seq . k) - (seq . k))) by RLVECT_1:29
.= (seq . n) - (((seq . m) - (seq . k)) + (seq . k)) by RLVECT_1:29
.= (seq . n) - ((seq . k) - ((seq . k) - (seq . m))) by RLVECT_1:33
.= ((seq . n) - (seq . k)) + ((seq . k) + (- (seq . m))) by RLVECT_1:29
.= (setsum Y01) + (- (setsum Y02)) by ;
then ||.((seq . n) - (seq . m)).|| <= ||.(setsum Y01).|| + ||.(- (setsum Y02)).|| by BHSP_1:30;
then A79: ||.((seq . n) - (seq . m)).|| <= ||.(setsum Y01).|| + ||.(setsum Y02).|| by BHSP_1:31;
||.(setsum Y01).|| < 1 / (k + 1) by A33, A44, A74, A75;
then ||.(setsum Y01).|| < e / 2 by ;
then ||.(setsum Y01).|| + ||.(setsum Y02).|| < (e / 2) + (e / 2) by ;
hence ||.((seq . n) - (seq . m)).|| < e by ; :: thesis: verum
end;
end;
end;
hence ||.((seq . n) - (seq . m)).|| < e ; :: thesis: verum
end;
end;
end;
hence ||.((seq . nn) - (seq . mm)).|| < e ; :: thesis: verum
end;
hence ex k being Nat st
for n, m being Nat st n >= k & m >= k holds
||.((seq . n) - (seq . m)).|| < e ; :: thesis: verum
end;
then seq is Cauchy by BHSP_3:2;
then seq is convergent by BHSP_3:def 4;
then consider x being Point of X such that
A80: for r being Real st r > 0 holds
ex m being Nat st
for n being Nat st n >= m holds
||.((seq . n) - x).|| < r by BHSP_2:9;
now :: thesis: for e being Real st e > 0 holds
ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st Y0 c= Y1 & Y1 c= Y holds
||.(x - (setsum Y1)).|| < e ) )
let e be Real; :: thesis: ( e > 0 implies ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st Y0 c= Y1 & Y1 c= Y holds
||.(x - (setsum Y1)).|| < e ) ) )

assume A81: e > 0 ; :: thesis: ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st Y0 c= Y1 & Y1 c= Y holds
||.(x - (setsum Y1)).|| < e ) )

A82: e / 2 > 0 / 2 by ;
then consider m being Nat such that
A83: for n being Nat st n >= m holds
||.((seq . n) - x).|| < e / 2 by A80;
ex i being Nat st
( 1 / (i + 1) < e / 2 & i >= m )
proof
set p = e / 2;
consider k1 being Nat such that
A84: (e / 2) " < k1 by SEQ_4:3;
take i = k1 + m; :: thesis: ( 1 / (i + 1) < e / 2 & i >= m )
k1 <= k1 + m by NAT_1:11;
then (e / 2) " < i by ;
then ((e / 2) ") + 0 < i + 1 by XREAL_1:8;
then 1 / (i + 1) < 1 / ((e / 2) ") by ;
then 1 / (i + 1) < 1 * (((e / 2) ") ") by XCMPLX_0:def 9;
hence ( 1 / (i + 1) < e / 2 & i >= m ) by NAT_1:11; :: thesis: verum
end;
then consider i being Nat such that
A85: 1 / (i + 1) < e / 2 and
A86: i >= m ;
reconsider i = i as Element of NAT by ORDINAL1:def 12;
consider Y0 being finite Subset of X such that
A87: not Y0 is empty and
A88: A . i = Y0 and
A89: seq . i = setsum Y0 by A35;
A90: ||.((setsum Y0) - x).|| < e / 2 by ;
now :: thesis: for Y1 being finite Subset of X st Y0 c= Y1 & Y1 c= Y holds
||.(x - (setsum Y1)).|| < e
let Y1 be finite Subset of X; :: thesis: ( Y0 c= Y1 & Y1 c= Y implies ||.(x - (setsum Y1)).|| < e )
assume that
A91: Y0 c= Y1 and
A92: Y1 c= Y ; :: thesis: ||.(x - (setsum Y1)).|| < e
now :: thesis: ( ( Y0 = Y1 & ||.(x - (setsum Y1)).|| < e ) or ( Y0 <> Y1 & ||.(x - (setsum Y1)).|| < e ) )
per cases ( Y0 = Y1 or Y0 <> Y1 ) ;
case Y0 = Y1 ; :: thesis: ||.(x - (setsum Y1)).|| < e
then ||.(x - (setsum Y1)).|| = ||.(- (x - (setsum Y0))).|| by BHSP_1:31
.= ||.((setsum Y0) - x).|| by RLVECT_1:33 ;
then ||.(x - (setsum Y1)).|| < e / 2 by ;
then ||.(x - (setsum Y1)).|| + 0 < (e / 2) + (e / 2) by ;
hence ||.(x - (setsum Y1)).|| < e ; :: thesis: verum
end;
case A93: Y0 <> Y1 ; :: thesis: ||.(x - (setsum Y1)).|| < e
ex Y2 being finite Subset of X st
( not Y2 is empty & Y2 c= Y & Y0 misses Y2 & Y0 \/ Y2 = Y1 )
proof
take Y2 = Y1 \ Y0; :: thesis: ( not Y2 is empty & Y2 c= Y & Y0 misses Y2 & Y0 \/ Y2 = Y1 )
A94: Y1 \ Y0 c= Y1 by XBOOLE_1:36;
Y0 \/ Y2 = Y0 \/ Y1 by XBOOLE_1:39
.= Y1 by ;
hence ( not Y2 is empty & Y2 c= Y & Y0 misses Y2 & Y0 \/ Y2 = Y1 ) by ; :: thesis: verum
end;
then consider Y2 being finite Subset of X such that
A95: ( not Y2 is empty & Y2 c= Y ) and
A96: Y0 misses Y2 and
A97: Y0 \/ Y2 = Y1 ;
(setsum Y1) - x = ((setsum Y0) + (setsum Y2)) - x by A1, A96, A97, Th2
.= ((setsum Y0) - x) + (setsum Y2) by RLVECT_1:def 3 ;
then ||.((setsum Y1) - x).|| <= ||.((setsum Y0) - x).|| + ||.(setsum Y2).|| by BHSP_1:30;
then ||.(- ((setsum Y1) - x)).|| <= ||.((setsum Y0) - x).|| + ||.(setsum Y2).|| by BHSP_1:31;
then A98: ||.(x + (- (setsum Y1))).|| <= ||.((setsum Y0) - x).|| + ||.(setsum Y2).|| by RLVECT_1:33;
||.(setsum Y2).|| < 1 / (i + 1) by A33, A88, A95, A96;
then ||.(setsum Y2).|| < e / 2 by ;
then ||.((setsum Y0) - x).|| + ||.(setsum Y2).|| < (e / 2) + (e / 2) by ;
hence ||.(x - (setsum Y1)).|| < e by ; :: thesis: verum
end;
end;
end;
hence ||.(x - (setsum Y1)).|| < e ; :: thesis: verum
end;
hence ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st Y0 c= Y1 & Y1 c= Y holds
||.(x - (setsum Y1)).|| < e ) ) by ; :: thesis: verum
end;
hence Y is summable_set ; :: thesis: verum
end;
now :: thesis: ( Y is summable_set implies for e being Real st e > 0 holds
ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < e ) ) )
assume Y is summable_set ; :: thesis: for e being Real st e > 0 holds
ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < e ) )

then consider x being Point of X such that
A99: for e being Real st e > 0 holds
ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st Y0 c= Y1 & Y1 c= Y holds
||.(x - (setsum Y1)).|| < e ) ) ;
now :: thesis: for e being Real st e > 0 holds
ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < e ) )
let e be Real; :: thesis: ( e > 0 implies ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < e ) ) )

assume e > 0 ; :: thesis: ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < e ) )

then consider Y0 being finite Subset of X such that
A100: not Y0 is empty and
A101: Y0 c= Y and
A102: for Y1 being finite Subset of X st Y0 c= Y1 & Y1 c= Y holds
||.(x - (setsum Y1)).|| < e / 2 by ;
reconsider Y0 = Y0 as non empty finite Subset of X by A100;
now :: thesis: for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < e
let Y1 be finite Subset of X; :: thesis: ( not Y1 is empty & Y1 c= Y & Y0 misses Y1 implies ||.(setsum Y1).|| < e )
assume that
not Y1 is empty and
A103: Y1 c= Y and
A104: Y0 misses Y1 ; :: thesis: ||.(setsum Y1).|| < e
set Z = Y0 \/ Y1;
Y0 c= Y0 \/ Y1 by XBOOLE_1:7;
then ||.(x - (setsum (Y0 \/ Y1))).|| < e / 2 by ;
then A105: ||.(x - (setsum (Y0 \/ Y1))).|| + ||.(x - (setsum Y0)).|| < (e / 2) + (e / 2) by ;
||.((x - (setsum (Y0 \/ Y1))) - (x - (setsum Y0))).|| <= ||.(x - (setsum (Y0 \/ Y1))).|| + ||.(- (x - (setsum Y0))).|| by BHSP_1:30;
then A106: ||.((x - (setsum (Y0 \/ Y1))) - (x - (setsum Y0))).|| <= ||.(x - (setsum (Y0 \/ Y1))).|| + ||.(x - (setsum Y0)).|| by BHSP_1:31;
(setsum (Y0 \/ Y1)) - (setsum Y0) = ((setsum Y1) + (setsum Y0)) - (setsum Y0) by
.= (setsum Y1) + ((setsum Y0) + (- (setsum Y0))) by RLVECT_1:def 3
.= (setsum Y1) + (0. X) by RLVECT_1:5
.= setsum Y1 by RLVECT_1:4 ;
then ||.(setsum Y1).|| = ||.(- ((setsum (Y0 \/ Y1)) - (setsum Y0))).|| by BHSP_1:31
.= ||.((setsum Y0) - (setsum (Y0 \/ Y1))).|| by RLVECT_1:33
.= ||.((0. X) + ((setsum Y0) - (setsum (Y0 \/ Y1)))).|| by RLVECT_1:4
.= ||.((x - x) + ((setsum Y0) - (setsum (Y0 \/ Y1)))).|| by RLVECT_1:5
.= ||.(x - (x - ((setsum Y0) - (setsum (Y0 \/ Y1))))).|| by RLVECT_1:29
.= ||.(x - ((x - (setsum Y0)) + (setsum (Y0 \/ Y1)))).|| by RLVECT_1:29
.= ||.((x - (setsum (Y0 \/ Y1))) - (x - (setsum Y0))).|| by RLVECT_1:27 ;
hence ||.(setsum Y1).|| < e by ; :: thesis: verum
end;
hence ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < e ) ) by A101; :: thesis: verum
end;
hence for e being Real st e > 0 holds
ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < e ) ) ; :: thesis: verum
end;
hence ( Y is summable_set iff for e being Real st e > 0 holds
ex Y0 being finite Subset of X st
( not Y0 is empty & Y0 c= Y & ( for Y1 being finite Subset of X st not Y1 is empty & Y1 c= Y & Y0 misses Y1 holds
||.(setsum Y1).|| < e ) ) ) by A2; :: thesis: verum