### The Mizar article:

### Convergent Real Sequences. Upper and Lower Bound of Sets of Real Numbers

**by****Jaroslaw Kotowicz**

- Received November 23, 1989
Copyright (c) 1989 Association of Mizar Users

- MML identifier: SEQ_4
- [ MML identifier index ]

environ vocabulary ARYTM, SEQ_1, ORDINAL2, SEQM_3, ARYTM_3, RELAT_1, ARYTM_1, SEQ_2, LATTICES, ABSVALUE, FUNCT_1, PROB_1, SEQ_4, MEMBERED; notation TARSKI, XBOOLE_0, SUBSET_1, ORDINAL1, NUMBERS, XCMPLX_0, XREAL_0, REAL_1, RELAT_1, SEQ_1, SEQ_2, NAT_1, FUNCT_2, SEQM_3, ABSVALUE, MEMBERED; constructors REAL_1, SEQ_2, NAT_1, SEQM_3, ABSVALUE, PARTFUN1, XCMPLX_0, XREAL_0, MEMBERED, XBOOLE_0; clusters XREAL_0, RELSET_1, SEQ_1, SEQM_3, ARYTM_3, REAL_1, XBOOLE_0, NUMBERS, SUBSET_1, MEMBERED, ZFMISC_1, ORDINAL2; requirements REAL, NUMERALS, SUBSET, BOOLE, ARITHM; definitions TARSKI, SEQ_2; theorems REAL_1, NAT_1, SEQ_1, SEQ_2, SEQM_3, AXIOMS, FUNCT_1, FUNCT_2, ABSVALUE, TARSKI, SQUARE_1, RELAT_1, RELSET_1, XREAL_0, XBOOLE_1, SUBSET_1, XCMPLX_0, XCMPLX_1, MEMBERED; schemes NAT_1, RECDEF_1, SEQ_1, REAL_1, SETWISEO; begin reserve n,k,k1,m,m1,n1,n2,l for Nat; reserve r,r1,r2,p,p1,g,g1,g2,s,s1,s2 for real number; reserve seq,seq1,seq2 for Real_Sequence; reserve Nseq for increasing Seq_of_Nat; reserve x for set; reserve X,Y for Subset of REAL; theorem Th1:0<r1 & r1<=r & 0<=g implies g/r<=g/r1 proof assume that A1: 0<r1 and A2: r1<=r and A3: 0<=g; 0<r by A1,A2; then A4: 0<=r" by REAL_1:72; A5: 0<=r1" by A1,REAL_1:72; r1*r"<=r*r" by A2,A4,AXIOMS:25; then r1*r"<=1 by A1,A2,XCMPLX_0:def 7; then r1"*(r1*r")<=r1"*1 by A5,AXIOMS:25; then r1"*r1*r"<=r1" by XCMPLX_1:4; then 1*r"<=r1" by A1,XCMPLX_0:def 7; then g*r"<=g*r1" by A3,AXIOMS:25; then g/r<=g*r1" by XCMPLX_0:def 9; hence thesis by XCMPLX_0:def 9; end; canceled 2; theorem Th4:0<s implies 0<s/3 proof assume 0<s; then 0/3<s/3 by REAL_1:73; hence thesis; end; canceled; theorem Th6:0<g & 0<=r & g<=g1 & r<r1 implies g*r<g1*r1 proof assume that A1: 0<g and A2: 0<=r and A3: g<=g1 and A4: r<r1; 0<=r1 by A2,A4,AXIOMS:22; then A5: g*r1<=g1*r1 by A3,AXIOMS:25; g*r<g*r1 by A1,A4,REAL_1:70; hence g*r<g1*r1 by A5,AXIOMS:22; end; theorem Th7:0<=g & 0<=r & g<=g1 & r<=r1 implies g*r<=g1*r1 proof assume that A1: 0<=g and A2: 0<=r and A3: g<=g1 and A4: r<=r1; A5: g*r<=g*r1 by A1,A4,AXIOMS:25; g*r1<=g1*r1 by A2,A3,A4,AXIOMS:25; hence g*r<=g1*r1 by A5,AXIOMS:22; end; theorem Th8: for X,Y st for r,p st r in X & p in Y holds r<p ex g st for r,p st r in X & p in Y holds r<=g & g<=p proof let X,Y such that A1: for r,p st r in X & p in Y holds r<p; for r,p st r in X & p in Y holds r<=p by A1; then consider g such that A2: for r,p st r in X & p in Y holds r <= g & g <= p by AXIOMS:26; reconsider g as Real by XREAL_0:def 1; take g; thus thesis by A2; end; theorem :: ARCHIMEDES LAW Th9:0<p & (ex r st r in X) & (for r st r in X holds r+p in X) implies for g ex r st r in X & g<r proof assume that A1: 0<p and A2: ex r st r in X and A3: for r st r in X holds r+p in X and A4: ex g st for r st r in X holds not g<r; consider g1 such that A5: for r st r in X holds not g1<r by A4; defpred X[Real] means for r st r in X holds not $1<r; consider Y such that A6: for p1 being Real holds p1 in Y iff X[p1] from SepReal; g1 is Real by XREAL_0:def 1; then A7: g1 in Y by A5,A6; now let r,p1 such that A8: r in X and A9: p1 in Y; A10: r+p in X by A3,A8; A11: r+0<r+p by A1,REAL_1:67; r+p<=p1 by A6,A9,A10; hence r<p1 by A11,AXIOMS:22; end; then consider g2 such that A12: for r,p1 st r in X & p1 in Y holds r<=g2 & g2<=p1 by Th8; A13: g2-p is Real by XREAL_0:def 1; A14: now assume not g2-p in Y; then consider r1 such that A15: r1 in X and A16: g2-p<r1 by A6,A13; g2-p+p<r1+p by A16,REAL_1:53; then g2-(p-p)<r1+p by XCMPLX_1:37; then A17: g2-0<r1+p by XCMPLX_1:14; r1+p in X by A3,A15; hence contradiction by A7,A12,A17; end; -p<0 by A1,REAL_1:26,50; then g2+-p<g2+0 by REAL_1:53; then g2-p<g2 by XCMPLX_0:def 8; hence contradiction by A2,A12,A14; end; theorem Th10:for r ex n st r<n proof let r; for r st r in NAT holds r+1 in NAT by AXIOMS:28; then consider p such that A1: p in NAT and A2: r<p by Th9; consider n1 such that A3: n1=p by A1; take n1; thus r<n1 by A2,A3; end; definition let X be real-membered set; attr X is bounded_above means :Def1: ex p st for r st r in X holds r<=p; attr X is bounded_below means :Def2: ex p st for r st r in X holds p<=r; end; definition let X; attr X is bounded means :Def3: X is bounded_below bounded_above; end; canceled 3; theorem Th14: X is bounded iff ex s st 0<s & for r st r in X holds abs(r)<s proof thus X is bounded implies ex s st 0<s & for r st r in X holds abs(r)<s proof assume A1: X is bounded; then X is bounded_above by Def3; then consider s1 such that A2: for r st r in X holds r<=s1 by Def1; X is bounded_below by A1,Def3; then consider s2 such that A3: for r st r in X holds s2<=r by Def2; take s=abs(s1)+abs(s2)+1; A4: 0<=abs(s1) by ABSVALUE:5; A5: 0<=abs(s2) by ABSVALUE:5; then 0+0<=abs(s1)+abs(s2) by A4,REAL_1:55; hence 0<s by REAL_1:67; let r such that A6: r in X; A7: s1<=abs(s1) by ABSVALUE:11; r<=s1 by A2,A6; then r<=abs(s1) by A7,AXIOMS:22; then r+0<=abs(s1)+abs(s2) by A5,REAL_1:55; then A8: r<s by REAL_1:67; A9: -abs(s2)<=s2 by ABSVALUE:11; A10: -abs(s1)<=0 by A4,REAL_1:26,50; s2<=r by A3,A6; then -abs(s2)<=r by A9,AXIOMS:22; then -abs(s1)+-abs(s2)<=0+r by A10,REAL_1:55; then -abs(s1)-abs(s2)<=r by XCMPLX_0:def 8; then -abs(s1)-abs(s2)+-1<r+0 by REAL_1:67; then A11: -abs(s1)-abs(s2)-1<r by XCMPLX_0:def 8; -abs(s1)-abs(s2)-1=-abs(s1)-(abs(s2)+1) by XCMPLX_1:36 .=-abs(s1)+-(1*(abs(s2)+1)) by XCMPLX_0:def 8 .=-(1*abs(s1))+(-1)*(abs(s2)+1) by XCMPLX_1:175 .=(-1)*abs(s1)+(-1)*(abs(s2)+1) by XCMPLX_1:175 .=(-1)*(abs(s1)+(abs(s2)+1)) by XCMPLX_1:8 .=-(1*(abs(s1)+(abs(s2)+1))) by XCMPLX_1:175 .=-(abs(s1)+abs(s2)+1) by XCMPLX_1:1; hence abs(r)<s by A8,A11,SEQ_2:9; end; given s such that 0<s and A12: for r st r in X holds abs(r)<s; now take g=s; let r; assume r in X; then A13: abs(r)<g by A12; r<=abs(r) by ABSVALUE:11; hence r<=g by A13,AXIOMS:22; end; then A14: X is bounded_above by Def1; now take g=-s; let r; assume r in X; then abs(r)<s by A12; then A15: -s<-abs(r) by REAL_1:50; -abs(r)<=r by ABSVALUE:11; hence g<=r by A15,AXIOMS:22; end; then X is bounded_below by Def2; hence thesis by A14,Def3; end; definition let r; redefine func {r} -> Subset of REAL; coherence proof {r} c= REAL proof let x be set; assume x in {r}; then x = r by TARSKI:def 1; hence thesis by XREAL_0:def 1; end; hence thesis; end; end; theorem Th15: {r} is bounded proof now take s=abs(r)+1; 0<=abs(r) by ABSVALUE:5; then 0+0<abs(r)+1 by REAL_1:67; hence 0<s; let r1 such that A1: r1 in {r}; abs(r)+0<s by REAL_1:67; hence abs(r1)<s by A1,TARSKI:def 1; end; hence thesis by Th14; end; theorem Th16: for X being real-membered set holds X is non empty bounded_above implies ex g st (for r st r in X holds r<=g) & for s st 0<s ex r st r in X & g-s<r proof let X be real-membered set; assume that A1:X is non empty and A2:X is bounded_above; consider r1 being real number such that A3:r1 in X by A1,MEMBERED:7; consider p1 such that A4:for r st r in X holds r<=p1 by A2,Def1; A5: X is Subset of REAL by MEMBERED:2; defpred X[Real] means for r st r in X holds r<=$1; consider Y such that A6:for p be Real holds p in Y iff X[p] from SepReal; p1 is Real by XREAL_0:def 1; then A7:p1 in Y by A4,A6; for r,p st r in X & p in Y holds r<=p by A6; then consider g1 such that A8:for r,p st r in X & p in Y holds r<=g1 & g1<=p by A5,AXIOMS:26; reconsider g1 as Real by XREAL_0:def 1; take g=g1; now given s1 such that A9:0<s1 and A10:for r st r in X holds not g-s1<r; g-s1 is Real by XREAL_0:def 1; then g-s1 in Y by A6,A10; then g<=g-s1 by A3,A8; then g-(g-s1)<=g-s1-(g-s1) by REAL_1:49; then g-(g-s1)<=0 by XCMPLX_1:14; hence contradiction by A9,XCMPLX_1:18; end; hence thesis by A7,A8; end; theorem Th17: for X being real-membered set holds (for r st r in X holds r<=g1) & (for s st 0<s ex r st (r in X & g1-s<r)) & (for r st r in X holds r<=g2) & (for s st 0<s ex r st (r in X & g2-s<r)) implies g1=g2 proof let X be real-membered set; assume that A1:for r st r in X holds r<=g1 and A2:for s st 0<s ex r st (r in X & g1-s<r) and A3:for r st r in X holds r<=g2 and A4:for s st 0<s ex r st (r in X & g2-s<r); A5:now assume g1<g2; then 0<g2-g1 by SQUARE_1:11; then consider r1 such that A6:r1 in X and A7:g2-(g2-g1)<r1 by A4; g1<r1 by A7,XCMPLX_1:18; hence contradiction by A1,A6; end; now assume g2<g1; then 0<g1-g2 by SQUARE_1:11; then consider r1 such that A8:r1 in X and A9:g1-(g1-g2)<r1 by A2; g2<r1 by A9,XCMPLX_1:18; hence contradiction by A3,A8; end; hence thesis by A5,AXIOMS:21; end; theorem Th18: for X being real-membered set holds X is non empty bounded_below implies ex g st (for r st r in X holds g<=r) & (for s st 0<s ex r st r in X & r<g+s) proof let X be real-membered set; assume that A1: X is non empty and A2:X is bounded_below; consider r1 being real number such that A3: r1 in X by A1,MEMBERED:7; consider p1 such that A4:for r st r in X holds p1<=r by A2,Def2; reconsider X as Subset of REAL by MEMBERED:2; defpred X[Real] means for r st r in X holds $1<=r; consider Y such that A5:for p be Real holds p in Y iff X[p] from SepReal; p1 is Real by XREAL_0:def 1; then A6:p1 in Y by A4,A5; for p,r st p in Y & r in X holds p<=r by A5; then consider g1 such that A7:for p,r st p in Y & r in X holds p<=g1 & g1<=r by AXIOMS:26; reconsider g1 as Real by XREAL_0:def 1; take g=g1; now given s1 such that A8:0<s1 and A9:for r st r in X holds not r<g+s1; g+s1 is Real by XREAL_0:def 1; then g+s1 in Y by A5,A9; then g+s1<=g by A3,A7; then g+s1-g<=g-g by REAL_1:49; then g+s1-g<=0 by XCMPLX_1:14; hence contradiction by A8,XCMPLX_1:26; end; hence thesis by A6,A7; end; theorem Th19: for X being real-membered set holds (for r st r in X holds g1<=r) & (for s st 0<s ex r st (r in X & r<g1+s)) & (for r st r in X holds g2<=r) & (for s st 0<s ex r st (r in X & r<g2+s)) implies g1=g2 proof let X be real-membered set; assume that A1:for r st r in X holds g1<=r and A2:for s st 0<s ex r st (r in X & r<g1+s) and A3:for r st r in X holds g2<=r and A4:for s st 0<s ex r st (r in X & r<g2+s); A5:now assume g1<g2; then 0<g2-g1 by SQUARE_1:11; then consider r1 such that A6:r1 in X and A7:r1<g1+(g2-g1) by A2; r1<g2 by A7,XCMPLX_1:27; hence contradiction by A3,A6; end; now assume g2<g1; then 0<g1-g2 by SQUARE_1:11; then consider r1 such that A8:r1 in X and A9:r1<g2+(g1-g2) by A4; r1<g1 by A9,XCMPLX_1:27; hence contradiction by A1,A8; end; hence thesis by A5,AXIOMS:21; end; definition let X be real-membered set; assume A1: X is non empty bounded_above; func upper_bound X -> real number means :Def4: (for r st r in X holds r<=it) & (for s st 0<s ex r st r in X & it-s<r); existence by A1,Th16; uniqueness by Th17; end; definition let X be real-membered set; assume A1: X is non empty bounded_below; func lower_bound X -> real number means :Def5: (for r st r in X holds it<=r) & (for s st 0<s ex r st r in X & r<it+s); existence by A1,Th18; uniqueness by Th19; end; definition let X; redefine func upper_bound X -> Real; coherence by XREAL_0:def 1; redefine func lower_bound X -> Real; coherence by XREAL_0:def 1; end; canceled 2; theorem Th22: lower_bound {r} = r & upper_bound {r} = r proof set X={r}; A1: X is bounded by Th15; then A2: X is bounded_below by Def3; A3: X is bounded_above by A1,Def3; now thus for p st p in X holds r<=p by TARSKI:def 1; now let s such that A4:0<s; take r1=r; thus r1 in X by TARSKI:def 1; r+0<r+s by A4,REAL_1:67; hence r1<r+s; end; hence for s st 0<s ex p st p in X & p<r+s; end; hence lower_bound X=r by A2,Def5; now thus for p st p in X holds p<=r by TARSKI:def 1; now let s such that A5:0<s; take r1=r; thus r1 in X by TARSKI:def 1; -s<0 by A5,REAL_1:26,50; then r+-s<r+0 by REAL_1:67; hence r-s<r1 by XCMPLX_0:def 8; end; hence for s st 0<s ex p st p in X & r-s<p; end; hence upper_bound X=r by A3,Def4; end; theorem Th23: lower_bound {r} = upper_bound {r} proof lower_bound {r}=r & upper_bound {r}=r by Th22; hence thesis; end; theorem X is bounded non empty implies lower_bound X <= upper_bound X proof assume that A1: X is bounded and A2: X is non empty; consider r being Element of REAL such that A3: r in X by A2,SUBSET_1:10; A4: X is bounded_below & X is bounded_above by A1,Def3; then A5: lower_bound X<=r by A3,Def5; r<=upper_bound X by A3,A4,Def4; hence thesis by A5,AXIOMS:22; end; theorem X is bounded non empty implies ((ex r,p st r in X & p in X & p<>r) iff lower_bound X < upper_bound X) proof assume that A1: X is bounded and A2: X is non empty; thus (ex r,p st r in X & p in X & p<>r) implies lower_bound X<upper_bound X proof given r,p such that A3: r in X and A4: p in X and A5: p<>r; A6: now assume A7:p<r; X is bounded_below by A1,Def3; then lower_bound X<=p by A4,Def5; then A8: lower_bound X<r by A7,AXIOMS:22; X is bounded_above by A1,Def3; then r<=upper_bound X by A3,Def4; hence lower_bound X<upper_bound X by A8,AXIOMS:22; end; now assume A9:r<p; X is bounded_below by A1,Def3; then lower_bound X<=r by A3,Def5; then A10: lower_bound X<p by A9,AXIOMS:22; X is bounded_above by A1,Def3; then p<=upper_bound X by A4,Def4; hence lower_bound X<upper_bound X by A10,AXIOMS:22; end; hence thesis by A5,A6,REAL_1:def 5; end; assume that A11: lower_bound X<upper_bound X and A12: for r,p st r in X & p in X holds p=r; consider r being Element of REAL such that A13: r in X by A2,SUBSET_1:10; now let x; now assume A14:x in X; then x is Real; hence x=r by A12,A13,A14; end; hence x in X iff x=r by A13; end; then X={r} by TARSKI:def 1; hence contradiction by A11,Th23; end; :: :: Theorems about the Convergence and the Limit :: theorem Th26:seq is convergent implies abs seq is convergent proof assume seq is convergent; then consider g1 such that A1:for p st 0<p ex n st for m st n<=m holds abs((seq.m)-g1)<p by SEQ_2:def 6; reconsider g1 as Real by XREAL_0:def 1; take g=abs(g1); let p; assume 0<p; then consider n1 such that A2:for m st n1<=m holds abs((seq.m)-g1)<p by A1; take n=n1; let m; assume n<=m; then A3:abs((seq.m)-g1)<p by A2; A4:abs(seq.m)-g<=abs((seq.m)-g1) by ABSVALUE:18; abs(g1-seq.m)=abs(-((seq.m)-g1)) by XCMPLX_1:143 .=abs((seq.m)-g1) by ABSVALUE:17; then g-abs(seq.m)<=abs((seq.m)-g1) by ABSVALUE:18; then -abs((seq.m)-g1)<= -(g-abs(seq.m)) by REAL_1:50; then -abs((seq.m)-g1)<=abs(seq.m)-g by XCMPLX_1:143; then abs(abs(seq.m)-g)<=abs((seq.m)-g1) by A4,ABSVALUE:12; then abs(((abs seq).m)-g) <=abs((seq.m)-g1) by SEQ_1:16; hence thesis by A3,AXIOMS:22; end; theorem seq is convergent implies lim abs seq = abs lim seq proof assume A1:seq is convergent; then A2:abs seq is convergent by Th26; now let p; assume 0<p; then consider n1 such that A3:for m st n1<=m holds abs((seq.m)-(lim seq))<p by A1,SEQ_2:def 7; take n=n1; let m; assume n<=m; then A4:abs((seq.m)-(lim seq))<p by A3; A5:abs(seq.m)-abs(lim seq)<=abs((seq.m)-(lim seq)) by ABSVALUE:18; abs((lim seq)-seq.m)=abs(-((seq.m)-(lim seq))) by XCMPLX_1:143 .=abs((seq.m)-(lim seq)) by ABSVALUE:17; then abs(lim seq)-abs(seq.m)<=abs((seq.m)-(lim seq)) by ABSVALUE:18; then -abs((seq.m)-(lim seq))<=-(abs(lim seq)-abs(seq.m)) by REAL_1:50; then -abs((seq.m)-(lim seq))<=abs(seq.m)-abs(lim seq) by XCMPLX_1:143; then abs(abs(seq.m)-abs(lim seq))<=abs((seq.m)-(lim seq)) by A5,ABSVALUE:12; then abs(((abs seq).m)-abs(lim seq)) <=abs((seq.m)-(lim seq)) by SEQ_1:16 ; hence abs(((abs seq).m)-abs(lim seq))<p by A4,AXIOMS:22; end; hence thesis by A2,SEQ_2:def 7; end; theorem abs seq is convergent & lim abs seq=0 implies seq is convergent & lim seq=0 proof assume that A1:abs seq is convergent and A2:lim abs seq=0; A3:-abs seq is convergent by A1,SEQ_2:23; A4:lim(-abs seq)=lim abs seq by A1,A2,REAL_1:26,SEQ_2:24; A5:now let n; -abs(seq.n)<=seq.n by ABSVALUE:11; then A6: -((abs seq).n)<=seq.n by SEQ_1:16; seq.n<=abs(seq.n) by ABSVALUE:11; hence (-abs seq).n<=seq.n & seq.n<=(abs seq).n by A6,SEQ_1:14,16; end; hence seq is convergent by A1,A3,A4,SEQ_2:33; thus lim seq=0 by A1,A2,A3,A4,A5,SEQ_2:34; end; theorem Th29:seq1 is_subsequence_of seq & seq is convergent implies seq1 is convergent proof assume that A1:seq1 is_subsequence_of seq and A2:seq is convergent; consider g1 such that A3:for p st 0<p ex n st for m st n<=m holds abs((seq.m)-g1)<p by A2,SEQ_2:def 6; consider Nseq such that A4:seq1=seq*Nseq by A1,SEQM_3:def 10; take g=g1; let p; assume 0<p; then consider n1 such that A5:for m st n1<=m holds abs((seq.m)-g1)<p by A3; take n=n1; let m such that A6:n<=m; m<=Nseq.m by SEQM_3:33; then A7:n<=Nseq.m by A6,AXIOMS:22; seq1.m=seq.(Nseq.m) by A4,SEQM_3:31; hence abs((seq1.m)-g)<p by A5,A7; end; theorem Th30:seq1 is_subsequence_of seq & seq is convergent implies lim seq1=lim seq proof assume that A1:seq1 is_subsequence_of seq and A2:seq is convergent; A3:seq1 is convergent by A1,A2,Th29; consider Nseq such that A4:seq1=seq*Nseq by A1,SEQM_3:def 10; now let p; assume 0<p; then consider n1 such that A5:for m st n1<=m holds abs((seq.m)-(lim seq))<p by A2,SEQ_2:def 7; take n=n1; let m such that A6:n<=m; m<=Nseq.m by SEQM_3:33; then A7:n<=Nseq.m by A6,AXIOMS:22; seq1.m=seq.(Nseq.m) by A4,SEQM_3:31; hence abs((seq1.m)-(lim seq))<p by A5,A7; end; hence thesis by A3,SEQ_2:def 7; end; theorem Th31:seq is convergent & (ex k st for n st k<=n holds seq1.n=seq.n) implies seq1 is convergent proof assume that A1:seq is convergent and A2:ex k st for n st k<=n holds seq1.n=seq.n; consider g1 such that A3:for p st 0<p ex n st for m st n<=m holds abs((seq.m)-g1)<p by A1,SEQ_2:def 6; consider k such that A4:for n st k<=n holds seq1.n=seq.n by A2; take g=g1; let p; assume 0<p; then consider n1 such that A5:for m st n1<=m holds abs((seq.m)-g1)<p by A3; A6:now assume A7:k<=n1; take n=n1; let m; assume A8:n<=m; then k<=m by A7,AXIOMS:22; then seq1.m=seq.m by A4; hence abs((seq1.m)-g)<p by A5,A8; end; now assume A9:n1<=k; take n=k; let m; assume A10:n<=m; then n1<=m by A9,AXIOMS:22; then abs((seq.m)-g1)<p by A5; hence abs((seq1.m)-g)<p by A4,A10; end; hence ex n st for m st n<=m holds abs(seq1.m-g)<p by A6; end; theorem seq is convergent & (ex k st for n st k<=n holds seq1.n=seq.n) implies lim seq=lim seq1 proof assume that A1:seq is convergent and A2:ex k st for n st k<=n holds seq1.n=seq.n; A3:seq1 is convergent by A1,A2,Th31; consider k such that A4:for n st k<=n holds seq1.n=seq.n by A2; now let p; assume 0<p; then consider n1 such that A5:for m st n1<=m holds abs((seq.m)-(lim seq))<p by A1,SEQ_2:def 7; A6:now assume A7:k<=n1; take n=n1; let m; assume A8:n<=m; then k<=m by A7,AXIOMS:22; then seq1.m=seq.m by A4;hence abs((seq1.m)-(lim seq))<p by A5,A8; end; now assume A9:n1<=k; take n=k; let m; assume A10:n<=m; then n1<=m by A9,AXIOMS:22; then abs((seq.m)-(lim seq))<p by A5; hence abs((seq1.m)-(lim seq))<p by A4,A10; end;hence ex n st for m st n<=m holds abs(seq1.m-(lim seq))<p by A6; end; hence thesis by A3,SEQ_2:def 7; end; theorem Th33: seq is convergent implies ((seq ^\k) is convergent & lim (seq ^\k)=lim seq) proof assume A1:seq is convergent; A2:seq^\k is_subsequence_of seq by SEQM_3:47; hence seq^\k is convergent by A1,Th29; thus lim (seq ^\k)=lim seq by A1,A2,Th30; end; canceled; theorem Th35: seq is convergent & (ex k st seq=seq1 ^\k) implies seq1 is convergent proof assume that A1: seq is convergent and A2: ex k st seq=seq1 ^\k; consider k such that A3: seq=seq1 ^\k by A2; consider g1 such that A4: for p st 0<p ex n st for m st n<=m holds abs(seq.m-g1)<p by A1,SEQ_2:def 6; take g=g1; let p; assume 0<p; then consider n1 such that A5: for m st n1<=m holds abs(seq.m-g1)<p by A4; take n=n1+k; let m; assume A6: n<=m; then consider l such that A7: m=n1+k+l by NAT_1:28; m-k=n1+k+l+-k by A7,XCMPLX_0:def 8; then m-k=n1+l+k+-k by XCMPLX_1:1; then m-k=n1+l+k-k by XCMPLX_0:def 8; then m-k=n1+l+(k-k) by XCMPLX_1:29; then m-k=n1+l+0 by XCMPLX_1:14; then consider m1 such that A8: m1=m-k; A9: now assume not n1<=m1; then m1+k<n1+k by REAL_1:53; then m-(k-k)<n1+k by A8,XCMPLX_1:37; then m-0<n1+k by XCMPLX_1:14; hence contradiction by A6; end; A10: m1+k=m-(k-k) by A8,XCMPLX_1:37 .=m-0 by XCMPLX_1:14 .=m; abs(seq.m1-g1)<p by A5,A9; hence abs(seq1.m-g)<p by A3,A10,SEQM_3:def 9; end; theorem seq is convergent & (ex k st seq=seq1 ^\k) implies lim seq1 =lim seq proof assume that A1: seq is convergent and A2: ex k st seq=seq1 ^\k; A3: seq1 is convergent by A1,A2,Th35; consider k such that A4: seq=seq1 ^\k by A2; now let p; assume 0<p; then consider n1 such that A5: for m st n1<=m holds abs(seq.m-(lim seq))<p by A1,SEQ_2:def 7; take n=n1+k; let m; assume A6: n<=m; then consider l such that A7: m=n1+k+l by NAT_1:28; m-k=n1+k+l+-k by A7,XCMPLX_0:def 8; then m-k=n1+l+k+-k by XCMPLX_1:1; then m-k=n1+l+k-k by XCMPLX_0:def 8; then m-k=n1+l+(k-k) by XCMPLX_1:29; then m-k=n1+l+0 by XCMPLX_1:14; then consider m1 such that A8: m1=m-k; A9: now assume not n1<=m1; then m1+k<n1+k by REAL_1:53; then m-(k-k)<n1+k by A8,XCMPLX_1:37; then m-0<n1+k by XCMPLX_1:14; hence contradiction by A6; end; A10: m1+k=m-(k-k) by A8,XCMPLX_1:37 .=m-0 by XCMPLX_1:14 .=m; abs(seq.m1-(lim seq))<p by A5,A9; hence abs(seq1.m-(lim seq))<p by A4,A10,SEQM_3:def 9; end; hence thesis by A3,SEQ_2:def 7; end; theorem Th37:seq is convergent & lim seq<>0 implies ex k st (seq ^\k) is_not_0 proof assume that A1:seq is convergent and A2:lim seq<>0; consider n1 such that A3:for m st n1<=m holds abs(lim seq)/2<abs((seq.m)) by A1,A2,SEQ_2:30; take k=n1; now let n; 0<=n by NAT_1:18; then 0+k<=n+k by REAL_1:55; then abs(lim seq)/2<abs((seq.(n+k))) by A3; then A4:abs(lim seq)/2<abs(((seq ^\k).n)) by SEQM_3:def 9; 0<abs(lim seq) by A2,ABSVALUE:6; then 0/2<abs(lim seq)/2 by REAL_1:73; hence (seq ^\k).n<>0 by A4,ABSVALUE: 7; end; hence thesis by SEQ_1:7; end; theorem seq is convergent & lim seq<>0 implies ex seq1 st seq1 is_subsequence_of seq & seq1 is_not_0 proof assume that A1:seq is convergent and A2:lim seq <>0; consider k such that A3:seq ^\k is_not_0 by A1,A2,Th37; take seq ^\k; thus thesis by A3,SEQM_3:47; end; theorem Th39: seq is constant implies seq is convergent proof assume seq is constant; then consider r such that A1:for n holds seq.n=r by SEQM_3:def 6; take g=r; let p such that A2:0<p; take n=0; let m such that n<=m; abs((seq.m)-g)=abs(r-g) by A1 .=abs(0) by XCMPLX_1:14 .=0 by ABSVALUE:7; hence abs((seq.m)-g)<p by A2; end; theorem Th40:(seq is constant & r in rng seq or seq is constant & (ex n st seq.n=r)) implies lim seq=r proof assume A1: seq is constant & r in rng seq or seq is constant & (ex n st seq.n=r); A2:now assume that A3:seq is constant and A4:r in rng seq; consider r1 such that A5:rng seq={r1} by A3,SEQM_3:15; consider r2 such that A6:for n holds seq.n=r2 by A3,SEQM_3:def 6; A7: r=r1 by A4,A5,TARSKI:def 1; A8: seq is convergent by A3,Th39; now let p such that A9:0<p; take n=0; let m such that n<=m; m in NAT; then m in dom seq by FUNCT_2:def 1; then seq.m in rng seq by FUNCT_1:def 5; then r2 in rng seq by A6; then r2=r by A5,A7,TARSKI:def 1; then seq.m=r by A6; then abs((seq.m)-r)=abs(0) by XCMPLX_1:14 .=0 by ABSVALUE:7; hence abs((seq.m)-r)<p by A9; end; hence lim seq =r by A8,SEQ_2:def 7; end; now assume that A10: seq is constant and A11: ex n st seq.n=r; consider n such that A12: seq.n=r by A11; n in NAT; then n in dom seq by FUNCT_2:def 1; hence lim seq= r by A2,A10,A12,FUNCT_1:def 5; end; hence thesis by A1,A2; end; theorem seq is constant implies for n holds lim seq=seq.n by Th40; theorem seq is convergent & lim seq<>0 implies for seq1 st seq1 is_subsequence_of seq & seq1 is_not_0 holds lim (seq1")=(lim seq)" proof assume that A1:seq is convergent and A2:lim seq<>0; let seq1 such that A3:seq1 is_subsequence_of seq and A4:seq1 is_not_0; A5:seq1 is convergent by A1,A3,Th29; lim seq1=lim seq by A1,A3,Th30; hence lim seq1"=(lim seq)" by A2,A4,A5,SEQ_2:36; end; theorem Th43: 0<r & (for n holds seq.n=1/(n+r)) implies seq is convergent proof assume that A1:0<r and A2:for n holds seq.n=1/(n+r); take g=0; let p; assume 0<p; then A3:0<p" by REAL_1:72; consider k1 such that A4:p"<k1 by Th10; take n=k1; let m such that A5:n<=m; p"+0<k1+r by A1,A4,REAL_1:67; then 1/(k1+r)<1/p" by A3,SEQ_2:10; then A6: 1/(k1+r)<1*p"" by XCMPLX_0:def 9; 0<n by A3,A4,AXIOMS:22; then A7: 0+0<n+r by A1,REAL_1:67; n+r<=m+r by A5,AXIOMS:24; then 1/(m+r)<=1/(n+r) by A7,Th1; then A8:1/(m+r)<p by A6,AXIOMS:22; A9:seq.m=1/(m+r) by A2; 0<=m by NAT_1:18; then 0+0<m+r by A1,REAL_1:67; then 0/(m+r)<1/(m+r) by REAL_1:73; hence abs(seq.m-g)<p by A8,A9,ABSVALUE:def 1; end; theorem Th44: 0<r & (for n holds seq.n=1/(n+r)) implies lim seq=0 proof assume that A1:0<r and A2:for n holds seq.n=1/(n+r); A3:seq is convergent by A1,A2,Th43; now let p; assume 0<p; then A4:0<p" by REAL_1:72; consider k1 such that A5:p"<k1 by Th10; take n=k1; let m such that A6:n<=m; p"+0<k1+r by A1,A5,REAL_1:67; then 1/(k1+r)<1/p" by A4,SEQ_2:10; then A7: 1/(k1+r)<1*p"" by XCMPLX_0:def 9; 0<n by A4,A5,AXIOMS:22; then A8: 0+0<n+r by A1,REAL_1:67; n+r<=m+r by A6,AXIOMS:24; then 1/(m+r)<=1/(n+r) by A8,Th1; then A9:1/(m+r)<p by A7,AXIOMS:22; A10:seq.m=1/(m+r) by A2; 0<=m by NAT_1:18; then 0+0<m+r by A1,REAL_1:67; then 0/(m+r)<1/(m+r) by REAL_1:73; hence abs(seq.m-0)<p by A9,A10, ABSVALUE:def 1; end; hence thesis by A3,SEQ_2:def 7; end; theorem (for n holds seq.n=1/(n+1)) implies seq is convergent & lim seq=0 by Th43,Th44; theorem 0<r & (for n holds seq.n=g/(n+r)) implies seq is convergent & lim seq=0 proof assume that A1:0<r and A2:for n holds seq.n=g/(n+r); reconsider r1 = r as Real by XREAL_0:def 1; deffunc U(Nat) = 1/($1+r1); consider seq1 such that A3:for n holds seq1.n=U(n) from ExRealSeq; A4:seq1 is convergent by A1,A3,Th43; then A5:g(#)seq1 is convergent by SEQ_2:21; A6:lim (g(#)seq1)=g*(lim seq1) by A4,SEQ_2:22 .=g*0 by A1,A3,Th44 .=0; now let n; thus (g(#)seq1).n=g*(seq1.n) by SEQ_1:13 .=g*(1/(n+r)) by A3 .=g*(1*(n+r)") by XCMPLX_0:def 9 .=g/(n+r) by XCMPLX_0:def 9 .=seq.n by A2; end; hence seq is convergent & lim seq=0 by A5,A6,FUNCT_2:113; end; theorem Th47: 0<r & (for n holds seq.n=1/(n*n+r)) implies seq is convergent proof assume that A1: 0<r and A2: for n holds seq.n=1/(n*n+r); take g=0; let p; assume 0<p; then A3: 0<p" by REAL_1:72; consider k1 such that A4: p"<k1 by Th10; take n=k1; let m such that A5: n<=m; A6: 0<k1 by A3,A4,AXIOMS:22; then 0+1<=k1 by NAT_1:38; then 1*p"<k1*k1 by A3,A4,Th6; then p"+0<k1*k1+r by A1,REAL_1:67; then 1/(k1*k1+r)<1/p" by A3,SEQ_2:10; then A7: 1/(k1*k1+r)<1*p"" by XCMPLX_0:def 9; A8: n*n<=m*m by A5,A6,Th7; 0<=n*n by NAT_1:18; then A9: 0+0<n*n+r by A1,REAL_1:67; n*n+r<=m*m+r by A8,AXIOMS:24; then 1/(m*m+r)<=1/(n*n+r) by A9,Th1; then A10: 1/(m*m+r)<p by A7,AXIOMS:22; A11: seq.m=1/(m*m+r) by A2; 0<=m*m by NAT_1:18; then 0+0<m*m+r by A1,REAL_1:67; then 0<(m*m+r)" by REAL_1:72; then 0<1/(m*m+r) by XCMPLX_1:217; hence abs(seq.m-g)<p by A10,A11, ABSVALUE:def 1; end; theorem Th48: 0<r & (for n holds seq.n=1/(n*n+r)) implies lim seq=0 proof assume that A1: 0<r and A2: for n holds seq.n=1/(n*n+r); A3: seq is convergent by A1,A2,Th47; now let p; assume 0<p; then A4: 0<p" by REAL_1:72; consider k1 such that A5: p"<k1 by Th10; take n=k1; let m such that A6: n<=m; A7: 0<k1 by A4,A5,AXIOMS:22; then 0+1<=k1 by NAT_1:38; then 1*p"<k1*k1 by A4,A5,Th6; then p"+0<k1*k1+r by A1,REAL_1:67; then 1/(k1*k1+r)<1/p" by A4,SEQ_2:10; then A8: 1/(k1*k1+r)<1*p"" by XCMPLX_0:def 9; A9: n*n<=m*m by A6,A7,Th7; 0<=n*n by NAT_1:18; then A10: 0+0<n*n+r by A1,REAL_1:67; n*n+r<=m*m+r by A9,AXIOMS:24; then 1/(m*m+r)<=1/(n*n+r) by A10,Th1; then A11: 1/(m*m+r)<p by A8,AXIOMS:22; A12: seq.m=1/(m*m+r) by A2; 0<=m*m by NAT_1:18; then 0+0<m*m+r by A1,REAL_1:67; then 0<(m*m+r)" by REAL_1:72; then 0<1/(m*m+r) by XCMPLX_1:217; hence abs(seq.m-0)<p by A11,A12,ABSVALUE:def 1; end; hence thesis by A3,SEQ_2:def 7; end; theorem (for n holds seq.n=1/(n*n+1)) implies seq is convergent & lim seq=0 by Th47,Th48; theorem 0<r & (for n holds seq.n=g/(n*n+r)) implies seq is convergent & lim seq=0 proof assume that A1:0<r and A2:for n holds seq.n=g/(n*n+r); reconsider r1 = r as Real by XREAL_0:def 1; deffunc U(Nat) = 1/($1*$1+r1); consider seq1 such that A3:for n holds seq1.n=U(n) from ExRealSeq; A4:seq1 is convergent by A1,A3,Th47; then A5:g(#)seq1 is convergent by SEQ_2:21; A6:lim (g(#)seq1)=g*(lim seq1) by A4,SEQ_2:22 .=g*0 by A1,A3,Th48 .=0; now let n; thus (g(#)seq1).n=g*(seq1.n) by SEQ_1:13 .=g*(1/(n*n+r)) by A3 .=g*(1*(n*n+r)") by XCMPLX_0:def 9 .=g/(n*n+r) by XCMPLX_0:def 9 .=seq.n by A2; end; hence seq is convergent & lim seq=0 by A5,A6,FUNCT_2:113; end; theorem Th51:seq is non-decreasing & seq is bounded_above implies seq is convergent proof assume that A1:seq is non-decreasing and A2:seq is bounded_above; consider r2 such that A3:for n holds seq.n<r2 by A2,SEQ_2:def 3; defpred X[Real] means ex n st $1=seq.n; consider X such that A4:for p be Real holds p in X iff X[p] from SepReal; A5: seq.0 in X by A4; A6:X is bounded_above iff ex r st for p st p in X holds p<=r by Def1; A7: now take r=r2; let p; assume p in X; then ex n1 st p=seq.n1 by A4; hence p<=r by A3; end; take g=upper_bound X; let s; assume A8:0<s; then consider p1 such that A9:p1 in X and A10:upper_bound X-s<p1 by A5,A6,Def4; consider n1 such that A11:p1=seq.n1 by A4,A9; take n=n1; let m; assume n<=m; then seq.n<=seq.m by A1,SEQM_3:12; then g-s<seq.m by A10,A11,AXIOMS:22; then g+ -s<seq.m by XCMPLX_0:def 8; then A12:-s<seq.m -g by REAL_1:86; seq.m in X by A4; then seq.m<=g by A6,A7,Def4; then seq.m +0<g+s by A8,REAL_1:67; then seq.m -g<s by REAL_1:84; hence abs(seq.m -g)<s by A12,SEQ_2:9; end; theorem Th52:seq is non-increasing & seq is bounded_below implies seq is convergent proof assume that A1:seq is non-increasing and A2: seq is bounded_below; consider r1 such that A3:for n holds r1< seq.n by A2,SEQ_2:def 4; defpred X[Real] means ex n st $1=seq.n; consider X such that A4:for p be Real holds p in X iff X[p] from SepReal; A5: seq.0 in X by A4; A6:X is bounded_below iff ex r st for p st p in X holds r<=p by Def2; A7: now take r=r1; let p; assume p in X; then ex n1 st p=seq.n1 by A4; hence r<=p by A3; end; take g=lower_bound X; let s; assume A8:0<s; then consider p1 such that A9:p1 in X and A10:p1<lower_bound X+s by A5,A6,Def5; consider n1 such that A11:p1=seq.n1 by A4,A9; take n=n1; let m; assume n<=m; then seq.m<=seq.n by A1,SEQM_3:14; then seq.m <g+s by A10,A11,AXIOMS:22; then A12:seq.m -g<s by REAL_1:84; seq.m in X by A4; then 0+g<=seq.m by A6,A7,Def5; then A13:0<=seq.m-g by REAL_1:84; -s<0 by A8,REAL_1:26,50; hence abs(seq.m -g)<s by A12,A13,SEQ_2:9; end; theorem Th53:seq is monotone & seq is bounded implies seq is convergent proof assume that A1:seq is monotone and A2:seq is bounded; A3:seq is bounded_below by A2,SEQ_2:def 5; seq is bounded_above by A2,SEQ_2:def 5; then A4: seq is non-decreasing implies thesis by Th51; seq is non-increasing implies thesis by A3,Th52; hence thesis by A1,A4,SEQM_3:def 7; end; theorem seq is bounded_above & seq is non-decreasing implies for n holds seq.n<=lim seq proof assume that A1:seq is bounded_above and A2:seq is non-decreasing; A3:seq is convergent by A1,A2,Th51; let m; deffunc U(Nat) = seq.m; consider seq1 such that A4:for n holds seq1.n=U(n) from ExRealSeq; A5:seq1 is constant by A4,SEQM_3:def 6; then A6:seq1 is convergent by Th39; seq1.0=seq.m by A4; then A7:lim seq1=seq.m by A5,Th40; deffunc U(Nat) = seq.($1+m); consider seq2 such that A8:for n holds seq2.n=U(n) from ExRealSeq; A9:seq2=seq^\m by A8,SEQM_3:def 9; then A10:lim seq2=lim seq by A3,Th33; A11: seq2 is convergent by A3,A9,Th33; now let n; A12:seq1.n=seq.m by A4; seq2.n=seq.(m+n) by A8; hence seq1.n<=seq2.n by A2,A12,SEQM_3:11; end; hence seq.m<=lim seq by A6,A7,A10,A11,SEQ_2:32; end; theorem seq is bounded_below & seq is non-increasing implies for n holds lim seq <= seq.n proof assume that A1:seq is bounded_below and A2:seq is non-increasing; A3: seq is convergent by A1,A2,Th52; let m; deffunc U(Nat) = seq.m; consider seq1 such that A4: for n holds seq1.n= U(n) from ExRealSeq; A5: seq1 is constant by A4,SEQM_3:def 6; then A6: seq1 is convergent by Th39; seq1.0=seq.m by A4; then A7: lim seq1=seq.m by A5,Th40; deffunc U(Nat) = seq.($1+m); consider seq2 such that A8: for n holds seq2.n=U(n) from ExRealSeq; A9: seq2=seq^\m by A8,SEQM_3:def 9; then A10: lim seq2=lim seq by A3,Th33; A11: seq2 is convergent by A3,A9,Th33; now let n; A12:seq1.n=seq.m by A4; seq2.n=seq.(m+n) by A8; hence seq2.n<=seq1.n by A2,A12,SEQM_3:13; end; hence lim seq<=seq.m by A6,A7,A10,A11,SEQ_2:32; end; theorem Th56:for seq ex Nseq st seq*Nseq is monotone proof let seq; defpred X[Nat] means for m st $1<m holds seq.$1<seq.m; consider XN being Subset of NAT such that A1:for n holds n in XN iff X[n] from SubsetEx; A2: now given k1 such that A3: for n st n in XN holds n<=k1; A4: now let k; assume k1<k; then not k in XN by A3; then consider m such that A5: k<m and A6: not seq.k<seq.m by A1; take n=m; thus k<n by A5; thus seq.n<=seq.k by A6; end; set seq1=seq ^\(1+k1); seq1 is_subsequence_of seq by SEQM_3:47; then consider Nseq such that A7: seq1=seq*Nseq by SEQM_3:def 10; A8: now let k; 0<=k by NAT_1:18; then 0<k+1 by NAT_1:38; then 0+k1<k+1+k1 by REAL_1:67; then consider n such that A9: k+1+k1<n and A10: seq.n<=seq.(k+1+k1) by A4; consider m such that A11: n=k+1+k1+m by A9,NAT_1:28; n-(1+k1)=k+1+k1+m+-(1+k1) by A11,XCMPLX_0:def 8; then n-(1+k1)=k+(1+k1)+m+-(1+k1) by XCMPLX_1:1; then n-(1+k1)=k+(1+k1)+-(1+k1)+m by XCMPLX_1:1; then n-(1+k1)=k+((1+k1)+-(1+k1))+m by XCMPLX_1:1; then n-(1+k1)=k+0+m by XCMPLX_0:def 6; then consider n1 such that A12: n1=n-(1+k1); take l=n1; now assume not k<l; then n-(1+k1)+(1+k1)<=k+(1+k1) by A12,AXIOMS:24; then n-(1+k1)+(1+k1)<=k+1+k1 by XCMPLX_1:1; then n-((1+k1)-(1+k1))<=k+1+k1 by XCMPLX_1:37; then n-0<=k+1+k1 by XCMPLX_1:14; hence contradiction by A9; end; hence k<l; A13: n=n+0 .=n+(-(1+k1)+(1+k1)) by XCMPLX_0:def 6 .=n+-(1+k1)+(1+k1) by XCMPLX_1:1 .=l+(1+k1) by A12,XCMPLX_0:def 8; seq.n<=seq.(k+(1+k1)) by A10,XCMPLX_1:1; then seq.n<=seq1.k by SEQM_3:def 9; hence seq1.l<=seq1.k by A13,SEQM_3:def 9; end; defpred P[Nat,set,set] means for k,l st k=$2 & l=$3 holds k<l & seq1.l<=seq1.k & for m st k<m & seq1.m<=seq1.k holds l<=m; A14: for n being Nat for x being Element of NAT ex y being Element of NAT st P[n,x,y] proof let n be Nat,x be Element of NAT; defpred X[Nat] means x<$1 & seq1.$1<=seq1.x; A15: ex n st X[n] by A8; ex l st X[l] & for m st X[m] holds l <= m from Min(A15); then consider l such that A16: x<l and A17: seq1.l<=seq1.x and A18: for m st x<m & seq1.m<=seq1.x holds l <= m; take l; thus thesis by A16,A17,A18; end; reconsider 0'=0 as Element of NAT qua non empty set; consider f being Function of NAT,NAT such that f.0 = 0' and A19: for n being Element of NAT holds P[n,f.n,f.(n+1)] from RecExD(A14); A20: rng f c= NAT by RELSET_1:12; then A21: rng f c= REAL by XBOOLE_1:1; A22: dom f =NAT by FUNCT_2:def 1; then reconsider f as Real_Sequence by A21,FUNCT_2:def 1,RELSET_1:11; A23: now let n; f.n in rng f by A22,FUNCT_1:def 5; hence f.n is Nat by A20; end; now let n; (f.n is Nat) & f.(n+1) is Nat by A23; hence f.n<f.(n+1) by A19; end; then reconsider f as increasing Seq_of_Nat by A20,SEQM_3:def 8,def 11; take Nseq1=Nseq*f; now let n; seq1.(f.(n+1))<=seq1.(f.n) by A19; then (seq1*f).(n+1)<=seq1.(f.n) by SEQM_3:31; then (seq*Nseq*f).(n+1)<=(seq1*f).n by A7,SEQM_3:31; then (seq*Nseq1).(n+1)<=(seq*Nseq*f).n by A7,RELAT_1:55; hence (seq*Nseq1).(n+1)<=(seq*Nseq1).n by RELAT_1:55; end; then seq*Nseq1 is non-increasing by SEQM_3:def 14; hence seq*Nseq1 is monotone by SEQM_3:def 7; end; now assume A24:for l ex n st n in XN & not n<=l; defpred P[Nat,set,set] means for k,l being Element of NAT st k=$2 & l=$3 holds l in XN & k<l & for m st m in XN & k<m holds l<=m; A25: for n being Nat for x being Element of NAT ex y being Element of NAT st P[n,x,y] proof let n be Nat,x be Element of NAT; defpred X[Nat] means $1 in XN & x<$1; A26: ex n st X[n] by A24; ex l st X[l] & for m st X[m] holds l <= m from Min(A26); then consider l such that A27: l in XN and A28: x<l and A29: for m st m in XN & x<m holds l <= m; take y=l; thus P[n,x,y] by A27,A28,A29; end; consider n1 such that A30:n1 in XN & not n1<=0 by A24; reconsider 0'=n1 as Element of NAT qua non empty set; consider f being Function of NAT,NAT such that A31: f.0 = 0' and A32: for n being Element of NAT holds P[n,f.n,f.(n+1)] from RecExD(A25); A33: rng f c= NAT by RELSET_1:12; then A34: rng f c= REAL by XBOOLE_1:1; A35: dom f =NAT by FUNCT_2:def 1; then reconsider f as Real_Sequence by A34,FUNCT_2:def 1,RELSET_1:11; A36: now let n; thus n is Element of NAT; A37: f.n in rng f by A35,FUNCT_1:def 5; hence f.n is Nat by A33; thus f.n is Element of NAT by A33,A37; end; A38: now let n; defpred X[Nat] means f.$1 in XN; A39: X[0] by A30,A31; A40: now let k such that X[k]; (f.k is Element of NAT) & f.(k+1) is Element of NAT by A36; hence X[k+1] by A32; end; thus for n holds X[n] from Ind(A39,A40); end; A41: now let n; (f.n is Element of NAT) & f.(n+1) is Element of NAT by A36; hence f.n<f.(n+1) by A32; end; then reconsider f as increasing Seq_of_Nat by A33,SEQM_3:def 8,def 11; take Nseq=f; now let n; A42: Nseq.n in XN by A38; Nseq.n<Nseq.(n+1) by A41; then seq.(Nseq.n)<seq.(Nseq.(n+1)) by A1,A42; then (seq*Nseq).n<seq.(Nseq.(n+1)) by SEQM_3:31; hence (seq*Nseq).n<(seq*Nseq).(n+1) by SEQM_3:31; end; then seq*Nseq is increasing by SEQM_3:def 11; then seq*Nseq is non-decreasing by SEQM_3:23; hence seq*Nseq is monotone by SEQM_3:def 7; end; hence thesis by A2; end; theorem :: BOLZANO-WEIERSTRASS THEOREM Th57:seq is bounded implies ex seq1 st seq1 is_subsequence_of seq & seq1 is convergent proof assume A1:seq is bounded; consider Nseq such that A2: seq*Nseq is monotone by Th56; take seq1=seq*Nseq; thus seq1 is_subsequence_of seq by SEQM_3:def 10; then seq1 is bounded by A1,SEQM_3:58; hence seq1 is convergent by A2,Th53; end; theorem :: CAUCHY THEOREM seq is convergent iff for s st 0<s ex n st for m st n<=m holds abs(seq.m -seq.n)<s proof thus seq is convergent implies for s st 0<s ex n st for m st n<=m holds abs(seq.m -seq.n)<s proof assume seq is convergent; then consider g1 such that A1: for s st 0<s ex n st for m st n<=m holds abs(seq.m -g1)<s by SEQ_2:def 6; let s; assume 0<s; then 0<s/2 by SEQ_2:3; then consider n1 such that A2: for m st n1<=m holds abs(seq.m -g1)<s/2 by A1; take n=n1; let m such that A3: n<=m; abs(seq.n -g1)<s/2 by A2; then abs(-(g1-seq.n))<s/2 by XCMPLX_1:143; then A4: abs(g1-seq.n)<s/2 by ABSVALUE:17; abs(seq.m -g1)<s/2 by A2,A3; then abs(seq.m -g1)+abs(g1-seq.n)<s/2+s/2 by A4,REAL_1:67; then A5: abs(seq.m -g1)+abs(g1-seq.n)<s by XCMPLX_1:66; A6: abs(seq.m -g1+(g1-seq.n))<=abs(seq.m -g1)+abs(g1-seq.n) by ABSVALUE:13; abs(seq.m -g1+(g1-seq.n))=abs(seq.m -(g1-(g1-seq.n))) by XCMPLX_1:37 .=abs(seq.m -(g1-g1+seq.n)) by XCMPLX_1:37 .=abs(seq.m -(0+seq.n)) by XCMPLX_1:14 .=abs(seq.m -seq.n); hence thesis by A5,A6,AXIOMS:22; end; assume A7: for s st 0<s ex n st for m st n<=m holds abs(seq.m -seq.n)<s; then consider n1 such that A8: for m st n1<=m holds abs(seq.m -seq.n1)<1; consider r1 such that A9: 0<r1 and A10: for m st m<=n1 holds abs(seq.m)<r1 by SEQ_2:16; now take r=r1+abs(seq.n1)+1; A11: 0<=abs(seq.n1) by ABSVALUE:5; then 0+0<r1+abs(seq.n1) by A9,REAL_1:67; hence 0<r by REAL_1:67; let m; A12: now assume m<=n1; then abs(seq.m)<r1 by A10; then abs(seq.m)+0<r1+abs(seq.n1) by A11,REAL_1:67; hence abs(seq.m)<r by REAL_1:67; end; now assume n1<=m; then A13: abs(seq.m -seq.n1)<1 by A8; abs(seq.m)-abs(seq.n1)<=abs(seq.m -seq.n1) by ABSVALUE:18; then abs(seq.m)-abs(seq.n1)<1 by A13,AXIOMS:22; then abs(seq.m)-abs(seq.n1)+abs(seq.n1)<1+abs(seq.n1) by REAL_1:53; then abs(seq.m)-(abs(seq.n1)-abs(seq.n1))<1+abs(seq.n1) by XCMPLX_1: 37 ; then abs(seq.m)-0<1+abs(seq.n1) by XCMPLX_1:14; then 0+abs(seq.m)<r1+(abs(seq.n1)+1) by A9,REAL_1:67; hence abs(seq.m)<r by XCMPLX_1:1; end; hence abs(seq.m)<r by A12; end; then seq is bounded by SEQ_2:15; then consider seq1 such that A14: seq1 is_subsequence_of seq and A15: seq1 is convergent by Th57; consider g1 such that A16: for s st 0<s ex n st for m st n<=m holds abs(seq1.m -g1)<s by A15,SEQ_2:def 6; consider Nseq such that A17: seq1=seq*Nseq by A14,SEQM_3:def 10; take g=g1; let s; assume 0<s; then A18: 0<s/3 by Th4; then consider n1 such that A19: for m st n1<=m holds abs(seq1.m -g1)<s/3 by A16; consider n2 such that A20: for m st n2<=m holds abs(seq.m -seq.n2)<s/3 by A7,A18; take n=n1+n2; let m such that A21: n<=m; n1<=n by NAT_1:37; then n1<=m by A21,AXIOMS:22; then abs((seq*Nseq).m -g1)<s/3 by A17,A19; then A22: abs(seq.(Nseq.m) -g1)<s/3 by SEQM_3:31; n2<=n by NAT_1:37; then A23: n2<=m by A21,AXIOMS:22; then A24: abs(seq.m -seq.n2)<s/3 by A20; m<=Nseq.m by SEQM_3:33; then n2<=Nseq.m by A23,AXIOMS:22; then abs(seq.(Nseq.m) -seq.n2)<s/3 by A20; then abs(-(seq.n2-seq.(Nseq.m)))<s/3 by XCMPLX_1:143; then abs(seq.n2-seq.(Nseq.m))<s/3 by ABSVALUE:17; then A25: abs(seq.m -seq.n2)+abs(seq.n2-seq.(Nseq.m))<s/3+s/3 by A24,REAL_1:67 ; A26: abs(seq.m -seq.n2+(seq.n2-seq.(Nseq.m)))<= abs(seq.m -seq.n2)+abs(seq.n2-seq.(Nseq.m)) by ABSVALUE:13; abs(seq.m -seq.n2+(seq.n2-seq.(Nseq.m))) =abs(seq.m -(seq.n2-(seq.n2-seq.(Nseq.m)))) by XCMPLX_1:37 .=abs(seq.m -(seq.n2-seq.n2+seq.(Nseq.m))) by XCMPLX_1:37 .=abs(seq.m -(0+seq.(Nseq.m))) by XCMPLX_1:14 .=abs(seq.m -seq.(Nseq.m)); then abs(seq.m -seq.(Nseq.m))<s/3+s/3 by A25,A26,AXIOMS:22; then A27: abs(seq.m -seq.(Nseq.m))+abs(seq.(Nseq.m) -g1)<s/3+s/3+s/3 by A22,REAL_1:67; A28: abs(seq.m -seq.(Nseq.m)+(seq.(Nseq.m) -g1))<= abs(seq.m -seq.(Nseq.m))+abs(seq.(Nseq.m) -g1) by ABSVALUE:13; abs(seq.m -seq.(Nseq.m)+(seq.(Nseq.m) -g1)) =abs(seq.m -(seq.(Nseq.m)-(seq.(Nseq.m) -g1))) by XCMPLX_1:37 .=abs(seq.m -(seq.(Nseq.m)-seq.(Nseq.m) +g1)) by XCMPLX_1:37 .=abs(seq.m -(0+g1)) by XCMPLX_1:14 .=abs(seq.m -g1); then abs(seq.m -g1)<s/3+s/3+s/3 by A27,A28,AXIOMS:22; hence abs(seq.m -g)<s by XCMPLX_1:69; end; theorem seq is constant & seq1 is convergent implies lim (seq+seq1) =(seq.0) + lim seq1 & lim (seq-seq1) =(seq.0) - lim seq1 & lim (seq1-seq) =(lim seq1) -seq.0 & lim (seq(#)seq1) =(seq.0) * (lim seq1) proof assume that A1: seq is constant and A2: seq1 is convergent; A3: seq is convergent by A1,Th39; hence lim (seq+seq1)=(lim seq)+(lim seq1) by A2,SEQ_2:20 .=(seq.0)+(lim seq1) by A1,Th40; thus lim (seq-seq1)=(lim seq)-(lim seq1) by A2,A3,SEQ_2:26 .=(seq.0)-(lim seq1) by A1,Th40; thus lim (seq1-seq)=(lim seq1)-(lim seq) by A2,A3,SEQ_2:26 .=(lim seq1)-(seq.0) by A1,Th40; thus lim (seq(#)seq1)=(lim seq)*(lim seq1) by A2,A3,SEQ_2:29 .=(seq.0)*(lim seq1) by A1,Th40; end;

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