reserve n,n1,m,k for Nat;
reserve x,y for set;
reserve s,g,g1,g2,r,p,p2,q,t for Real;
reserve s1,s2,s3 for Real_Sequence;
reserve Nseq for increasing sequence of NAT;
reserve X for Subset of REAL;

theorem Th9:
  (for p st p in X ex r,n st 0<r & (for m st n<m holds r<|.s1.m-p.|))
  implies for s2 st s2 is subsequence of s1 holds not (s2 is convergent &
  lim s2 in X)
proof
  assume that
A1: for p st p in X ex r,n st 0<r & for m st n<m holds r<|.s1.m-p.| and
A2: not for s2 st s2 is subsequence of s1 holds not (s2 is convergent &
  lim s2 in X);
  consider s2 such that
A3: s2 is subsequence of s1 and
A4: s2 is convergent and
A5: lim s2 in X by A2;
  consider r,n such that
A6: 0<r and
A7: for m st n<m holds r<|.s1.m - lim s2.| by A1,A5;
  consider n1 such that
A8: for m st n1<=m holds |.s2.m-(lim s2).|<r by A4,A6,SEQ_2:def 7;
  consider NS being increasing sequence of NAT such that
A9: s2 = s1*NS by A3,VALUED_0:def 17;
  reconsider k = n + 1 + n1 as Element of NAT by ORDINAL1:def 12;
  |.(s1*NS).k-(lim s2).|<r by A8,A9,NAT_1:11;
  then
A10: |.s1.(NS.k)-(lim s2).|<r by FUNCT_2:15;
  n + 1 <= k by NAT_1:11;
  then
A11: n < k by NAT_1:13;
  k<=NS.k by SEQM_3:14;
  then n<NS.k by A11,XXREAL_0:2;
  hence contradiction by A7,A10;
end;
