reserve r, s, t for Real;
reserve seq for Real_Sequence,
  X, Y for Subset of REAL;

theorem Th38: ::: SEQ_2
  seq is convergent & seq is non-zero & lim seq = 0 implies seq" is non bounded
proof
  assume that
A1: seq is convergent and
A2: seq is non-zero and
A3: lim seq = 0;
  given r such that
A4: for n being Nat holds (seq").n<r;
  given s such that
A5: for n being Nat holds s<(seq").n;
  set aa = |.r.|, ab = |.s.|;
  set rab = 1/(aa+ab);
A6: 0 <= aa & 0 <= ab by COMPLEX1:46;
A7: now
    let n be Element of NAT;
    set g = seq.n, t = (seq").n;
    set At = |.t.|;
    t= 1/g by VALUED_1:10;
    then
A8: At = 1/|.g.| by ABSVALUE:7;
    t" = g " " by VALUED_1:10;
    then t <> 0 by A2,SEQ_1:5,XCMPLX_1:202; then
A9: 0 < At by COMPLEX1:47;
    s<t & t<r by A4,A5;
    then At < aa+ab by Th37;
    then At" > (aa+ab)" by A9,XREAL_1:88;
    hence |.seq.n.| > rab by A8;
  end;
A10: (seq").1<r by A4;
A11: s<(seq").1 by A5;
  now
    assume 0 >= aa+ab; then
A12: aa+ab = 0 by A6;
    then r = 0 by Th36;
    hence contradiction by A11,A10,A12,Th36;
  end;
  then consider n being Nat such that
A13: for m being Nat st n<=m holds |.seq.m-0 .|<rab by A1,A3,
SEQ_2:def 7;
A14: n in NAT by ORDINAL1:def 12;
  |.seq.n-0 .|<rab by A13;
  hence contradiction by A7,A14;
end;
