reserve            x for object,
               X,Y,Z for set,
         i,j,k,l,m,n for Nat,
                 r,s for Real,
                  no for Element of OrderedNAT,
                   A for Subset of [:NAT,NAT:];
reserve X,Y,X1,X2 for non empty set,
          cA1,cB1 for filter_base of X1,
          cA2,cB2 for filter_base of X2,
              cF1 for Filter of X1,
              cF2 for Filter of X2,
             cBa1 for basis of cF1,
             cBa2 for basis of cF2;
reserve T for non empty TopSpace,
        s for Function of [:NAT,NAT:], the carrier of T,
        M for Subset of the carrier of T;
reserve cF3,cF4 for Filter of the carrier of T;
reserve Rseq for Function of [:NAT,NAT:],REAL;

theorem Th63:
  for m being non zero Nat ex n being Nat st for n1,n2 being Nat st
    n <= n1 & n <= n2 holds |. dblseq_ex_1.(n1,n2) - 0 .| < 1/m
  proof
    let m be non zero Nat;
    now
      let n1,n2 be Nat;
      assume that
A1:   m <= n1 and
      m <= n2;
      m + 0 < n1 + 1 by A1,XREAL_1:8;
      then 1/(n1 + 1) < 1/m by XREAL_1:76;
      then |.1/(n1 + 1) - 0.| < 1/m by ABSVALUE:def 1;
      hence |.dblseq_ex_1.(n1,n2) - 0.| < 1/m by Def5;
    end;
    hence ex n being Nat st for n1,n2 being Nat st
    n <= n1 & n <= n2 holds |. dblseq_ex_1.(n1,n2) - 0 .| < 1/m;
  end;
