Is the sequence 1 n Cauchy?
Claim: The sequence { 1 n } is Cauchy. Proof: Let ? > 0 be given and let N > 2 ?. Then for any n, m > N, one has 0 < 1 n, 1 m < ? 2. Therefore, ? > 1 n + 1 m = | 1 n | + | 1 m | ? | 1 n ? 1 m |. Thus, the sequence is Cauchy as was to be shown. Everything you wrote is correct, but I think your point would be better illustrated by = ? 1.
Is x n Cauchy in R?
Examples: 1. (X;d) = Q, as a subspace of R with the usual metric. Take x 0= 2 and defne x n+1= xn 2 +1 xn The sequence continues 3=2, 17=12, 577=408;:::and indeed x n!xwhere x=x 2 +1 x , i.e., x2= 2. But this isn’t in Q. Thus (x n) is Cauchy in R, since it converges to p 2 when we think of it as a sequence in R.
How can we guarantee that (x n) will be Cauchy?
n 1) for n1. This gives a sequence (x n); if it is Cauchy and (X;d) is complete, then x= lim n!1x nexists and xshould solve x= ?(x). How can we guarantee that (x n) will be Cauchy? Note that d(x n;x n+1) = d(?(x n 1);?(x n)), so to get (x n) Cauchy we want ?to shrink distances.
How do you prove that a metric space is Cauchy?
Prove directly that it’s Cauchy, by showing how the nin the de nition depends upon . De nition: A metric space (X;d) is complete if every Cauchy sequence in Xconverges in X (i.e., to a limit that’s in X).
1.4 Cauchy Sequence in R