Are there non-convergent cauchy sequences?

Question

By non convergent I mean DIVERGENT and NOT that it converges to a value outside the space. Examples are appreciated. Thanks

http://math.stackexchange.com/questions/471577/a-non-trivial-example-of-a-cauchy-sequence-that-does-not-converge — njguliyev, Aug 20 '13 at 14:41
It is possible to have different metrics on the same set, each metric gives rise to a different completion (I'm thinking of p-adic numbers). It is quite possible for a Cauchy Sequence in one metric to be divergent in an alternative metric. http://en.wikipedia.org/wiki/P-adic_number — Mark Bennet, Aug 20 '13 at 15:01

score 0 · Answer 1 · answered Aug 20 '13 at 14:56

If you talk only about real sequence you have: Let $a_{n}$ Cauchy sequence then $a_{n}$ is bounded.

infact for all $\varepsilon>0$ exists $n_{0}$ such that $$|a_{n}-a_{m}|<\varepsilon $$ $\forall n,m\geq n_{0}$

if $n\geq n_{0}$ $$|a_{n}|\leq |a_{n}-a_{n_{0}}|+|a_{n_{0}}|<C$$ if $n\leq n_{0}$ $$|a_{n}|\leq C$$

score 0 · Answer 2 · answered Aug 20 '13 at 14:58

0

I think I found an example: (1/x)sin(1/(x-floorx) is convergent i R but not in Z.

answered Aug 20 '13 at 14:58

user1118686

sorry, my bad. That function is not defined from neither N to R neither N to Z... – user1118686 Aug 20 '13 at 15:07

2 Answers2