# A “non-trivial” example of a Cauchy sequence that does not converge?

A Cauchy sequence doesn't necessarily converge, e.g. take the sequence $(1/n)$ in the space $(0,1)$.

Maybe my intuition is wrong but I tend to think of this as, "it does converge but what it converges to is not in the space". Are there any examples of a Cauchy sequence that does not converge and avoids this type of saying?

• proofwiki.org/wiki/Completion_Theorem_(Metric_Space) – anon Aug 19 '13 at 22:23
• You can construct the completion of $X$ for every metric space $X$, in which every Cauchy sequence converges. So in a way, every example is of the type "it converges, but the limit is not in the space". – Daniel Fischer Aug 19 '13 at 22:23
• It's better to think of a Cauchy sequence as one that ought to converge but doesn't, owing to the absence of the point to which it 'wants' to converge. The completion supplies the missing point. – Brian M. Scott Aug 20 '13 at 0:40
• Argh. I just noticed an important omission in that last comment: I meant to write such a Cauchy sequence, meaning one that does not converge. – Brian M. Scott Aug 20 '13 at 3:25
• It is possible to have different metrics on the same set, so that 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 (not a Cauchy Sequence) in an alternative metric. en.wikipedia.org/wiki/P-adic_number – Mark Bennet Aug 20 '13 at 15:08

For any metric space $Q$ we can define the completion, that is a (bigger) metric space $R$ such that $Q$ is a (dense) subspace of $R$ and all Cauchy sequences in $Q$ have a limit in $R$. So the Cauchy sequences in $Q$ "do converge but what they converge to is not in the space". This is precisely one way of defining $\mathbb R$ from $\mathbb Q$.