# 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". 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. 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. 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 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 $\mathbb Q$.