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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?

user41728
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    http://www.proofwiki.org/wiki/Completion_Theorem_(Metric_Space) – anon Aug 19 '13 at 22:23
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    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
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    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 00:40
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    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 03:25
  • 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. http://en.wikipedia.org/wiki/P-adic_number – Mark Bennet Aug 20 '13 at 15:08
  • Related. – Andrés E. Caicedo Sep 12 '13 at 22:29

2 Answers2

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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 $\mathbb Q$.

Hagen von Eitzen
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    Bingo-it took me several years to understand this conception of Cauchy completion and it's extremely intuitive and powerful. All teachers of analysis should use it. – Mathemagician1234 Nov 08 '14 at 22:20
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In fact, there are no examples at all. Any metric space admits a completion, and every Cauchy sequence in the original space is again Cauchy in the completion, where it converges.

For your given example, $[0,1]$ is a completion of $(0,1).$

Cameron Buie
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