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The Wikipedia page on complete metric spaces gives various examples of metric spaces that are and are not complete - http://en.wikipedia.org/wiki/Complete_metric_space

Here's a few lines in particular -

The open interval $(0, 1)$, again with the absolute value metric, is not complete either. The sequence defined by $x_n = \frac{1}{n}$ is Cauchy, but does not have a limit in the given space. However the closed interval $[0, 1]$ is complete; the given sequence does have a limit in this interval and the limit is zero.

I know that a metric space M is complete if every Cauchy sequence of points in M has a limit that is also in M, but that example above just considers one Cauchy sequence and then announces that the interval is complete.

How can they say it is complete without considering all possible Cauchy sequences in the interval which is what the definition demands..and for that matter, how would it be possible to consider all Cauchy sequences in an interval given that, I presume, there are an infinite number of them?

Can anyone clear this up for me...I have a feeling I'm overlooking something straightforward.

  • 3
    In the passage you quote, they haven't proved that $[0,1]$ is complete.2012-11-01
  • 2
    I think they are just saying that that particular sequence converges in $[0,1]$, whereas it did not in $(0,1)$. They are no claiming that the covergence of that particular sequence is in itself sufficient to make the space complete. There is a general theorem that a close subspace of a complete metric space is complete, which may be applied here as $\mathbb{R}$ is complete.2012-11-01
  • 0
    They do say it is complete, they literally say "the closed interval $[0, 1]$ is complete"..?2012-11-01
  • 3
    Yes, they say that it is complete; they do **not** say that they are proving this statement, and indeed they are not.2012-11-01
  • 1
    People often say things without proving it, and they don't mean to imply they have proved it.2012-11-01
  • 0
    You're probably aware that $\mathbb{R}$ is complete. What can you say about a closed subspace of a complete metric space?2012-11-01
  • 0
    A subspace $(Y,d)$ of a complete metric space $(X,d)$ is complete iff $Y$ is closed in $X$.2015-12-16

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