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The Wikipedia page on complete metric spaces gives various examples of metric spaces that are and are not complete -

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.

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In the passage you quote, they haven't proved that $[0,1]$ is complete. – littleO Nov 1 '12 at 0:58
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. – Geoff Robinson Nov 1 '12 at 1:01
They do say it is complete, they literally say "the closed interval $[0, 1]$ is complete"..? – dukenukem Nov 1 '12 at 1:07
Yes, they say that it is complete; they do not say that they are proving this statement, and indeed they are not. – Brian M. Scott Nov 1 '12 at 1:13
People often say things without proving it, and they don't mean to imply they have proved it. – littleO Nov 1 '12 at 1:16

To show that a closed interval is complete, you can do the following: You first show that every Cauchy-sequence is bounded. It therefore has a convergent subsequence and the limit is in the interval since it is closed. Then you verify that if a sub-sequence of a Cauchy sequence converges to a point, the sequence itself converges to the same point.

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If the set of points of the sequence is finite, the sequence clearly converges. If not, the closed interval is compact. Hence, every infinite subset has a limit point. It is easy to show that the sequence converges to this limit point:

Let the limit point be $L$. Fix $\varepsilon > 0$. Pick $N \in \mathbb N$ so that $\forall n, m > N : |a_n - a_m| < \frac{\varepsilon}{2}$. Pick $a_{p}$ so that $p > N, |a_p - L| < \frac{\varepsilon}{2}$. This is always possible since $L$ is a limit point. We have: $$ \forall n > N : |a_n - L| \le |a_n - a_p| + |a_p - L| < \varepsilon $$

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The closed interval $[a, b]$ is complete.

Proof: Observe that $[a, b] \subseteq \mathbb R$ where $\mathbb R$ is a complete metric space. Consider $[a, b]^C = (-\infty, a) \cup (b, \infty)$. Now

  • If $x_0 \in (-\infty, a)$, choose $\displaystyle \delta = \frac{a + x_0}{2}$, then $x_0 \in B(x_0, \delta) \subseteq (-\infty, a) \subseteq [a, b]^C$.

  • If $x_0 \in (b, \infty)$, choose $\displaystyle \delta = \frac{b + x_0}{2}$, then $x_0 \in B(x_0, \delta) \subseteq (b, \infty) \subseteq [a, b]^C$.

Thus $[a, b]^C$ contains a ball about each of its points, so $[a, b]^C$ is open and thus $[a, b]$ is closed.

Suppose $(x_n)$ is a Cauchy sequence in $[a, b]$. Since every Cauchy sequence in $\mathbb R$ converges, it follows that $x_n \to x \in \overline{[a, b]}$ where $\overline{[a, b]}$ is the closure of $[a, b]$ ($\overline{[a, b]} = [a, b] \cup \{\text{accumulation points}\}$) which is the smallest closed set containing $[a, b]$. But observe that the smallest closed set containing $[a, b]$ is $[a, b]$. Hence $x \in \overline{[a, b]} = [a, b]$ Thus $x_n \to x \in [a, b]$ and we can conclude that $[a, b]$ is a complete metric space.

In particular, $a=0$ and $b=1$ implies $[0, 1]$ is complete.

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