Showing that a metric space is complete 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.
 A: 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.
A: 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:
$$
\forall n > N : |a_n - L| \le |a_n - a_p| + |a_p - L| < \varepsilon
$$
A: 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.
