Metric that makes interval $[a,b]$ not complete The interval $[a,b]$ is complete under the Euclidean metric. Is there a metric that makes $[a,b]$ not complete?
Could you give general means to solve this question?
 A: Let $f : [a,b] \to [a,b)$ be a bijection. Then, define $d(x,y)=|f(x)-f(y)|$. Now, $([a,b],d)$ is complete iff $[a,b)$ is complete.
A: Every metric on $[a,b]$ compatible with the usual topology is complete.  
Suppose $d$ is a metric on $[a,b]$ compatible with the usual topology (i.e., generates the same open sets).  Since $[a,b]$ is compact (and metrizable) it is also sequentially compact, meaning that every sequence has a convergent subsequence.  If $\langle x_n \rangle_n$ is a $d$-Cauchy sequence (i.e., Cauchy with respect to the metric $d$) then by sequential compactness it has a convergent subsequence $\langle x_{n_i} \rangle_i$, say $\lim_{i \rightarrow \infty} x_{n_i} = x$.  It is relatively straightforward to show that $\lim_{n \rightarrow \infty} x_n = x$.
Note: This does not contradict Seirios's answer; it just indicates that the topology induced by that metric is different than the usual one.
A: Seiros has given lots of examples, one for each bijection $f:[a,b]\to[a,b)$, but those bijections tend to be a bit complicated.  Here's a simpler-looking example.  Define $d(x,y)$ to be the usual distance $|x-y|$ except that, if one of $x$ and $y$ is $b$ while the other is not, then you define the distance $d(x,y)$ to be $|x-y|+1$.  Intuitively, this just detaches the point $b$ from the interval $[a,b]$ and moves it one unit to the right.  A sequence that, in the usual metric, approaches $b$ from the left is, in the new metric, still a Cauchy sequence but it doesn't converge because its intended limit has moved away.  (As Arthur Fischer points out, the new metric must induce a different topology than the usual one; indeed, my new metric makes $\{b\}$ open.)
