A not complete metric space? Please ,how to prove that the space $\mathbb{R}$ endowed with the metric $d(x,y)=|e^x-e^y|$ is  not a complete space?
I don't find a Cauchy sequence   but not convergent 
Please
Thank you.
 A: HINT: Is $d(-100,-101)$ a large, medium, or small number? Look at the graph of $y=e^x$.
A: Consider the sequence $x_n=-n$.  Clearly it doesn't converge. For, suppose $x_n \to x$ for some real number $x$. Then $|e^{-n}−e^x|→0$. Hence, $e^x=0$, which cannot be. 
However, $N<n<m \implies |e^{x_n}-e^{x_m}|=e^{-n}-e^{-m}<e^{-N} \to 0$ as $N \to \infty$.  So the sequence is Cauchy.
A: Here is a less direct way of viewing the problem. The function $f(x) = e^x$ defines an isometry from your metric space to the set $\mathbb{R}^+$ of positive real numbers with the usual metric. This fact is straightforward: $f$ is a bijection (the inverse function is the natural logarithm), and if $d$ is your metric and $d_1$ is the usual metric, then for $x,y \in \mathbb{R}$, we have $$d(x,y) = |e^x - e^y| = |(f(x) - f(y)| = d_1(f(x),f(y)).$$
So, $(\mathbb{R}, d)$ and $(\mathbb{R}^+, d_1)$ are isomorphic as metric spaces. We can see that the latter space is not complete by considering any sequence in $\mathbb{R}^+$ converging to 0. Such a sequence is Cauchy, but does not converge to a point of $\mathbb{R}^+$.
