How to show the metric space is complete? the space is the Real line with bounded metric (i.e. $d/(1+d)$, $d$: euclidean).
We thought that since nd the space real line with euclidean metric is complete and the bounded metric is smaller than d, then any cauchy sequence that converges with eucliedan metric will also converge with bounded metric. but there could be more cauchy seq. with bounded metric, so we have to show that these also converge. this is the part we couldn't do.

Write $e(x,y) = d(x,y)/(1+d(x,y))$. Let $\{x_n\}$ be a Cauchy sequence in the metric $e$. 
Let $\epsilon > 0$ be given. Let $f(t) = \dfrac{t}{1+t}$. Since $f$ is increasing on $[0,\infty)$ and $\lim_{t \to 0^+} f(t) = 0$, there exists $\delta$ with the property that $e(x,y) < \delta$ implies $d(x,y) < \epsilon$. Let $N$ be an index with the property that $n,m \ge N$ implies $e(x_n,x_m) < \delta$.  We conclude that $n,m \ge N$ implies $d(x_m,x_n) < \epsilon$. It follows that $\{x_n\}$ is Cauchy in the metric $d$ and converges to a number $x$.
Since $e(x_n,n) \le d(x_n,x)$ you get $x_n \to x$ in the metric $e$. Thus $e$ is a complete metric.
