Not Complete Metric Space Suppose I have a metric $\rho: R^n$ x $R^n \rightarrow R$ defined by $\rho (x, y) = \frac{d(x, y)}{1+d(x, y)}$ and I need to show that $(R^n, \rho)$ is not a complete metric space?
Can anyone suggest a Cauchy sequence that does not converge in $(R^n, \rho)$?
 A: Let $(X,d)$ be any metric space, and define $\rho(x,y) := \dfrac{d(x,y)}{1+d(x,y)}$. Then $\rho$ is a metric on $X$ (proof omitted) that induces the same topology as $d$, and $(X,\rho)$ is a complete metric space if and only if $(X,d)$ is complete.
To see that $\rho$ induces the same topology as $d$, note that for $r > 0$ we have
$$B_r^d(x) = \{y \in X : d(x,y) < r\} = \left\{ y \in X : \rho(x,y) < \frac{r}{1+r}\right\} = B_{\frac{r}{1+r}}^\rho(x),$$
and for $0 < s < 1$, we have
$$B_s^\rho(x) = B_{\frac{s}{1-s}}^d(x),$$
for all $x \in X$.
By a similar reasoning, we can see that $\rho$ has the same Cauchy-sequences as $d$. Since $\rho \leqslant d$, every $d$-Cauchy sequence is also a $\rho$-Cauchy sequence. Conversely, let $(x_n)$ be a $\rho$-Cauchy sequence. Let $\varepsilon > 0$ be given. Then, since $(x_n)$ is a $\rho$-Cauchy sequence, there is an $N_\varepsilon \in \mathbb{N}$ with
$$m,n \geqslant N_\varepsilon \Rightarrow \rho(x_n,x_m) < \frac{\varepsilon}{1+\varepsilon}.$$
But that means $d(x_n,x_m) < \varepsilon$ for $n,m \geqslant N_\varepsilon$.
