Prove that $(\Bbb R^n,\overline{d})$ is a complete metric space

Question

Consider the metric space $(\Bbb R^n,\overline{d})$ where $\overline{d}=\frac{d(x,y)}{1+ d(x,y)}$. Prove that it is complete.

That is, to conclude that all the Cauchy sequence in $(\Bbb R^n,\overline{d})$ converge in $(\Bbb R^n,\overline{d})$.

To do this, I aim to explicitly conclude that:

$1.$ All the Cauchy sequence in $(\Bbb R^n,\overline{d})$ are also Cauchy in $(\Bbb R^n,{d})$.

$2.$By completeness of $(\Bbb R^n,{d})$, all the Cauchy sequence converges in $(\Bbb R^n,{d})$.

$3.$And if by $d(x,y)<\epsilon$ we can conclude that $\overline{d}(x,y)<\epsilon$, we can show that the point the sequence in $(\Bbb R^n,{d})$ convergesto is also where the sequence in $(\Bbb R^n,\overline{d})$ converges to. Then every sequence in $(\Bbb R^n,\overline{d})$ converges.

I have difficulty of proving every Cauchy sequence in $(\Bbb R^n,\overline{d})$ is also a Cauchy sequence in $(\Bbb R^n,\overline{d})$ and the final part of concluding that the sequence with same terms but under different metric converges to the same point.

Could someone please give a proof? Thanks!

Answer 1

Hint: Assume that $(\mathbb R^n, d)$ is a complete metric space. Denote $e(x, y)=\min\{1, d(x, y)\}$ and you can prove that $e$ is also a metric on $\mathbb R^n$. Next, obeserve that $\overline{d}(x, y) \leqslant e(x, y) \leqslant 2\overline{d}(x, y)$, i.e. metrices $e$ and $\overline{d}$ are equivalent. Hence, in order to prove $\overline{d}$ is complete, it remains to prove that $e$ is complete and this is not difficult to see.

Answer 2

If $\{x_n\}$ is Cauchy in $(\mathbb{R}^d,\bar{d})$, fix $\epsilon > 0$. Choose $N$ large enough that for $n>m>N$ we have $\bar{d}(x_n,x_m) < \epsilon$. That is,

$$\frac{d(x_n,x_m)}{1+d(x_n,x_m)} < \epsilon $$

Rearranging,

$$d(x_n,x_m) \le \frac{\epsilon}{1-\epsilon} $$

Note that the rhs tends to $0$ as $\epsilon \to 0$