Cauchy sequence in non-complete metric space

Question

I am considering the metric space ($X, d$) , in which $X = \mathbb{R}$ and $d(x, y) = \big | \frac{x|x|}{1 + x^2} - \frac{y |y|}{1+ y^2} \big |$.

To prove this is not complete, I came up with a sequence $x_n = n$, $n \in \mathbb{N}$ whose limit is $+\infty$ and therefore, not in $X$. I want to show this is a Cauchy sequence with respect to the given metric. Without loss of generality, we can suppose $n \geq m$ so that:

$d(x_n, x_m) = \frac{n^2}{1 + n^2} - \frac{m^2}{1 + m^2} = \frac{n^2 - m^2}{(1 + n^2) (1 + m^2)} < \frac{n^2 - m^2}{n^2 m^2}$.

I want to show that $\frac{n^2 - m^2}{n^2 m^2} < \epsilon$ for some $N$ such that $n \geq m > N$.

Can anyone help me proceed?

Thanks.

If the aforementioned sequence is not Cauchy, then please give an example of one which is Cauchy.

Answer 1

I think I figured this out.

Let $n \geq m > N$ so that

$\frac{n^2 - m^2}{n^2 m^2} < \frac{n^2 - m^2}{n^2 N^2} = \frac{1}{N^2} \left (1 - \frac{m^2}{n^2} \right ) < \frac{1}{N^2}$.

Therefore, if $N > \frac{1}{\sqrt{\epsilon}}$, we have $d(x_n, d_m) < \epsilon$ for all $n \geq m > N$.

The case for $m \geq n > N$ is the same.

Answer 2

This is very easy. $\frac {n|n|} {1+n^{2}} \to 1$ as $ n \to \infty$ in the usual metric. Hence $|\frac {n|n|} {1+n^{2}}-\frac {m|m|} {1+m^{2}}| \to 0$ as $n,m \to \infty$ (because any convergent sequence (in the usual metric) is Cauchy). Thus $d(n,m) \to 0$ as $n,m \to \infty$.

If this Cauchy sequence converges the there exists $x$ such that $d(n,x) \to 0$ but this gives $\frac {x|x|} {1+x^{2}}=1$ or $x^{2}=1+x^{2}$ (because $x >0$ necessarily) which is impossible.