proving completness of a metric space Given $$d(a,b):=\left|\frac{a}{1+|a|}-\frac b{1+|b|}\right|$$
I want to know if $(\mathbb R,d)$ is complete.
My attempt: I think it's complete because given a Cauchy sequence the sequence is bounded. By Bolzano-Weierstrass the sequence has a limit point and and I know a Cauchy sequence has not more than one limit point so the sequence converge.
But this seems too easy for me and I first guess this space isn't complete by intuition but I couldn't find any counterexample. So is this correct?
 A: Take $(n)_{n\in\mathbb N}$ . Can you check this a Cauchy-sequence? And (obviously) doesn't converge?
A: Your issue is that you think that bounded under the standard metric and bounded under the metric $d$ you described is different. A sequence which is Cauchy under the metric $d$ that you provided will be bounded in the sense that 
$$
d(x,0) = \frac{|x|}{1+|x|} < 1
$$ 
for every real $x$ for instance (the space ($\mathbb R$,$d$) is a bounded space! So of course every sequence will be bounded...). 
You need not confuse the usual metric and the metric you are given!... One reason that you should "feel" that the space is not complete is because of this : 
$$
\{ d(x,0) \, | \, x \in \mathbb R \} = [0,1[
$$
but the function $d(-,0) : (\mathbb R,d) \to (\mathbb R, |\cdot|)$ is continuous, hence if a sequence with $d(x_n,0) \nearrow 1$ would be convergent, you would have $x_n \to x$ and $d(x,0) = 1$, which is an element that doesn't exist in your space... can you find such a sequence $x_n$?
Hope that helps,
