Show that $(X, d_2)$ is incomplete

I have a set $X = [0, \infty)$ and two metrics:

$$d_1(x, y) = |x-y|$$ $$d_2(x, y) = \left| \frac{x}{1+x} - \frac{y}{1+y} \right|$$

I already showed that $d_1$ is equivalent to $d_2$. Now I have to show that $(X, d_2)$ is incomplete. As far as I have understood it I have to find a sequence that converges in both (because that depends on the topology that is the same) but not cauchy (since that depend on the metric).

What would such a sequence be?

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It is actually the other way around. You need to find a Cauchy sequence (in $(X,d_2)$) that is not convergent. –  azarel Apr 15 '12 at 18:56
So I need to find a sequence that is Cauchy in $(x, d_2)$ but does not converge with respect to $d_2$? –  queueoverflow Apr 15 '12 at 19:01

TO show that $(X,d_2)$ is not complete, just show that there is a sequence contained in $X$, which is Cauchy for $d_2$ but not convergent for $d_2$ (forget $d_1$). Take $x_n=n$. Since $$d_2(x_n,x_m)=\left|\frac n{n+1}-\frac m{m+1}\right|=\left|\frac{n+1-1}{n+1}-\frac {m+1-1}{m+1}\right|=\left|\frac 1{n+1}-\frac 1{m+1}\right|\leq \frac 1n+\frac 1m$$ which converges to $0$ as $m,n\to+\infty$, the sequence $\{x_n\}$ is Cauchy for $d_2$. If it was convergent to $l\in X$ then we would have $\lim_{n\to+\infty}\left|\frac n{n+1}-\frac l{1+l}\right|=0$ hence $\lim_{n\to+\infty}\left|\frac 1{n+1}-\frac 1{1+l}\right|=0$ so $\frac 1{1+l}=0$. It's not possible.

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Awesome, this makes a lot of sense. Thanks! –  queueoverflow Apr 15 '12 at 19:03

Hint: The sequence $x_n=n$ is Cauchy but does not converges.

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How do I learn to come up with answers like that? I stared onto that problem and I could not think of any sequence … but it seems to trivial now. –  queueoverflow Apr 15 '12 at 19:11
@queueoverflow You sort of need a sequence that converges to a point not in your space (a point in the completion ). So $\infty$ looked like a reasonable to point to which you could converge. –  azarel Apr 15 '12 at 19:18
I think there is something wrong in the text. Indeed, $X=[0,+\infty)$ is a closed subset in $\mathbb R$, hence it is complete with respect to standard metric (the one you call $d_1$).
If $d_2$ is equivalent to $d_1$, then also $(X,d_2)$ is complete.