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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 $d_2$? – Martin Ueding Apr 15 '12 at 19:01
up vote 1 down vote accepted

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! – Martin Ueding 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. – Martin Ueding 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.

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@Davide: There are various notions that are called equivalence of metrics and two of the ones listed in that question indeed have the property that completeness is invariant under equivalence. Of course, the metrics considered here are not equivalent in these senses. – t.b. Apr 15 '12 at 19:05
They are equivalent in the sense of open sense they define, that is what we have defined as “equivalent“. – Martin Ueding Apr 15 '12 at 19:17

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