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Let $X=(0,1]$ and

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

be two metric on $X$. Show that $(X,d_1), (X,d_2)$ have the same convergent sequences, but $(X,d_2)$ is complete while $(X,d_1)$ is not.

Attempt: Let $x_n$ be convergent in $(X,d_1)$ $\Longrightarrow |x_n-x|\rightarrow 0,n\rightarrow\infty $ $\Longrightarrow |\frac{1}{x_n}-\frac{1}{x}|=\frac{|x-x_n|}{x_n*x}=d_2(x_n,x)\rightarrow 0, n\rightarrow\infty$ [is this enough to prove they have the same convergent sequences?]

To show that $(X,d_1)$ is not complete, consider sequence $\frac{1}{n}$, which is Cauchy, but converges to $0\notin X, n\rightarrow\infty$

[This is the primary source of my troubles:] I am trying to use an isometry: $\phi:(0,1]\to[1,\infty)$ defined by $\phi(t)=\frac{1}{t}$ to show that $(X,d_2)$ is complete.

I think I can show that $\phi(x_n)$ converges to y in $[1,\infty)$ and therefore, $x_n$ converges to $\frac{1}{y} \in X$. but, I am having trouble in the execution.

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Denote the usual metric on ${\mathbb R}$ by $\|$. Then $(X,d_1)=\bigl(\>]0,1],\|\bigr)$, and $(X,d_2)$ is isometric with $\bigl(\>[1,\infty[\>,\|\bigr)$ via the map $$\phi:\quad ]0,1]\to[1,\infty[\ ,\qquad x\mapsto \phi(x):={1\over x}\ .$$ Since $\phi$ is $\|$-continuous in both directions it maps convergent sequences in $\bigl(\>]0,1],\|\bigr)$ into convergent sequences in $\bigl(\>[1,\infty[\>,\|\bigr)\sim(X,d_2)$, and conversely.

The statement in the title is only seemingly paradoxical: Convergent sequences in $X$ are of course $d_1$-Cauchy as well as $d_2$-Cauchy. On the other hand, among the non-convergent sequences there are no $d_2$-Cauchy ones, since $(X,d_2)\sim \bigl(\>[1,\infty[\>,\|\bigr)$ is complete, but some are $d_1$-Cauchy, e.g. the sequence $\bigl({1\over n}\bigr)_{n\geq1}$. This proves the claim in the title.

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  • $\begingroup$ is saying $\phi(x_n)\rightarrow y\in [1,\infty)$ implies $x_n\rightarrow \frac{1}{y}\in (0,1]$ a correct representation? $\endgroup$ – JKK Sep 16 '15 at 16:55
  • $\begingroup$ Yes: $x_n\to\xi={1\over\eta}\in\ ]01]$ iff $\phi(x_n)\to \eta ={1\over\xi}\in[1,\infty[\ $. That's what "continuous in both directions" means. $\endgroup$ – Christian Blatter Sep 16 '15 at 17:45
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The issue is that the sequence of points $\phi(1/n)=n$ is unbounded and so is not Cauchy in $[1,\infty)$ with the usual metric, so it's certainly not convergent. Likewise, the sequence of points $1/n$ is not $d_2$-Cauchy, since for example, $$d_2\left(\frac1n,\frac1{n+1}\right)=1$$ for all $n$.

Instead, start with an arbitrary $d_2$-Cauchy sequence of points of $(0,1],$ then use the isometry and proceed as you intended.

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  • $\begingroup$ So, what is proper method for proving (X,d2) is complete? $\endgroup$ – JKK Sep 16 '15 at 5:23
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    $\begingroup$ Start by taking a Cauchy sequence, and proving that there must be some $0<c<1$ such that the sequence is bounded below by $c.$ At that point, you can show that the sequence must have a convergent subsequence, since it's bounded. Can you take it from there? $\endgroup$ – Cameron Buie Sep 16 '15 at 5:32
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    $\begingroup$ You can still use the isometry $\phi$ as you did before, you just need to start with an arbitrary Cauchy sequence. $\endgroup$ – Cameron Buie Sep 16 '15 at 10:51

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