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I want to prove the following statements: For $X:=(0,1),$ prove the following:

(a) $d(x,y):=\big|(1/x)-(1/y)\big|$ is a metric on $X.$

(b) The natural metric and $d$ are equivalent.

(c) There is no metric on $\mathbb{R}$ which is equivalent to the natural metric and which induces $d$ on $X.$

I found this problem in page 140 of Amann's book Analysis, Vol I, which is the 10th exercise. And I have proved the first two statements, but was stuck by (c).

I have tried to prove by contradiction, but did not know how to derive a contradiction. I guess it may be helpful to consider the ball $\mathbb{B}(0,\epsilon):=\{y\in\mathbb{R}\mid |y|<\epsilon\},$ since in $d,$ $\mathbb{B}_d(0,\epsilon):=\{y\in\mathbb{R}\mid |1/0-1/y|<\epsilon\},$ which is impossible, for the denominator is $0.$ But I do not know how to use this. Can anyone help me?

PS: Two metrics $d_1$ and $d_2$ on a set $X$ are called equivalent if, for each $x\in X$ and $\epsilon>0,$ there are positive numbers $r_1, r_2$ such that $\mathbb{B}_1(x,r_1)\subset \mathbb{B}_2(x,\epsilon)$ and $\mathbb{B}_2(x,r_2)\subset \mathbb{B}_1(x,\epsilon),$ where $\mathbb{B}_j$ denotes the open ball in $(X,d_j), j=1,2.$

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  • $\begingroup$ Please can you provide your definition of when two metrics are equivalent? $\endgroup$ – Rudy the Reindeer Oct 16 '14 at 10:45
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Suppose that $e$ is a metric on $\mathbb R$ equivalent to the standard metric with the property that $e(x,y) = d(x,y)$ if $x,y \in (0,1)$. Then there must exist $\delta > 0$ with the property that $(-\delta,\delta) \subset B_e(0,1)$.

If $n > 1/\delta$, then $\dfrac 1n$ and $\dfrac 1{n+3}$ both belong to $(-\delta,\delta)$ so that $\dfrac 1n$ and $\dfrac 1{n+3}$ both belong to $B_e(0,1)$.

If $(x,y) \in (0,1) \cap B_e(0,1)$, then $d(x,y) = e(x,y) \le e(x,0) + e(y,0) < 1 + 1$ so that $d(x,y) < 2$.

But $d(1/n,1/{n+3}) = 3$.

This contradiction means that no such metric exists.

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  • $\begingroup$ Excellent! Thanks a lot. $\endgroup$ – azc Oct 17 '14 at 2:21
  • $\begingroup$ @Umberto P. I think there is a small mistake here. 0 is not in $(0,1)$. So you can't use 0 to construct open ball $\endgroup$ – john Jun 15 '15 at 13:24
  • $\begingroup$ @john $e$ is a metric on $\mathbb R$. $\endgroup$ – Umberto P. Jun 15 '15 at 13:36
  • $\begingroup$ @UmbertoP. oops. thanks for pointing out. $\endgroup$ – john Jun 15 '15 at 17:55

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