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I can't find a metric $\delta$ in $\mathbb{R}^2\setminus\{0\}$ such that be equivalent to euclidean metric, be equal to euclidean metric in the unitary circle and for all $r>0$ the set $\{(x,y)\in\mathbb{R}^2\setminus\{0\}; 0 be $\delta$-unbounded and the set $\{(x,y)\in\mathbb{R}^2\setminus\{0\}; x^2+y^2>r\}$ be $\delta$-bounded.

I tried defining $\delta$ by cases, but is really dificult to obtain a metric in this form.

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    @Phira Ty for clarifications2012-05-22

3 Answers 3

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Consider the inversion $I : \Bbb{R}^2\setminus \{0\} \to \Bbb{R}^2 \setminus \{0\},\ I(x)=\frac{x}{\|x\|^2}$.

The define $d(x,y)=d_E(I(x),I(y))$, where $d_E$ is the Euclidean distance.

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    Nice pullback. Ty2012-05-22
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Do you remember how I used the function $f$ to define the metric $\delta$ in this answer? You can use the same idea here with the function $f(x)=\dfrac{x}{\|x\|^2}$: let $\delta(x,y)=d(f(x),f(y))$, where $d$ is the Euclidean metric.

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    @Gastó$n$: Yes, that’s right. (And no, they’re not metriza$b$le, since they’re not even H$a$usdorff.)2012-05-22
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The simplest choice is $d(x,y)= \left|\frac 1x - \frac 1y\right|.$

This metric is identical to the metric found by using the inversion, because the difference between the involution $\frac 1 z$ and the inversion $\frac 1 {\overline z}$ is just complex conjugation which is a distance-preserving reflection about the $x$-axis.

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    Ty for clarification2012-05-22