# A metric in $\mathbb{R}^2\setminus\{0\}$

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 $\{(x,y)\in\mathbb{R}^2\setminus\{0\}; 0<x^2+y^2<r\}$ 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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I don't understand the question. What does $\delta$-bounded mean? –  Olivier Bégassat May 22 '12 at 22:22
@GastónBurrull $1/x$ is exactly what you need, but remember that you want to apply it to the points, not the metric. –  Phira May 22 '12 at 22:37
@GastónBurrull: That is not an inversion. See my answer. –  Beni Bogosel May 22 '12 at 22:42
@GastónBurrull Yes, I am sure. But you should check it, as it is quite easy to check. Also, $1/z$ is an involution that differs from the inversion $1/\overline z$ just by a distance-preserving reflection. For this kind of calculation, it is preferable because it is simpler. –  Phira May 22 '12 at 22:46
@BeniBogosel Since, I didn't say that 1/x is an inversion, it is hardly an argument against my metric that it isn't. –  Phira May 22 '12 at 22:51

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. Ty –  Gastón Burrull May 22 '12 at 22:49

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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Yes is a good pullback =). Ty. –  Gastón Burrull May 22 '12 at 22:49
$f$ is homeomorphism, then metric are equivalent right? –  Gastón Burrull May 22 '12 at 23:04
The fact that $f$ is a homeomorphism doesn’t guarantee that the metrics are equivalent $-$ we saw that in the earlier problem $-$ but in this case they are equivalent, because unlike the function in that example, $f$ is a homeomorphism with the same topology in domain and range, not just homeomorphic topologies. –  Brian M. Scott May 22 '12 at 23:10
Yes, youre right, I must say if $f$ is homeomorphism the induced metric (or pullback) is equivalent with domain metric. –  Gastón Burrull May 22 '12 at 23:14
Your comment was very clarificator. Is important the distinction between homeo topologies and the same topologies (equivalent metrics iff same topologies not homeo). For example in the set $\{a,b,c\}$ the topology given by $\{c\}$ and whole and empty set may be homeomorphic to topology given by $\{b\}$ but if both topologies were metrizable (probably not) metrics cant be equivalent right? –  Gastón Burrull May 22 '12 at 23:39

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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I've proved that is a metric. Is equivalent to euclidean metric, and must be the same metric that given by the pullback of Benis post –  Gastón Burrull May 22 '12 at 22:55
Ty for clarification –  Gastón Burrull May 22 '12 at 22:57