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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 $\{(x,y)\in\mathbb{R}^2\setminus\{0\}; 0$\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?2012-05-22
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    $X$ $\delta$-bounded mean that $\delta|_X$ is a bounded function.2012-05-22
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    Is this question homework? There's a rather simple solution, but I don't want to solve everything if it is homework.2012-05-22
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    @LoganMaingi Is not homework, Im doing all exercises of "Espacos Metricos" of Elon Lages Lima and this exercise I tried for a week.2012-05-22
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    @GastónBurrull So, what you want to do is to exchange the role of the inside and the outside of the unit circle?2012-05-22
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    @Phira exactly. I tried defining equivalent metrics such that satisfies this conditions each one in a different subset but when I define metric by parts, triangular innequality doesnt works2012-05-22
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    @GastónBurrull Forget about the metric for a moment and just consider if you know a function that exchanges inside and outside of the unit circle.2012-05-22
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    1/x, but 1/d(x,y) doesnt works2012-05-22
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    Have you tried an inversion? I'm not sure it works, but it might.2012-05-22
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    @GastónBurrull $1/x$ is exactly what you need, but remember that you want to apply it to the points, not the metric.2012-05-22
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    @BeniBogosel dont works. If d is euclidean metric I tried to use d/(d+1) out of unitary circle and min{1,d} inside unitary circle, you can prove easly that two metric are equivalent to euclidean metric. And satisfies bounded and unbounded conditions. But doesnt works.2012-05-22
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    @Phira, you mean defining $d(x,y)=|1/x-1/y|$ are you sure that is a metric and equivalent?2012-05-22
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    @GastónBurrull: That is not an inversion. See my answer.2012-05-22
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    @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.2012-05-22
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    @GastónBurrull No, 1/x is not continuous at 0 which is the point.2012-05-22
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    @BeniBogosel Since, I didn't say that 1/x is an inversion, it is hardly an argument against my metric that it isn't.2012-05-22
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    @Phira Ty for clarifications2012-05-22

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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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    Yes is a good pullback =). Ty.2012-05-22
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    $f$ is homeomorphism, then metric are equivalent right?2012-05-22
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    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.2012-05-22
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    Yes, youre right, I must say **if $f$ is homeomorphism the induced metric (or pullback) is equivalent with domain metric**.2012-05-22
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    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?2012-05-22
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    @Gastón: Yes, that’s right. (And no, they’re not metrizable, since they’re not even Hausdorff.)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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    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 post2012-05-22
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    Ty for clarification2012-05-22