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Sorry for such question, But I don't know what exactly is the Euclidean metric on $\Bbb R^n\setminus\{0\}$? Induced by $\iota:\Bbb R^n\setminus\{0\}\to \Bbb R^n$ or $\varphi :\Bbb R^n\setminus\{0\}\to\Bbb R\times \Bbb S^{n-1}$ or something else?

Update: It seems that my problem is misunderstanding the "induced" maybe. This is my thought: because $x,y\in \Bbb R^n\setminus\{0\}$ are also in $\Bbb R^n$ so $d(x,y)=d_{can}(x,y)$. but this do not work for $x=(1,0,0...,0)$ and $y=(−1,0,0...,0)$!!

C.F.G
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    It is an open set in $\Bbb R^n$ with the Euclidean metric, see here. – Dietrich Burde Jul 26 '19 at 11:14
  • It's the one induced by $\iota$. Did you mean $\varphi$ as $\Bbb R^n\setminus{0} \to \Bbb R_{>0}\times \Bbb S^{n-1}$? By the way, I guess they generate equivalent metrics. – Berci Jul 26 '19 at 11:26
  • Depending on the context, perhaps what is meant is simply the restriction to $\Bbb R^n\setminus{0}$ of the usual Euclidean metric on $\Bbb R^n$. This is what I would have thought of first, if the answer was not necessarily supposed to have something to do with tag riemannian-geometry – Mirko Jul 26 '19 at 14:06
  • @Mirko, My question is exactly what you explained. if it is just a restriction, so what is the $d(x,y)$ for antipodal points $x,y$ i.e. $x=-y$? – C.F.G Jul 26 '19 at 14:29
  • it is the same as what it was before $0$ was removed. $d(-1,1)=2$.What do you denote by $d_{can}$ in your post? $d$ is a metric, formally a function from $\Bbb R^n\times\Bbb R^n$ into $\Bbb R$, and then we restrict the domain of this function to $(\Bbb R^n\times\Bbb R^n)\setminus((\Bbb R^n\times{0})\cup({0}\times\Bbb R^n))$ – Mirko Jul 26 '19 at 14:32
  • Are you sure? but I think $d(x,y):= \inf{ \text{curves from $x$ to $y$ on $M$}}$ that that line is not on $\Bbb R^n\setminus{0}$. – C.F.G Jul 26 '19 at 14:36
  • well, I am not sure, it depends on the context. If the context was not necessarily tag riemanian-geometry, then I would have been pretty sure. But even if you take $\inf$, don't you get again that $d(-1,1)=2$? – Mirko Jul 26 '19 at 14:39
  • No. Then we must have two minimizing curve in $\Bbb R^n$ that is impossible. – C.F.G Jul 26 '19 at 14:57
  • Why nobody answer to my question precisely? – C.F.G Aug 11 '19 at 10:00

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By definition, a Euclidean space is an affine space, often confused with a vector space. Since $\mathbb{R}^n \setminus \{0\}$ is no longer an affine space, there is no standard definition of the Euclidean distance on it.

Siminore
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