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I have a quick question. Doing some research in Natural Language Processing I am coming across the space of rays in $\mathbb{R}^n$ coming out of the origin, in other words, the space $\mathbb{R}^n/\sim$ where $\sim$ is the equivalence relation defined by $a\sim b$ $\Leftrightarrow$ $\lambda>0$ such that $a=\lambda b$.

This space is kind of similar to the projective real space $\mathbb{RP}^n$ but it is not the same, does this space have a name? I have look for it but I have found nothing.

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    The topology might be different because of weirdness happening at the origin, but this seems like a sphere to me. There's a unique unit vector in each equivalence class, as opposed to the projective space which has two unit vectors in each equivalence class. – Andrew Maurer May 11 '21 at 18:56
  • Don’t you want to exclude 0? I think otherwise it won’t be a sphere, because it has an additional dense point (the equivalence class of 0). – Jonas Linssen May 11 '21 at 19:35
  • Yes, I forgot to exclude the origin. I will not edit my original question because the answer considers the two cases. – gmauricio May 12 '21 at 23:36

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It can be two very different things, depending on whether you consider $(\Bbb R^n\setminus \{0\})/\sim$ or $\Bbb R^n/\sim$.

In the former situation, it can indeed be identified with the sphere, as Andrew Maurer pointed out in the comments. One quick way (hit me up, if you want me to elaborate) to see that would be for example by noting that $$\begin{array}{rcl} \Bbb S^n & \rightarrow &(\Bbb R^n\setminus\{0\})/\sim\\ x & \mapsto & [x] \end{array}$$ defines a continuous bijection from a compact space $\Bbb S^n$ to the Hausdorff space $(\Bbb R^n\setminus\{0\})/\sim$, hence a homeomorphism. Continuity is detected by noting that the map is just the composite $$\Bbb S^n \rightarrow \Bbb R^n\setminus\{0\} \rightarrow (\Bbb R^n\setminus\{0\})/\sim$$ It is bijective by definition and is Hausdorff given the fact that the equivalence relation is induced by a good group action $\Bbb R_{+} \curvearrowright \Bbb R^n\setminus\{0\}$.

In the case of having $\Bbb R^n/\sim$ things get a bit more complicated. Note that by definition of the quotient topology every open subset contains the equivalence class of 0, making it a dense point. In particular this quotient is not Hausdorff, so it won’t be homeomorphic to any of the usual nice spaces.

Jonas Linssen
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