Timeline for Represent the metric of $n$ points in the Euclidean space [closed]
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 7, 2020 at 14:52 | history | closed |
user44191 Gerry Myerson Alex M. ARG Emil Jeřábek |
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| Apr 5, 2020 at 19:00 | review | Close votes | |||
| Apr 7, 2020 at 14:55 | |||||
| Apr 5, 2020 at 17:18 | comment | added | Taras Banakh | I hope that this means that any $n$ points of a Euclidean space are contained in an $(n-1)$ affine subspace (which is isometric to the $(n-1)$-dimensional Euclidean space), so we can think of these $n$ points as points of the Euclidean space $\mathbb R^{n-1}$. | |
| Apr 5, 2020 at 16:01 | history | edited | keyboardAnt | CC BY-SA 4.0 |
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| Apr 5, 2020 at 15:21 | comment | added | Ben McKay | Isometric embedding means an isometry of metric spaces to its image. So you represent a metric space $X$ in Euclidean space $\mathbb{E}^n$ by mapping $\phi \colon X \to \mathbb{E}^n$ with a map $\phi$ which is an isometry to its image, in the induced metric on its image. | |
| Apr 5, 2020 at 15:11 | comment | added | keyboardAnt | @BenMcKay, thanks for your fast reply. I'm not familiar with a few definitions you mentioned. Would you mind to elaborate a little bit more about "isometrically embedding" and the example you presented? I understood that we have 3 points in an Euclidean space that create a triangle; and a 4th point which isn't collinear with them creates a tetrahedron. How's that related to the question above? Best regards | |
| Apr 5, 2020 at 15:04 | comment | added | Ben McKay | It just means to isometrically embed. Think about embedding a 3 point metric space as a triangle, using the triangle inequality. A 4th point can sit in space above those 3 to make a tetrahedron. | |
| Apr 5, 2020 at 14:57 | history | asked | keyboardAnt | CC BY-SA 4.0 |