Timeline for Mutual metric projection
Current License: CC BY-SA 4.0
12 events
when toggle format | what | by | license | comment | |
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Jul 9 at 13:37 | history | edited | David Gao | CC BY-SA 4.0 |
deleted 2 characters in body
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Jul 9 at 5:43 | history | edited | Erel Segal-Halevi | CC BY-SA 4.0 |
Shorten the proof using Helly's theorem
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Jul 8 at 11:56 | comment | added | David Gao | @ErelSegal-Halevi Yeah, I’m aware. I thought about the case for higher dimensions, but unless one imposes additional assumptions, there’s not much that can be done. This is just an ad hoc and very much brute force approach to one specific case, that’s all. | |
Jul 8 at 10:13 | comment | added | Erel Segal-Halevi | OK! Unfortunately, for two dimensions we would need intersections of three sets to be non-empty, so this argument cannot be used.. | |
Jul 8 at 8:28 | comment | added | David Gao | @ErelSegal-Halevi Ah, I didn’t know about Helly’s theorem before. This is not really my field. But yes, it looks to be just Helly’s theorem applied to $d = 1$. | |
Jul 8 at 8:26 | comment | added | David Gao | @ErelSegal-Halevi Yes, it should have been $c_0$. Fixed. | |
Jul 8 at 8:25 | history | edited | David Gao | CC BY-SA 4.0 |
edited body
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Jul 8 at 5:40 | comment | added | Erel Segal-Halevi | In the proof of Lemma 3, $c_2$ should be $c_0$? | |
Jul 8 at 5:34 | comment | added | Erel Segal-Halevi | Are your Lemmas 1 and 2 special cases of Helly's theorem for $d=1$? | |
Jul 4 at 7:49 | history | edited | David Gao | CC BY-SA 4.0 |
added 8 characters in body
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Jul 3 at 19:18 | comment | added | David Gao | A remark here: since the OP mentioned, under Alex’s answer, that an especially interesting case is that $S(x)$ are all bounded and $x \in S(x)$ for all $x$, the above classification shows that in the interval case, this is only possible if there is a fixed $c$ s.t. $S(x)$ is the closed interval between $x$ and $c$ for all $x$, which the third example in the OP’s post. | |
Jul 3 at 18:40 | history | answered | David Gao | CC BY-SA 4.0 |