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Timeline for Mutual metric projection

Current License: CC BY-SA 4.0

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Jul 9 at 13:37 history edited David Gao CC BY-SA 4.0
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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
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
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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
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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