Timeline for Short selection in the space of subsets
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| S Dec 17, 2022 at 14:04 | history | bounty ended | CommunityBot | ||
| S Dec 17, 2022 at 14:04 | history | notice removed | CommunityBot | ||
| S Dec 9, 2022 at 12:13 | history | bounty started | Anton Petrunin | ||
| S Dec 9, 2022 at 12:13 | history | notice added | Anton Petrunin | Draw attention | |
| Nov 29, 2022 at 0:33 | comment | added | anon | Of course, I missed the request that the space be geodesic! It is an interesting question. | |
| Nov 28, 2022 at 10:36 | history | edited | Anton Petrunin | CC BY-SA 4.0 |
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| Nov 28, 2022 at 10:33 | comment | added | Anton Petrunin | @anon I asked for geodesic space, and snowflake is not geodesic. But it is a nice observation. | |
| Nov 27, 2022 at 1:50 | comment | added | anon | If $(X,d)$ is a metric space and $\epsilon \in (0,1)$, then $(X, d^\epsilon)$ is also a metric space (called a ``snowflake'' of $(X,d)$). It seems to me that if the former space is nice, then so is the latter, trivially. The snowflake will not be injective. For a concrete example, take $X=[0,1]$ and $\epsilon=\frac{1}{2}$. Does this work or have I made a mistake? | |
| Nov 24, 2022 at 9:27 | history | edited | Anton Petrunin | CC BY-SA 4.0 |
added 144 characters in body
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| Nov 23, 2022 at 15:43 | comment | added | user44143 | This is just another way of saying that the Hausdorff distance between $X$ and $Y$ is greater than that between $f(X)$ and $f(Y)$ — so the $x\in X,y\in Y$ there just comes from the definition of Hausdorff distance. Without that clause the condition I wrote would be trivial; as it stands it is equivalent to the condition in the post. | |
| Nov 23, 2022 at 15:32 | comment | added | Anton Petrunin | @MattF. No, I do not assume $r(A)\in A$. $r(A)$ might be any point such that $r(\{x\})=x$ and $|r(A)-r(B)|\leqslant|A-B|_H$. | |
| Nov 23, 2022 at 15:27 | comment | added | user44143 | The Lipschitz niceness condition can also be written as: for any $X,Y$, there are $x\in X,\,y\in Y$ with $d(x,y)\ge d(f(X),f(Y))$ and $d(x,y)$ equal to either $d(x,Y)$ or $d(X,y)$. | |
| Nov 23, 2022 at 10:41 | history | asked | Anton Petrunin | CC BY-SA 4.0 |