Timeline for Conditions for a set being closed under taking complement of a ball twice
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 23, 2015 at 16:19 | answer | added | Yoav Kallus | timeline score: 2 | |
| May 23, 2015 at 15:44 | comment | added | Yoav Kallus | I did not say "the ball", but "a ball". In your example every singleton is a delta ball, and every point not in S is in such a ball. | |
| May 23, 2015 at 13:24 | comment | added | user74022 | Not sure I understand this statement -- clearly there are examples when the equality holds and there are points which are not in the ball of radius $\delta$. E.g. take $S = {x}$, $\delta < \inf_{y,z} d(y,z)$ so that $N_\delta(S) = F \setminus {x}$. | |
| May 23, 2015 at 13:19 | history | edited | user74022 | CC BY-SA 3.0 |
added 4 characters in body
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| May 23, 2015 at 6:27 | comment | added | Yoav Kallus | $N_\delta(N_\delta(S))=S$ if and only if every point $x\not\in S$ is in a ball of radius $\delta$ disjoint from $S$. | |
| May 23, 2015 at 4:07 | comment | added | user74022 | How should one define curvature for a finite metric space? | |
| May 23, 2015 at 3:56 | review | First posts | |||
| May 23, 2015 at 4:44 | |||||
| May 23, 2015 at 3:51 | history | asked | user74022 | CC BY-SA 3.0 |