Timeline for Function and distance on bounded set [closed]
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 23, 2018 at 1:30 | comment | added | Check drummer | To Mateusz: but $|x_1−y_1|≤|x_0−y_0|$ is not genrally true, | |
| Jul 21, 2018 at 14:26 | comment | added | Mateusz Kwaśnicki | Choose $x_0 \ne y_0$, and let $x_{n+1} = f(x_n)$ and $y_{n+1} = f(y_n)$. Next, choose a convergent subsequence $(x_{k_n},y_{k_n})$, and set $l_n=k_{n+1}-k_n$. Since $|x_0-x_{l_n}| \le |x_{k_n}-x_{k_{n+1}}|$, $(x_{l_n})$ converges to $x_0$. Similarly, $y_{l_n}$ converges to $y_0$. Therefore, $|x_{l_n}-y_{l_n}| \to |x_0 - y_0|$. On the other hand, $|x_{l_n}-y_{l_n}|\ge|x_1-y_1|$, so $|f(x_0)-f(y_0)| = |x_1-y_1|\le|x_0-y_0|$, a contradiction. See here. | |
| Jul 21, 2018 at 9:35 | comment | added | Alec Rhea | It's possible for a moderator to move it and that would preserve the rep boost I believe, but probably faster to copy then delete it yourself and repost. No problem. | |
| Jul 20, 2018 at 23:23 | comment | added | Check drummer | Can it be moved automatically or can I do it somehow? Thank you | |
| Jul 18, 2018 at 22:20 | history | closed |
David Roberts♦ Emil Jeřábek Wolfgang Nik Weaver Yemon Choi |
Not suitable for this site | |
| Jul 18, 2018 at 10:48 | history | edited | Andrés E. Caicedo |
edited tags
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| S Jul 18, 2018 at 7:41 | history | suggested | Amir Sagiv | CC BY-SA 4.0 |
English and Tex
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| Jul 18, 2018 at 6:59 | review | Suggested edits | |||
| S Jul 18, 2018 at 7:41 | |||||
| Jul 18, 2018 at 6:33 | review | Close votes | |||
| Jul 18, 2018 at 22:24 | |||||
| Jul 18, 2018 at 5:58 | comment | added | Alec Rhea | This should be moved over to/deleted and re-asked at math.stackexchange. | |
| Jul 18, 2018 at 5:56 | review | First posts | |||
| Jul 18, 2018 at 6:59 | |||||
| Jul 18, 2018 at 5:56 | history | asked | Check drummer | CC BY-SA 4.0 |