Timeline for The closure of all periodic homeomorphisms of circle
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| S Feb 26, 2016 at 15:24 | history | suggested | Johannes Huisman | CC BY-SA 3.0 |
corrected index of T, and some trivial characters in order to satisfy the stupid bound of minimum 6 character edits
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| Feb 26, 2016 at 15:10 | review | Suggested edits | |||
| S Feb 26, 2016 at 15:24 | |||||
| Jan 17, 2015 at 7:54 | vote | accept | Ali Taghavi | ||
| Jan 17, 2015 at 7:54 | comment | added | Ali Taghavi | I appologize for my misunderestanding. Now I underestand your counter example. | |
| Jan 17, 2015 at 0:13 | comment | added | Victor Kleptsyn | Sorry, but yes, you are: the length of $I_n$ is $2^n$, not $2^{-n}$, so this is almost half-circle arc, not very small one. On the other hand, $I_0$ has length 1 in a circle of length $2^{n+1}-1$, so it is a very small arc. | |
| Jan 16, 2015 at 9:46 | comment | added | Ali Taghavi | In fact the last acr $I_{n}$ is very small for large n, and is mapped to $I_{0}$ which has a fixed size. so we loose equicontinuity.Am I mistaken? | |
| Jan 15, 2015 at 13:24 | comment | added | Ali Taghavi | In the other words, are you sure that $T_{n}$, $C^{0}-$ converges to $T$? | |
| Jan 15, 2015 at 13:16 | comment | added | Ali Taghavi | thank you very much for your answer. Lets imagine these $I_{j}$ as arcs in the unit circle. So it seems that you fix $I_{0}$ at whole upper hemi circle, $I_{1}$ approachs to the whole lower hemi circle, and the remaining $I_{j}$'s are limited to the small remainig part. So it seems that the familly $T_{n}$'s is not an equicontinuous familly of maps. So this contradicts to uniforms convergence. So could you please more explain about your construction? | |
| Jan 14, 2015 at 20:11 | history | answered | Victor Kleptsyn | CC BY-SA 3.0 |