Timeline for Can Cantor set be the zero set of a continuous function?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 11, 2010 at 17:36 | comment | added | Robin Chapman | So we back in one dimension again. | |
| May 11, 2010 at 7:13 | comment | added | Roland Bacher | Sorry, part of the correction got lost du to a missing space between $<$ and $a$. | |
| May 11, 2010 at 7:11 | history | edited | Roland Bacher | CC BY-SA 2.5 |
added 134 characters in body
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| May 10, 2010 at 18:30 | comment | added | Robin Chapman | Of course, a closed set need not have an upper bouned Also what is $\beta$? | |
| May 10, 2010 at 17:54 | comment | added | Roland Bacher | Of course. I have fixed this stupid error. | |
| May 10, 2010 at 17:53 | history | edited | Roland Bacher | CC BY-SA 2.5 |
stupid error fixed
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| May 10, 2010 at 16:47 | comment | added | André Henriques | This is not C^inf. Example: take the closed set to be {-1,1}, then your function is not C^1 at 0. | |
| May 10, 2010 at 15:41 | history | answered | Roland Bacher | CC BY-SA 2.5 |