Timeline for A Borel perfectly everywhere surjective function on the Cantor set
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 7, 2021 at 16:37 | vote | accept | Iian Smythe | ||
| Jun 29, 2021 at 6:13 | answer | added | 喻 良 | timeline score: 2 | |
| Jun 29, 2021 at 0:13 | answer | added | Iian Smythe | timeline score: 3 | |
| Jun 28, 2021 at 23:42 | comment | added | Iian Smythe | Ah, yes. I suspect then if $f$ is Borel (or measurable), then since it must be continuous on a large set, it cannot have this property either. | |
| Jun 28, 2021 at 23:41 | comment | added | Anonymous | There is no continuous example. If $p$ is any element of the Cantor set, then if $f^{-1}(p)$ is uncountable, it contains a closed perfect set which maps only onto $p$, and if $f^{-1}(p)$ is countable, its complement contains a closed perfect set whose image does not contain $p$. | |
| Jun 28, 2021 at 23:31 | history | asked | Iian Smythe | CC BY-SA 4.0 |