Timeline for A non-Borel union of unit half-open squares
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 1, 2019 at 20:08 | history | undeleted | Nate Eldredge | ||
| May 1, 2019 at 20:08 | history | edited | Nate Eldredge | CC BY-SA 4.0 |
added 103 characters in body
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| May 1, 2019 at 20:00 | history | deleted | Nate Eldredge | via Vote | |
| May 1, 2019 at 19:55 | comment | added | Taras Banakh | @YCor The problem was to find a non-Borel union for ANY function $p$, not for SOME $p$. | |
| May 1, 2019 at 19:54 | comment | added | YCor | @TarasBanakh I don't know if you're addressing Nate's answer or my comment, but in both we have $p=0$ as assumption. | |
| May 1, 2019 at 19:53 | comment | added | YCor | Remark: assuming (as we can) that $Z$ is dense in $L$, then $E$ is equal to the open strip $\{x+iy:0<x+y<2\}$ union $Z$. | |
| May 1, 2019 at 19:51 | comment | added | Taras Banakh | It is not that easy: note that we have the function $p$, which rotates the squares. It may happen that on this diagonal line $p(z)=1$ then the union of such squares will be the open strip. | |
| May 1, 2019 at 19:33 | history | answered | Nate Eldredge | CC BY-SA 4.0 |