Timeline for Is there a set that intersects every line twice which is Lebesgue measurable or Borel?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 9, 2021 at 18:01 | comment | added | Alessandro Codenotti | @Accumulation Since it is consistent with $\mathsf{ZF}$ that all subsets of $\Bbb R$ are Borel and the existence of such a Borel set is open, it must be open whether $\mathsf{ZF}$ proves that such a set exists as well | |
| Nov 8, 2021 at 20:56 | comment | added | Wojowu | @Acccumulation Yes, more specifically on a well-ordering of $\mathbb R$. I'm not aware of a choiceless construction. | |
| Nov 8, 2021 at 20:51 | comment | added | Acccumulation | Does this construction depend on Choice? | |
| Nov 8, 2021 at 3:22 | comment | added | LMP | Thanks for your answers. | |
| Nov 7, 2021 at 17:07 | history | edited | Wojowu | CC BY-SA 4.0 |
added 13 characters in body
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| Nov 7, 2021 at 15:52 | vote | accept | LMP | ||
| Nov 7, 2021 at 14:37 | history | answered | Wojowu | CC BY-SA 4.0 |