Timeline for Measurability and Axiom of choice
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 14, 2015 at 13:00 | comment | added | Joel David Hamkins | For example, it is consistent with ZF that $\mathbb{R}$ is a countable union of countable sets, and in this case every set of reals is Borel. So one needs at least some AC to prove even that there are any non-Borel sets at all, let alone non-Borel analytic sets. | |
| Jul 14, 2015 at 12:19 | comment | added | Matthias Ludewig | Sorry, I meant to write Lebesgue-measurable. | |
| Jul 14, 2015 at 12:12 | history | answered | Gerald Edgar | CC BY-SA 3.0 |