Timeline for The difference between Baire 2 and 'effectively Baire 2'
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 20, 2022 at 10:59 | vote | accept | Sam Sanders | ||
| Dec 20, 2022 at 10:08 | answer | added | Elliot Glazer | timeline score: 5 | |
| Dec 20, 2022 at 6:54 | comment | added | Patrick Lutz | @ElliotGlazer You're right; I misunderstood what you meant by "coded by a real" in your first comment. | |
| Dec 20, 2022 at 5:15 | comment | added | Elliot Glazer | @PatrickLutz If $\mathbb{R}$ is a countable union of countable sets then every subset of $\mathbb{R}$ is also a countable union of countable sets. My point is that there is then $|\mathcal{P}(\mathbb{R})|$ many functions in Baire class 3 so they can't all be coded by a real. | |
| Dec 20, 2022 at 5:08 | comment | added | Asaf Karagila♦ | @PatrickLutz: Related. | |
| Dec 20, 2022 at 4:52 | comment | added | Patrick Lutz | @ElliotGlazer that doesn't quite work. For example, if the countable union of countable sets is all of $\mathbb{R}$ then the indicator function is effectively Baire class 3 so there's no contradiction there. However, using the effective perfect set theorem you can get a contradiction if you have a countable union of countable sets whose cardinality is neither countable nor continuum. | |
| Dec 20, 2022 at 3:58 | comment | added | Elliot Glazer | @AsafKaragila It's easy to get a negative result for Baire 3. The indicator function of any countable union of countable sets is Baire 3, so if $\mathbb{R}$ is a countable union of countable sets, not all such functions will be codable by a real. | |
| Dec 19, 2022 at 18:25 | comment | added | Asaf Karagila♦ | I wouldn't be too surprised if the answer is positive, but then again, it might be that AC comes into play only for Baire 3, or Baire $\omega$. These things can be finicky. | |
| Dec 19, 2022 at 16:23 | history | asked | Sam Sanders | CC BY-SA 4.0 |