Timeline for Existence of a non-$Q$-set without the perfect set property
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 20, 2023 at 15:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 22, 2022 at 14:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 25, 2022 at 14:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 25, 2022 at 13:08 | answer | added | Corey Bacal Switzer | timeline score: 1 | |
| Apr 22, 2022 at 11:16 | comment | added | 喻 良 | Sorry, should be this:mathoverflow.net/questions/420656/co-analytic-q-sets | |
| Apr 22, 2022 at 9:47 | comment | added | Lorenzo | @喻良 the link sends me back to this question... is the question's link correct? | |
| Apr 20, 2022 at 18:57 | comment | added | Jason Zesheng Chen | In fact, the hypothesis is not necessary for the existence of a largest $\Pi^1_1$ set without a perfect subset. Provably in ZF+DC, the set $\{x\in \mathbb{R} \mid x\in L_{\omega_1^x} \}$ is such a set. | |
| Apr 20, 2022 at 18:20 | comment | added | Will Brian | @NoahSchweber: Oh right! Thanks -- that makes sense now. | |
| Apr 20, 2022 at 18:03 | comment | added | Noah Schweber | @WillBrian I think the issue is lightface vs. boldface - adding a single point to a $\Pi^1_1$ set may indeed make it no longer $\Pi^1_1$, although it will of course still be ${\bf \Pi^1_1}$. Similarly, if $\omega^\omega\cap L$ is countable then it is the largest countable lightface $\Sigma^1_2$ set. And there are various other results along these lines. | |
| Apr 20, 2022 at 15:57 | comment | added | Will Brian | I'm confused by your claim that there can exist a "greatest" $\Pi^1_1$ set without the perfect set property. Given any $\Pi^1_1$ set without the perfect set property, you can add countably many points to it, and it will still be a $\Pi^1_1$ set without the perfect set property. Right? | |
| Apr 20, 2022 at 15:12 | history | undeleted | Lorenzo | ||
| Apr 20, 2022 at 15:08 | history | deleted | Lorenzo | via Vote | |
| Apr 20, 2022 at 15:00 | history | asked | Lorenzo | CC BY-SA 4.0 |