Timeline for A $\Pi^0_1$ class of $\alpha$-subgenerics for all $\alpha < w_1^{CK}$
Current License: CC BY-SA 4.0
7 events
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| Jan 29 at 7:09 | 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. | |
| Oct 1, 2023 at 7:01 | 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 1, 2023 at 6:36 | comment | added | Peter Gerdes | @DanTuretsky The problem I was really having was to figure out how to make everything work at those non-standard notations since you can't actually extend transfinite induction up to them nor even assume you have a unique set $0^{(\delta)}$. However, I realized you can stuff everything into a single $\Sigma^1_1$ formula which asserts the construction worked up to $\alpha$ and lets you read off the standard part from the set witness (some non-standard $\alpha$ must satisfy lest we give a $\Sigma^1_1$ def of $\mathscr{O}$). | |
| Aug 31, 2023 at 20:57 | answer | added | Peter Gerdes | timeline score: 1 | |
| Aug 12, 2023 at 13:30 | comment | added | Dan Turetsky | I haven't read the notes, but it sounds like you're describing a worker argument. Those can generally be run to a pseudo-ordinal length: fix some pseudo-ordinal $\delta$, a version of $0^{(\delta)}$, and run the same construction there. Then since $\delta$ bounds a notation for every computable ordinal, what you get will be a computable tree whose paths are $\alpha$-subgeneric for every computable $\alpha$. Presumably one also ensures that none of the paths are computable from $0^{(\delta)}$, so not hyperarithmetic, which guarantees the tree is perfect. | |
| Aug 12, 2023 at 13:24 | history | edited | Dan Turetsky | CC BY-SA 4.0 |
edited body; edited title
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| Aug 11, 2023 at 8:48 | history | asked | Peter Gerdes | CC BY-SA 4.0 |