Timeline for Is $0^{(\omega)}$ a minimal cover in the arithmetic degrees
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 26, 2023 at 8:09 | answer | added | Peter Gerdes | timeline score: 1 | |
| Mar 30, 2023 at 17:55 | answer | added | Noah Schweber | timeline score: 3 | |
| Mar 24, 2023 at 0:09 | history | edited | Peter Gerdes | CC BY-SA 4.0 |
Added question mark
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| Mar 24, 2023 at 0:01 | history | edited | Peter Gerdes | CC BY-SA 4.0 |
Clarified the question.
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| Mar 23, 2023 at 19:17 | comment | added | Joel David Hamkins | Ah, that makes sense. Sorry for the misunderstanding. | |
| Mar 23, 2023 at 18:17 | comment | added | François G. Dorais | @Joel: I think Peter means in the degree structure associated to the relation "A is arithmetic in B". | |
| Mar 23, 2023 at 15:42 | comment | added | Joel David Hamkins | Oh, I realized I may have misunderstood your question. Are you asking whether $0^{(\omega)}$ is a minimal cover of some degree $B$? Obviously $B$ cannot be arithmetic (so not in the arithmetic degrees--perhaps you should have meant hyperarithmetic?). | |
| Mar 23, 2023 at 15:24 | comment | added | Joel David Hamkins | I posted a solution based on that idea. | |
| Mar 23, 2023 at 15:24 | answer | added | Joel David Hamkins | timeline score: 5 | |
| Mar 23, 2023 at 10:14 | comment | added | Joel David Hamkins | I wonder whether one can make a smaller upper bound by taking sum of suitably generically modified copies of $0^{(n)}$? That is, put them together and make a finite change to each one, so as to prevent the whole thing from computing $0^{(\omega)}$. | |
| Mar 23, 2023 at 4:30 | history | asked | Peter Gerdes | CC BY-SA 4.0 |