Timeline for Sets $A$ such that $A$-maximal sets are $\Delta^0_2$
Current License: CC BY-SA 4.0
24 events
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| Mar 22, 2022 at 1:04 | vote | accept | David J. Webb | ||
| Mar 22, 2022 at 1:04 | answer | added | David J. Webb | timeline score: 0 | |
| Mar 19, 2022 at 4:04 | history | edited | David J. Webb | CC BY-SA 4.0 |
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| Mar 12, 2022 at 21:07 | history | edited | David J. Webb | CC BY-SA 4.0 |
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| Mar 12, 2022 at 6:39 | history | edited | David J. Webb | CC BY-SA 4.0 |
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| Mar 12, 2022 at 6:09 | history | edited | David J. Webb | CC BY-SA 4.0 |
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| Mar 12, 2022 at 5:59 | comment | added | David J. Webb | I now suspect the relativization of Martin is not as nice as I hoped, which nixes my comments about an equivalent characterization. | |
| Mar 12, 2022 at 5:58 | history | edited | David J. Webb | CC BY-SA 4.0 |
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| Mar 12, 2022 at 4:00 | history | edited | David J. Webb | CC BY-SA 4.0 |
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| Mar 12, 2022 at 3:31 | comment | added | Bjørn Kjos-Hanssen | @NoahSchweber ah okay. The interesting question is about maximality anyway | |
| Mar 12, 2022 at 3:25 | history | edited | David J. Webb | CC BY-SA 4.0 |
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| Mar 12, 2022 at 3:23 | comment | added | David J. Webb | Mm, I did delete that since the question was settled, so I suppose it comes down to whether I understand correctly how to relativize Martin's result that every high c.e. degree contains a maximal set - it should be that every A-high, A-c.e. degree contains an A-maximal set, but I haven't worked through the details | |
| Mar 12, 2022 at 3:20 | comment | added | Noah Schweber | @BjørnKjos-Hanssen I'm not thinking about maximality at all, I'm addressing the OP's (now-deleted?) variant question: "By Martin's high domination theorem, this is equivalent to asking for a set $A$ such that all $A$-high, $A$-c.e. sets are $\Delta^0_2$ and high." I'm merely claiming that there is no non-$\Delta^0_2$ set $A$ such that all $A$-high $A$-c.e. sets are $\Delta^0_2$. | |
| Mar 12, 2022 at 3:18 | comment | added | Bjørn Kjos-Hanssen | @NoahSchweber if $A$ is precoded then that conflicts with maximality does it not? | |
| Mar 12, 2022 at 3:15 | comment | added | Noah Schweber | @BjørnKjos-Hanssen If $B$ is $A$-c.e. and $A$-high, then so is $B\oplus A$, right? (Or: just start the usual construction of an $A$-high $A$-c.e. set but with $A$ itself "pre-coded" into one of the rows of the set you're building. This should amount to the same thing I think.) | |
| Mar 12, 2022 at 3:12 | comment | added | Bjørn Kjos-Hanssen | @NoahSchweber but how do you make it above $A$? | |
| Mar 12, 2022 at 2:36 | history | edited | David J. Webb | CC BY-SA 4.0 |
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| Mar 12, 2022 at 2:30 | comment | added | David J. Webb | Ah, okay, Bjorn and I somehow missed that. That settles it! | |
| Mar 12, 2022 at 2:30 | comment | added | Noah Schweber | Sure, but I'm saying we can always construct an $A$-high $A$-c.e. which is itself above $A$. | |
| Mar 12, 2022 at 2:29 | comment | added | David J. Webb | When we say it's $A$-high, doesn't that only give that $(B\oplus A)'$ is above $A$, not necessarily $B$ itself? | |
| Mar 12, 2022 at 2:15 | comment | added | Noah Schweber | If $B\ge_TA$ is $A$-c.e. and $A$-high (so just relativize the usual high c.e. construction to $A$) we must have $B\not\in\Delta^0_2$ since $\Delta^0_2$ is closed downwards. Or am I misunderstanding your question? | |
| Mar 12, 2022 at 1:45 | history | edited | David J. Webb | CC BY-SA 4.0 |
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| S Mar 12, 2022 at 1:20 | review | First questions | |||
| Mar 12, 2022 at 2:47 | |||||
| S Mar 12, 2022 at 1:20 | history | asked | David J. Webb | CC BY-SA 4.0 |