Timeline for Class-theoretic sentences that are $\Pi^1_1$ or $\Pi^1_2$
Current License: CC BY-SA 4.0
17 events
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| Sep 16, 2021 at 20:02 | vote | accept | Paul Blain Levy | ||
| Sep 10, 2021 at 6:48 | history | edited | Paul Blain Levy | CC BY-SA 4.0 |
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| Sep 6, 2021 at 11:37 | answer | added | Corey Bacal Switzer | timeline score: 4 | |
| Sep 3, 2021 at 12:04 | answer | added | Farmer S | timeline score: 4 | |
| Sep 3, 2021 at 10:14 | comment | added | Farmer S | What about „There is a wellorder of the classes which is $\Pi^1_1$-definable in a set parameter“, where „wellorder“ is just in the internal sense that there is no class coding a strictly descending $\omega$-sequence? This is $\Pi^1_2$, and I don’t see an easy way to express it in $\Sigma^1_2$, though I‘m not sure. It is consistent with multiple Woodin cardinals, as it holds in the context mentioned by @ElliotGlazer mentioned above, if the background universe is a short extender mouse, and $\kappa$ inaccessible there. | |
| Sep 3, 2021 at 3:34 | comment | added | Elliot Glazer | To remove ambiguity about what choice, reflection, and large cardinal principles are, it might be easier to work in an ambient model of ZFC and ask about models of the form $(V_{\kappa}, V_{\kappa+1}),$ where $\kappa$ is assumed to have large cardinal properties. | |
| Sep 3, 2021 at 0:12 | history | edited | Paul Blain Levy | CC BY-SA 4.0 |
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| Sep 2, 2021 at 23:49 | history | edited | Paul Blain Levy | CC BY-SA 4.0 |
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| Sep 2, 2021 at 23:35 | history | edited | Paul Blain Levy | CC BY-SA 4.0 |
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| Sep 2, 2021 at 22:59 | history | edited | Paul Blain Levy | CC BY-SA 4.0 |
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| Sep 2, 2021 at 22:56 | comment | added | Paul Blain Levy | Thanks @Hanul Jeon, but I want to assume all forms of choice. I've edited the question to make that clear. | |
| Sep 2, 2021 at 22:32 | comment | added | Hanul Jeon | Global Choice is $\Sigma^1_1$. If $\mathsf{GBc}$ proves it is $\Pi^1_1$, then it is absolute between models of $\mathsf{GBc}$ with the same sets. However, every model of $\mathsf{GBc}$ is contained in a model, whose sets are the same with the original model, of $\mathsf{GBc}$ with Global Choice. Hence every model of $\mathsf{GBc}$ satisfies Global Choice, and so $\mathsf{GBc}$ proves Global Choice, a contradiction. | |
| Sep 2, 2021 at 22:30 | history | edited | Paul Blain Levy | CC BY-SA 4.0 |
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| Sep 2, 2021 at 22:22 | history | edited | Paul Blain Levy | CC BY-SA 4.0 |
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| Sep 2, 2021 at 22:14 | history | edited | Paul Blain Levy | CC BY-SA 4.0 |
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| Sep 2, 2021 at 22:09 | comment | added | Paul Blain Levy | This question is about second-order set theory. I recently asked a similar question about second-order arithmetic, and another one about third-order arithmetic. | |
| Sep 2, 2021 at 22:07 | history | asked | Paul Blain Levy | CC BY-SA 4.0 |