Timeline for Can Vopenka's principle be violated definably?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2016 at 11:40 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
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| Jun 14, 2016 at 11:32 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Updated answer with an account of my 2016 article
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| Nov 19, 2010 at 18:36 | comment | added | Joel David Hamkins | Another way to describe the argument is: if there were such a definable counterexample, then VP would be first order expressible---you just have to say that the counterexample really does define a counterexample. But neither the full second-order version of VP nor the scheme version of VP is first order expressible by a single sentence, for the reasons I explain in my answer. | |
| Nov 19, 2010 at 16:57 | vote | accept | Mike Shulman | ||
| Nov 19, 2010 at 16:55 | comment | added | Mike Shulman | This is very beautiful, thank you! I suspected that the answer might have something to do with whether one formulates VP as first- or second-order. | |
| Nov 18, 2010 at 21:11 | comment | added | Joel David Hamkins | The whole argument can be given quickly by the last paragraph: your formula $\varphi$ will have some complexity $\Sigma_n$, but it can happen that $\Sigma_n$-Vopenka holds without $\Sigma_{n+1}$-Vopenka holding, and if your background universe is like that, your formula won't give the counterexample that is needed. | |
| Nov 18, 2010 at 21:01 | comment | added | Joel David Hamkins | Thanks, Andres. I definitely learned something about Vopenka doing this. | |
| Nov 18, 2010 at 20:59 | comment | added | Andrés E. Caicedo | Oh, please say hi to Andrew from me. | |
| Nov 18, 2010 at 20:57 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
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| Nov 18, 2010 at 20:56 | comment | added | Andrés E. Caicedo | Joel: Very nice! | |
| Nov 18, 2010 at 20:52 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |