Timeline for Let's keep adding once undecidable statements
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2016 at 19:36 | vote | accept | asdfusername | ||
| Jun 13, 2016 at 19:55 | comment | added | Asaf Karagila♦ | I disagree with your last statement. Saying "a bit" is a bit of an understatement. | |
| Jun 13, 2016 at 19:44 | comment | added | Noah Schweber | @JoelDavidHamkins Yes, but of course comparing two $\le_\mathcal{O}$-incomparable notations is in general not effective. | |
| Jun 13, 2016 at 19:43 | comment | added | Noah Schweber | @FrançoisG.Dorais Indeed - and if I recall correctly, every true $\Pi^0_1$ is true in some version of $ZFC_{\omega+1}$. | |
| Jun 13, 2016 at 19:37 | comment | added | François G. Dorais | Note that different notations for $\omega+1$ might disagree on what "$ZFC_{\omega+1}$" means. Choosing paths works but, as you suggest, only to a limited extent. For what it's worth: working inside a model of set theory, it makes perfect sense to consider $ZFC_\alpha$ (with the obvious straight-line definition) for any ordinal $\alpha$. Of course, this stops being recusively axiomatizable pretty quickly (in the internal sense) and then $Con(ZFC_\alpha)$ loses its arithmetic meaning. | |
| Jun 13, 2016 at 19:36 | comment | added | Joel David Hamkins | Regarding the incomputability of the order relation, I believe that the underlying partial order relation of $\mathcal{O}$ the restriction of a computable relation to the elements of $\mathcal{O}$. And furthermore, for every element $b\in\mathcal{O}$, the order relation on the cone below $b$ is a computable order. Basically, if $b\in\mathcal{O}$, then you can be sure that the programs involved in the $\mathcal{O}$ relation will all halt, and so you can simply unravel the entire order below $b$. | |
| Jun 13, 2016 at 19:22 | history | answered | Noah Schweber | CC BY-SA 3.0 |