Timeline for Class-theoretic sentences that are $\Pi^1_1$ or $\Pi^1_2$
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 10, 2021 at 7:30 | comment | added | Corey Bacal Switzer | Sorry, I'm seeing now the OP asked for KM, not GBC. Don't know why I missed that. | |
| Sep 10, 2021 at 7:28 | comment | added | Corey Bacal Switzer | Yes, I should have been more clear, the above argument only works for GBC, not KM. KM, in fact the weaker ETR_0, proves that every class has a satisfaction class. However, GBC does not since given any model M of ZFC, (M, def(M)) is a model of GBC. In particular, GBC is conservative over ZFC for first order sentences so no first order definable property of ORD can imply that there is a satisfaction class, though second order large cardinal properties of ORD do imply there is a satisfaction class. For example, I think "every ORD tree has a cofinal branch" does. | |
| Sep 9, 2021 at 21:05 | comment | added | Farmer S | True, it looks like KM is enough. | |
| Sep 9, 2021 at 20:08 | comment | added | Elliot Glazer | I'm pretty sure KM proves that for every $X$ there is a satisfaction class for the language expanded by $X.$ It certainly holds for $(V_{\kappa}, V_{\kappa+1}),$ $\kappa$ inaccessible. | |
| Sep 9, 2021 at 17:24 | comment | added | Farmer S | But don't large cardinal properties of $\mathrm{Ord}$ prove there is such a satisfaction class? | |
| Sep 9, 2021 at 10:35 | comment | added | Corey Bacal Switzer | It can't be $\Sigma^1_2$ for suppose it were equivalent to a sentence of the form $\exists Y \forall Z \psi(Y, Z)$. Let $(M, \mathcal X) \models \exists Y \forall Z \psi(Y, Z)$ be a model of GBC. Pick a witness $Y \in \mathcal X$. The point is that the model $(M, def(M, Y))$ is still a model of GBC, it has to satisfy $\forall Z \psi (Y, Z)$ but there is no satisfaction class for $Y$ again by undefinability of truth. 2/2 | |
| Sep 9, 2021 at 10:35 | comment | added | Corey Bacal Switzer | Actually the above also gives another example of a properly $\Pi^1_2$ sentence: ``for all X there is a satisfaction class for the first order sentences in the expanded language with X as a predicate". This is $\Pi^1_2$ clearly. 1/2 | |
| Sep 9, 2021 at 8:30 | comment | added | Corey Bacal Switzer | @PaulBlainLevy, here is a much simpler, parameter free, example for (1), I don't know why I didn't think of this before. Simply say ``there is a satisfaction class for the first order sentences". This is obviously a $\Sigma^1_1$ sentence. Any $\omega$-standard or recursively saturated model of ZFC has a satisfaction class so it has a GBC realization satisfying this sentence. However, the sentence can't be equivalent to a $\Pi^1_1$ sentence since any model of the form $(M, def(M))$ will not have a satisfaction class by the undefinability of truth. | |
| Sep 7, 2021 at 22:52 | comment | added | Paul Blain Levy | But (1) is still unresolved. I'm looking for a sentence, so parameters are excluded. | |
| Sep 7, 2021 at 22:50 | comment | added | Paul Blain Levy | Thanks! Another nice answer for (2). And it has the advantage of not referring to syntax. | |
| Sep 6, 2021 at 11:37 | history | answered | Corey Bacal Switzer | CC BY-SA 4.0 |