Timeline for Lowenheim-Skolem numbers for SOL + correctness quantifiers
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 1, 2021 at 16:39 | vote | accept | Noah Schweber | ||
| May 1, 2021 at 11:45 | answer | added | Farmer S | timeline score: 4 | |
| Sep 29, 2020 at 20:58 | comment | added | Noah Schweber | @YairHayut I see the issue, I was unclear in my definition (and comment): I mean that the cardinality of the set in question is appropriately correct. So it's not coding-dependent and is genuinely a quantifier. | |
| Sep 29, 2020 at 20:57 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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| Sep 29, 2020 at 16:31 | comment | added | Yair Hayut | Obviously I am missing something. The definition of correctness seems to refer only to cardinals, and depends only on the language. So I don't understand your example - it seems like the truth value of the statement depends on the set theoretical coding of the group. | |
| Sep 29, 2020 at 13:00 | comment | added | Noah Schweber | @AsafKaragila Isn't a third-order predicate just a quantifier? | |
| Sep 29, 2020 at 12:59 | comment | added | Asaf Karagila♦ | I guess @Yair is suggesting a 3rd-order predicate? | |
| Sep 29, 2020 at 12:50 | comment | added | Noah Schweber | @YairHayut I don't understand how that would work. Suppose I have a group $G$. With the logic I've described, we can say e.g. that the center of $G$ is $\mathcal{L}^2_0$-correct. How exactly would that be done with a unary predicate? In no sense do particular elements of $G$ correspond to large sets; "of correct size" is a property of subsets, not elements, of the model and so is a quantifier rather than a predicate. | |
| Sep 29, 2020 at 12:41 | comment | added | Yair Hayut | Why can't we consider $C_n$ to be just a unary predicate which is evaluated as the set of all $\mathcal{L}^2_n$-correct cardinals? I don't understand why $C_n$ is a quantifier and not simply an additional predicate. | |
| Sep 29, 2020 at 3:48 | comment | added | Asaf Karagila♦ | If beer isn't helping, try another one. In any case, argue by induction that if $\delta$ is extendible, then the $n$th logic still takes place below it. If it wasn't, for every theory, pick an elementary embedding that moves $\delta$ high enough, and then by elementarity we can pull back to below $\delta$. But that was just off the cuff thought before falling asleep. I must be missing something. | |
| Sep 29, 2020 at 2:13 | comment | added | Noah Schweber | @AsafKaragila I'd love it if that were the case, but I don't see how to get that bound. (Maybe it's obvious and beer isn't helping my set theory.) | |
| Sep 29, 2020 at 0:48 | comment | added | Asaf Karagila♦ | Whatever it is, it looks like it should be at most an extendible. | |
| Sep 29, 2020 at 0:31 | history | asked | Noah Schweber | CC BY-SA 4.0 |