Timeline for Does ACA prove categoricity of the reals?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 16, 2021 at 4:01 | vote | accept | user21820 | ||
| Oct 16, 2021 at 4:01 | comment | added | user21820 | Ah ok thanks for clarifying (✻). For my second question I see I've to think more carefully. Thanks again! | |
| Oct 16, 2021 at 3:40 | comment | added | Noah Schweber | (To clarify, we can talk in the language of $\mathsf{ACA_0}$ about whether a second-order property in the standard sense holds of a countably coded structure, but this doesn't buy us anything here since $\mathsf{RCA_0}$ already proves that there are no countable complete ordered fields.) | |
| Oct 16, 2021 at 3:35 | comment | added | Noah Schweber | @user21820 "could you clarify what you meant by "ACA0 is enough [for (✻)]?" By "$\mathsf{ACA_0}$ is enough" I meant "the answer to $(*)$ is yes," that is, that $\mathsf{ACA_0}$ alone is enough to ensure the "interpretable categoricity" phenomenon I describe. "is it then correct that if I have full-semantics second-order completeness then ACA does in fact prove the categoricity?" The problem is that the language of second-order arithmetic can't even talk about genuine second-order semantics in the way it would need to. I don't see a way to avoid interpretations or something equivalent here. | |
| Oct 16, 2021 at 3:20 | comment | added | user21820 | Oh, gah, now I see; I accidentally applied "internal completeness" wrongly, to a set that was at the meta-level... The rationals are forced to be identical by the "$ω$-model" condition, but the interpretation of $S$ can lack the set I wanted to apply completeness to. Sigh, what a silly mistake. So is it then correct that if I have full-semantics second-order completeness then ACA does in fact prove the categoricity? (I won't change my question but I just wanted to confirm what I was missing.) Also, could you clarify what you meant by "ACA0 is enough [for (✻)]? Thanks! =) | |
| Oct 16, 2021 at 1:42 | comment | added | Noah Schweber | @user21820 I've added a brief recipe explaining why the version you pose must be false. | |
| Oct 16, 2021 at 1:42 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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| Oct 15, 2021 at 20:58 | comment | added | Noah Schweber | @user21820 That version is now false - note that classically your theory $T$ has many non-isomorphic countable $\omega$-models. I've edited to say a bit about this. | |
| Oct 15, 2021 at 20:58 | history | edited | Noah Schweber | CC BY-SA 4.0 |
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| Oct 15, 2021 at 20:34 | comment | added | user21820 | Hmm, could you please check out my current version of the question? I think it's the precise form of what I had in mind. | |
| Oct 15, 2021 at 20:15 | comment | added | Noah Schweber | @user21820 Some serious circumlocution will be required to get a syntactic phrasing of this, since it's quantifying over arbitrary-complexity interpretations and arbitrary-complexity definable isomorphisms and the completeness criterion itself involves quantifying over arbitrary-complexity sequences. Basically, you'll get a family of $\mathsf{ACA_0}$-theorems like "Every $\Sigma_{17}$-interpretation of a $\Sigma_{38}$-complete ordered field is $\Sigma_{42}$-isomorphic to $\mathbb{R}$." It's nasty. | |
| Oct 15, 2021 at 19:58 | comment | added | Noah Schweber | @user21820 I don't actually think "precisiate" is an English word, it's something I saw a speaker say frequently (Lofti Zadeh, in the context of fuzzy logic) and liked so much I stole it. :P | |
| Oct 15, 2021 at 19:58 | comment | added | user21820 | I believe your answer would be a semantic analogue of the syntactic version I actually had in mind, which I am in the midst of 'precising' (I didn't know it is an English word haha). Sorry to make you wait. =) | |
| Oct 15, 2021 at 19:54 | history | answered | Noah Schweber | CC BY-SA 4.0 |