Timeline for How special is first-order $\mathsf{PA}$?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 3, 2021 at 18:58 | vote | accept | Noah Schweber | ||
| Mar 3, 2021 at 18:55 | comment | added | James E Hanson | Yes. I can write it up as an answer to your MSE question. | |
| Mar 3, 2021 at 18:29 | comment | added | Noah Schweber | By "full second-order arithmetic" I presume you mean the first-order theory $\mathsf{Z}_2$, right? (God I hate us sometimes ... :P) If that's right, that would be awesome - I am entirely hooked on $Q_{\mathsf{Brch}}$ now. | |
| Mar 3, 2021 at 18:26 | comment | added | James E Hanson | @NoahSchweber By modifying the proof of compactness of $\mathcal{L}(Q_{\mathrm{Brch}})$ in the Mekler-Shelah paper, I think you can show that any model of full second-order arithmetic has a (first-order) elementary extension whose number sort is a model of $\mathsf{PA}(Q_{\mathrm{Brch}})$, which would imply that it doesn't entail true arithmetic. | |
| Mar 3, 2021 at 16:40 | comment | added | Noah Schweber | Right you are, I just realized my mistake! | |
| Mar 3, 2021 at 16:39 | comment | added | James E Hanson | Right, but does the branch corresponding to the initial $\omega^M$ not always exist? | |
| Mar 3, 2021 at 16:38 | comment | added | James E Hanson | Infiniteness matters because as written the tree has no branches if either of $A$ or $B$ is finite. | |
| Mar 3, 2021 at 16:35 | comment | added | James E Hanson | I'm implicitly assuming $A$ and $B$ are infinite. One way to make sure that there's always a branch would be to allow repetition in the isomorphism, but anyways you can always define a map between the initial $M$ part of two infinite well-orderings, so that branch is always there. | |
| Mar 3, 2021 at 16:35 | comment | added | Noah Schweber | (The point about compact logics is a great one, and I did some looking there as well - I just didn't see the promise of this one!) | |
| Mar 3, 2021 at 16:34 | comment | added | James E Hanson | The newer paper I linked is about other fully compact logics. (Incidentally, it was pretty lucky that one of those has such natural computability/reverse mathematical implications. I was just looking at fully compact logics, because I knew those couldn't pin down $\mathbb{N}$.) Some of these logics might be intermediate in the stronger sense. These allow for stuff like quantifying over isomorphisms between definable Boolean algebras or definable ordered fields. | |
| Mar 3, 2021 at 16:26 | comment | added | Noah Schweber | The former - I think the argument is right (although I haven't accepted yet since I'm too tired to do a thorough check at the moment). Basically, I'm now trying to figure out of $Q_{\mathsf{Brch}}$ is an answer to the stronger MSE question too. | |
| Mar 3, 2021 at 16:24 | comment | added | James E Hanson | @NoahSchweber Are you just saying that you think $\mathsf{PA}(Q_{\mathrm{Brch}})$ entails true arithmetic, or are you saying that you think the argument is somehow wrong? | |
| Mar 3, 2021 at 11:33 | comment | added | Noah Schweber | Hm, I guess I need $M$ to admit a full satisfaction class. I suspect it does, though. | |
| Mar 3, 2021 at 11:09 | comment | added | Noah Schweber | I think your claim $1$ already gives true first-order arithmetic, the idea being that $(i)$ we can form trees of approximations to Skolem functions and $(ii)$ code bounded initial segments of sets in $M_2$ by elements of $M$ appropriately. But it's early so I'll need to think about that. | |
| Mar 3, 2021 at 7:56 | history | edited | James E Hanson | CC BY-SA 4.0 |
Typos
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| Mar 3, 2021 at 7:03 | history | answered | James E Hanson | CC BY-SA 4.0 |