Timeline for Extensions of the Ackermann interpretation to nonstandard theories of arithmetic
Current License: CC BY-SA 4.0
29 events
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| S Mar 13, 2023 at 2:16 | history | suggested | C7X | CC BY-SA 4.0 |
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| Mar 13, 2023 at 2:13 | review | Suggested edits | |||
| S Mar 13, 2023 at 2:16 | |||||
| Dec 29, 2016 at 23:43 | vote | accept | Thomas Benjamin | ||
| Dec 29, 2016 at 17:14 | comment | added | Noah Schweber | @ThomasBenjamin I've edited my answer to incorporated some of my comments. Feel free to roll it back to the previous version if you think that one was better, but I think the version now is a bit clearer. | |
| Dec 29, 2016 at 17:03 | comment | added | Thomas Benjamin | @NoahSchweber: This is very helpful. Thanks | |
| Dec 29, 2016 at 16:55 | comment | added | Noah Schweber | The point is: definable phenomena aren't going to grant you surprising power, even in nonstandard models. And Parikh's exponentiations, being undefinable, play no role in the analysis of the Ackermann interpretation. | |
| Dec 29, 2016 at 16:53 | comment | added | Noah Schweber | This is false. Think about how - even in a nonstandard model of PA - there are no definable cuts. If $M$ is a nonstandard model of $PA$, then $Ack(M)$ is a nonstandard model of $ZF-Inf+\neg Inf$, but it still satisfies $\neg Inf$ internally. So in particular it doesn't help you define $\omega$. I think the key point you are missing is that the Ackermann interpretation, being definable, makes sense for all elements of any model of PA, and the relevant facts about it, being PA-expressible (by the definability of the Ackermann interpretation) and PA-provable, are true in all models. | |
| Dec 29, 2016 at 16:51 | comment | added | Noah Schweber | even though of course the Ackermann structure of $M$, $Ack(M)$, thinks that every set is finite. This is just the usual phenomenon of a theory PA or ZF-Inf+$\neg$Inf thinking that every object is finite, but having models with externally infinite objects. This addresses the "appears finite" issue. The other point is that the Ackermann interpretation is already extended to all of the model of PA, including the nonstandard elements! Parikh's model doesn't gain us anything here. You then write that the Ackermann interpretation of a nonstandard number should help define $\omega$; (cont'd) | |
| Dec 29, 2016 at 16:49 | comment | added | Noah Schweber | The Ackermann interpretation is definable. That is, there is a formula $\varphi(x, y)$ - interpreted as "the set represented by $x$ is an element of the set represented by $y$ - such that PA proves that the structure $(\mathbb{N}, \varphi)$ is a model of ZF-Inf+$\neg$Inf. Note that "$\mathbb{N}$" here refers to the set of natural numbers as understood by the model of PA - this is what I meant by "appear finite": a nonstandard element of $M\models PA$ appears finite to $M$, even though it is externally infinite. Such an element will code an externally infinite set (cont'd) | |
| Dec 29, 2016 at 13:33 | review | Close votes | |||
| Dec 29, 2016 at 15:32 | |||||
| Dec 29, 2016 at 13:04 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
corrected spelling
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| Dec 29, 2016 at 12:48 | comment | added | Thomas Benjamin | @MattF.: Sorry my presentation is not entirely clear. I will work on it. | |
| Dec 29, 2016 at 12:45 | comment | added | Thomas Benjamin | (cont.) This is why I believe the Ackermann interpretation interacts with Ressayre's substructures and Parikh's exponentiations (rightly or wrongly) and why I asked the questions I asked. Hope this helps. | |
| Dec 29, 2016 at 12:41 | comment | added | Thomas Benjamin | (cont.) extensionality, empty set, sum set, foundation, transitive closure, power set, pair set, and the negation of the axiom of infinity. In $M_{Parikh}$, by his theorem in Prof Buss's paper and Ressayre's theorem, it would seem one can extend the Ackermann interpretation to nonstandard (infinite, in the external venue) integers, which seems to allow one to be able to define the infinite set $\omega$ (the sum set of $V_{\omega}$), and Ressayre's theorem seems to allow for an Ackermann interpretation of the axiom of infinity in $M_{Parikh}$ (what would that Ackermann interpretation be?). | |
| Dec 29, 2016 at 12:20 | comment | added | Thomas Benjamin | @NoahSchweber: In your comment, "Now that's certainly false for $PA$: the only sets that a model of $PA$ sees via the Ackermann interpretation are (or rather, appear) finite!", could you define term 'appear'? This seems to go back to Prof. Enayat's comment to Mirco regarding the distinction between internal' and 'external' venues. In the Kay and Wong paper, $I{\Delta_0}$ + ($\forall$$x$)($\exists$$y$)($y$=$2^x$) ("the totality of the function $x$$\mapsto$$2^x$") one can prove ( what does this mean in this context?) the Ackermann interpretation of the following set theory axioms: | |
| Dec 29, 2016 at 8:30 | comment | added | user44143 | Will you simplify this? There are 11 paragraphs of introduction here before any direction to a question. Then there are 6 questions, of which 4 use the word "know" in quotes. It would help to either ask directly for a formalization of "know", in which case the four questions here are secondary; or suggest a definition so you can ask the questions without the quote marks. | |
| Dec 29, 2016 at 6:49 | comment | added | Noah Schweber | (What do I mean by "knows the Henkin recipe"? Well, think of Henkinization as a process: given a consistent theory $T$, I build an increasing sequence of finite sets of sentences whose union is the elementary diagram of a model of $T$. The point is that the statement "$n$ codes the $s$th step in the Henkin process for $T$" is expressible in $PA$ (so long as $T$ is appropriately definable).) | |
| Dec 29, 2016 at 6:46 | comment | added | Noah Schweber | So what can PA do? Well, PA knows the Henkin recipe for building a model of a consistent theory, even if it doesn't "have time to do the whole recipe". What PA can do is get a definable model of the theory, but not a "set" model (not finite, that is). For instance, if $M$ is a model of PA which thinks ZFC is consistent, then there is a formula defining a binary relation $E$ on $M$ such that $(M, E)\models ZFC$. Think of this as a class model, rather than a set model. PA is too "small" to have enough sets for the completeness theorem, but we still get the arithmetized completeness phenomenon. | |
| Dec 29, 2016 at 6:42 | comment | added | Noah Schweber | This is the arithmetized completeness theorem; see page 83 of this paper. Here's a brief summary. ZF proves the completeness theorem: if W is a model of ZF and T is a theory in W, and W thinks T is consistent, then W thinks T has a set model. Now, that's certainly false for PA: the only sets that a model of PA sees via the Ackermann interpretation are (or rather, appear) finite! (cont'd) | |
| Dec 29, 2016 at 6:34 | comment | added | Thomas Benjamin | @NoahSchweber: Perhaps it would help if you would clarify what Prof. Enayat means when he says that "$PA$ knows about the completeness theorem". The words 'know' and 'believe', when it comes to formal theories and models of formal theories, are often used without any precise definitions of these terms. Is it possible for you to provide these precise definitions? It would help me a lot....(thanks in advance). | |
| Dec 29, 2016 at 6:25 | comment | added | Noah Schweber | Ah, sorry, that was a bad read on my part. I'm a little unclear on what you're asking, though. Could you clarify what you mean by "know about", and how Parikh's model is relevant (see my answercomment for why I'm asking this)? | |
| Dec 29, 2016 at 6:25 | comment | added | Thomas Benjamin | @NoahSchweber: No. All I am trying to do here is cover all the bases (syntactic and semantic). I just didn't space properly. Sorry. (Note: I added extra spacing between '$PA$' and '($M_{Parikh}$)' but it didn't seem to help much. You can read the last question as pertaining to $PA$ or to '$M_{Parikh}$' as you choose. Prof. Enayat seems, in his comment to Mirco that I quoted, to jump between $PA$ and "internal models" (submodels of $M_{parikh}$?) so I put both in.Should I Leave out $PA$ and opt for $M_{Parikh}$, perhaps? | |
| Dec 29, 2016 at 6:14 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
added better spacing for clarity
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| Dec 29, 2016 at 6:07 | answer | added | Noah Schweber | timeline score: 6 | |
| Dec 29, 2016 at 6:07 | comment | added | Noah Schweber | What is $PA(M_{Parikh})$? Is that the thing $M_{Parikh}$ thinks is $PA$? | |
| Dec 29, 2016 at 6:04 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
finished last question
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| Dec 29, 2016 at 6:00 | comment | added | Thomas Benjamin | @NoahSchweber: You are so right! I am in the process of finishing. | |
| Dec 29, 2016 at 5:59 | history | edited | Thomas Benjamin | CC BY-SA 3.0 |
finished last question
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| Dec 29, 2016 at 4:59 | history | asked | Thomas Benjamin | CC BY-SA 3.0 |