Timeline for Had this attempt to salvage naïve comprehension been studied before?
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24 events
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| Aug 8 at 13:41 | comment | added | Joel David Hamkins | OK, so the thing to do is to prove upper and lower bounds on the interpretative strength of the theory, not to throw it all out by going to finite fragments. | |
| Aug 8 at 13:21 | comment | added | paste bee | It might be possible but I think we have to be careful about it. The sets defined by formulas that don't contain $\in$ are exactly the finite and cofinite sets, which means that by $\in_1$ we can easily interpret arithmetic truth, defining the natural numbers as von Neumann ordinals, and $+$ and $\cdot$ by finding finite sets that code sufficient pieces of them. If we want to keep the strength down, we have to avoid giving the structure the real notion of finiteness. | |
| Aug 8 at 12:54 | comment | added | Joel David Hamkins | OK, I see. I believe you can get even the full theory, if you start only with a naked set, since equivalence of formulas will be relatively weak. For example, on an infinite set with no structure (or with infinitely many distinct constants) we can decide equivalence of formulas, and thereby define suitable $\in_0$ relations, and then iterate this idea I think in PA. | |
| Aug 8 at 12:48 | comment | added | paste bee | If, for example, we take the theory that only contains $\in_0$, $\in_1$, $\in_2$, and which only allows Comprehension with five formulas $\varphi_0$, $\varphi_1$, $\varphi_2$, $\varphi_3$, $\varphi_4$, then PA interprets this theory - since we only have to deal with five formulas, we don't run into the issue that PA can't define arithmetic truth, because we only need a "truth predicate" that works on these formulas in particular. | |
| Aug 8 at 12:31 | comment | added | Joel David Hamkins | Yes, the Parsons construction take place over any first-order model, and the formulas allow the base language. If you just start with a naked set, then I agree it is very weak, but I still don't understand your proposal exactly. | |
| Aug 8 at 12:20 | comment | added | paste bee | But anyway, the key piece is that I said a finite fragment - if we fix finitely many formulas (in the metatheory), and finitely many levels of $\in$, then we don't need the entirety of a truth predicate, we can just handle each formula individually (or if you prefer, we can make it all work with just truth predicates for $\Sigma^0_n$ sentences for various $n$). This doesn't quite prove in PA that the entire theory is consistent, but it shows that if PA is consistent then the theory is consistent. | |
| Aug 8 at 12:16 | comment | added | paste bee | I don't see why it makes a difference whether we start with a model of arithmetic or just a set? Or do you mean that the formulas can also include $+$ and $\cdot$? | |
| Aug 7 at 14:52 | comment | added | Joel David Hamkins | If we just start with a naked set, however, then truth would be trivial, and so we could do the whole construction in PA I think. Perhaps that is what you meant. | |
| Aug 7 at 1:51 | comment | added | Joel David Hamkins | @pastebee I was imagining starting with a model of arithmetic, and then defining the $\in_n$ relations as an expansion of this. I don't quite see how to define the $\in_n$ relations in PA, since we'd need to decide which formulas are equivalent, which is why I think a truth predicate is involved. Can you explain your idea a bit more? | |
| Aug 6 at 21:54 | comment | added | paste bee | If we just want a finite fragment of the theory instead of the whole thing, then we don't even need a truth predicate, and so this argument works in PA. So the theory is at most the strength of first-order arithmetic. I'm not convinced it even interprets PA though - we can make things that look sort of like $\mathbb{N}$, but then I don't know how to make induction work. | |
| Aug 6 at 17:53 | comment | added | Joel David Hamkins | No, it is not strong enough (but I don't know anything about TST). If you start with $X$ as the standard model of arithmetic and do my construction on top, then this is rather low in the hyperarithmetic hierarchy, since you are just iterating a truth predicate $\omega$ many times. There is a model computable from $0^{(\omega^2)}$. | |
| Aug 6 at 17:12 | vote | accept | Zuhair Al-Johar | ||
| Aug 6 at 17:12 | comment | added | Zuhair Al-Johar | Do you think this theory can interpret each $n$-order arithmetic? So, it might be at the strength of Mac Lane set theory? Which is the same of $\sf TST + Infinity$. | |
| Aug 6 at 15:27 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Aug 6 at 13:47 | history | undeleted | Joel David Hamkins | ||
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| Aug 6 at 13:34 | history | deleted | Joel David Hamkins | via Vote | |
| Aug 6 at 13:32 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| Aug 6 at 13:22 | history | answered | Joel David Hamkins | CC BY-SA 4.0 |