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Dec 31, 2016 at 6:12 comment added Thomas Benjamin Any suggestions (or any references I might avail myself of)?
Dec 30, 2016 at 20:05 comment added Noah Schweber @ThomasBenjamin Ah. Well, it sounds to me that what you're really interested in are interpretations of strong set theories in strong theories of arithmetic. These exist, but not via an Ackermann-style construction. The reason is that the Ackermann coding has useful, but in this case frustrating, monotonicity property: if $y$ codes a set with $x$ as an element, then $y>x$, and moreover this is provable in PA. This, together with induction, proves that the Ackermann structure will never satisfy Infinity. If you want to code in a model of Infinity, you need a more complicated coding system.
Dec 30, 2016 at 12:27 comment added Thomas Benjamin (cont.) consistent extension of $PA$ so that the consistency of $ZF$ could be proved in the extension. Since $ZF$ with the negation of the axiom of infinity proves (?) the existence of all finite ordinals, there certainly must be a consistent way (so to speak) to put curly brackets around this collection and call it a 'whole'....
Dec 30, 2016 at 12:18 comment added Thomas Benjamin (cont.) on the existence of a completed infinity. Having noticed (from the Kaye and Wong paper I quoted) that the axioms of $ZF$ with the axiom of infinity negated (under the Ackermann interpretation) are all theorems of $PA$, I was hoping to somehow develop an extension of the Ackermann interpretation in this extension of $PA$ I wished to develop that would derive the axiom of infinity as a theorem via its Ackermann interpretation. My questions were preliminary questions to get a feel for what could be done. (Note: The extension of $PA$ I wished to develop, was, of course, to be a
Dec 30, 2016 at 11:42 comment added Thomas Benjamin What I was hoping to do was develop a formal theory of arithmetic, slightly stronger than $PA$, whose models satisfy the axiom of infinity. I find it interesting that, in point of fact, $PA$ and $ZF$ with the axiom of infinity negated are bi-interpretable, and that $PRA$ + $TI({\epsilon_0})$ can prove the consistency of $ZF$ with the axiom of infinity negated (the bi-interpretability of $PA$ and $ZF$ with the negation of the axiom of infinity seems, superficially, to fly in the face of the late Edward Nelson's contention that $PA$'s induction axiom is somehow dependent
Dec 30, 2016 at 1:15 comment added Noah Schweber Besides, I don't really see how having a different exponential function around helps you - what exactly do you plan on doing with it?
Dec 30, 2016 at 1:09 comment added Noah Schweber @ThomasBenjamin First of all, let's make the question more precise. How do you express the existence of an undefinable map? What you could do is add a new function symbol $f$ to the language, and axioms asserting that it behaves like exponentiation yet is different from the actual (defined-as-usual) exponentiation function. The problem, now, is that the induction scheme of PA can't apply to formulas which use the symbol "$f$": otherwise consider the point of least difference between $f$ and "true" exponentiation. (continued)
Dec 30, 2016 at 1:06 comment added Thomas Benjamin Could the existence of Parikh's undefinable maps which satisfy the axioms for exponentiation be added as extra axioms to $PA$? Could this new system define $\omega$ and negate the negation of the axiom of infinity (if this strategem is silly I will delete the comment)?
Dec 29, 2016 at 23:43 vote accept Thomas Benjamin
Dec 29, 2016 at 17:14 history edited Noah Schweber CC BY-SA 3.0
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Dec 29, 2016 at 6:46 history edited Noah Schweber CC BY-SA 3.0
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Dec 29, 2016 at 6:25 history edited Noah Schweber CC BY-SA 3.0
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Dec 29, 2016 at 6:07 history answered Noah Schweber CC BY-SA 3.0