In their paper, " On Interpretations of Arithmetic and Set Theory" (*Notre Dame Journal of Formal Logic*, Vol. 8, No. 4 (2007), pp. 497-510) in section 7, "Fragments of Arithmetic and Set Theory", Kaye and Wong make the following (interesting) remarks:

It is fairly straightforward to see that $I{\Delta_0}$ proves the Ackermann interpretation of many set theory axioms including extensionality, empty set, sum set, foundation, transitive closure, and the negation of the axiom of infinity.

Adding the further arithmetic axiom $exp$, ($\forall$$x$)($\exists$$y$)($y$=$2^x$), stating the totality of of the function $x$$\mapsto$$2^x$, we may prove the Ackermann interpretation of of the pair axiom and the power set axiom. Moreover, $exp$ is necessary as well as sufficient for these: to show $I{\Delta_0}$$\vdash$$Pair^{\mathfrak a}$$\rightarrow$$exp$ and $I{\Delta_0}$$\vdash$$Pow^{\mathfrak a}$$\rightarrow$$exp$, take a number $y$ and $2^x$, the smallest power of two greater than $y$. Then as $2^x$ codes the set {$x$}, by pair set or power set the set {$x$, {}} must also be coded, and this code can only be $2^{2^x}$+1. Thus $2^{2^x}$ exists and so therefore does $2^y$.

Furthermore, consider the following theorem of Ressayre as quoted by Ali Enayat in his answer to Mirco Mannucci's mathoverflow question, "Set Theory inside Arithmetic via the Ackermann Yoga":

Suppose $($ $M$, $+$, $\cdot$ $)$ is a nonstandard model of $PA$, and $\in_{Ack}$ is the Ackermann epsilon on $M$, i.e., $a$$\in_{Ack}$$b$ iff $M$ satisfies "the $a$-th digit of the binary expansion of $b$ is 1". Then for every consistent recursive extension $T$ of $ZF$ there is a subset $A$ of $M$ such that $($ $A$, $\in_{Ack}$ $)$ is a model of $T$.

Prof. Enayat then makes the following comment to Mirco in the the 'comments' section for the question:

$PA$ proves the negation of the axiom of infinity for the Ackermann interpretation, so far as the Ackermann interpretation is concerned, every set is "doomed" to be finite from $PA$'s point of view. So there indeed is a severe limit to the "Yoga", unless--as in Ressayre's theorem--one moves to an

externalvenue.

It should be noted that Sayre and Wong, in their above remarks, sem to be working in that external venue.

Finally, consider the following statement Samuel Buss made regarding Rohit Parikh's 1971 paper "Existence and feasibility in arithmetic" in his own paper, "Bounded arithmetic, proof complexity and two papers of Parikh" (*Annals of Pure and Applied Logic* 96 (1999), 43-55):

(i) In section 3 of the "feasibility" paper, Parikh presents a model-theoretic result illustrating the gap between exponentiation and the feasible operations of addition and multiplication in models of arithmetic. Specifically, consider the axioms

$f$($x$,0)=1, $f$($x$, $y$+$z$)=$f$($x$,$y$) $\cdot$ $f$($x$,$z$)

$f$($x$,$S$($y$))=$x$ $\cdot$ $f$($x$,$y$), $f$($x$, $y$$\cdot$$z$)=$f$($f$($x$,$y$), $z$)

which uniquely characterize $f$($x$,$y$) as the exponentiation function $x^y$ in the standard integers. Parikh proved, however, that there is a non-standard model of

Th($\mathbb N$) and two distinct function $f_1$ and $f_2$ on $M$ both of which satisfy the above four axioms for all values of $x$, $y$, $z$ $\in$ $M$. [Parikh seems to also be working in Ressayre's 'external venue'--my comment.]

Consider now Parikh's non-standard model $M_{Parikh}$. By Ressayre's theorem, there is a subset $A$ of $M_{Parikh}$ such that $($ $A$, $\in_{Ack}$ $)$ is a model of $ZF$, including the Axiom of Infinity.

Question: Does $M_{Parikh}$ 'know' that the Ackermann interpretation of the axiom of infinity *is* the axiom of infinity (in the 'external venue')?

Question: Is the set {0,1,2,3,...}=$\omega$ representable in $($ $A$, $\in_{AcK}$ $)$, and does $M_{Parikh}$ 'know' this?

Question: In a comment to Mirco regarding his question, Prof. Enayat writes: "Since $PA$ 'knows' about the completeness theorem (thanks to a theorem of Hilbert-Bernays), arithmetical models of such consistency statements can define an epsilon relation on themselves that turns them into models of set theory; moreover, the model of arithmetic has even a definable truth-predicate for such internal models of set theory. But such internal models, from the point of view of the ambient model of arithmetic, have nonstandard integers." Apparently, this is is also true of $M_{Parikh}$. In what way does $PA$ ($M_{Parikh}$) 'know' about the completeness theorem? Does $PA$ ($M_{Parikh}$) 'know' that $($ $A$, $\in_{Ack}$ $)$ is a model of $ZF$? How does Goedel's second incompleteness theorem apply here (how does $PA$ ($M_{Parikh}$) avoid inconsistency in this situation)?

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) $\endgroup$ – Noah Schweber Dec 29 '16 at 16:49is alreadyextended 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) $\endgroup$ – Noah Schweber Dec 29 '16 at 16:51but 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 forallelements ofanymodel 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. $\endgroup$ – Noah Schweber Dec 29 '16 at 16:5314more comments