Among the basic results of logic which, simple as they are, never fail to intrigue me, is Ackermann's interpretation of ZF-Infinity in PA (see for refs this MO [question](https://mathoverflow.net/questions/63887/non-standard-models-of-finite-set-theory) and [here](http://www.math.nus.edu.sg/%7Ematwtl/papers/finitesettheory.pdf) for an excellent overview)


Of course, the minus here is critical: PA does not know anything about infinite objects.


Yet, my curiosity is by no means $\aleph\_0$-bound:one could add to some fragment of PA strong enough to verify all the axioms of ZF-infinity another axiom stating the existence of a number encoding an infinite countable set. That number would be non-standard in all models, but so what?


Perhaps the Ackermann Yoga can be pushed even further, attempting *to add higher infinity axioms* to the arithmetical theory, to "catch up" with Set Theory. The Ackermann's Yoga could give some insights on non standard models of arithmetics, by thinking of some nonstandard elements as large sets.


Has anything been investigated along these lines?



Set theory inside arithmetics via the Ackermann yoga
This question reminds me of a magical little-known theorem of Jean Pierre Ressayre that shows that every nonstandard model of $PA$ has a model of $ZF$ as a *submodel* of its Ackermann intepretation, more specifically:


**Theorem.** [Ressayre] 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 $\mathcal{M}$ satisfies "the $a$-th digit in 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 $T$.


**Proof Outline:** By Löwenheim-Skolem, it suffices to consider the case when $M$ is countable. Choose a nonstandard integer $k$ in $M$, and consider the submodel $M\_k$ of $(M,\in\_{Ack})$ consisting of sets of ordinal rank less than $k$ [as computed within $(M,\in\_{Ack})$]. "Usual arguments" show that $(M,\in\_{Ack})$ has a Tarskian truth-definition for $M\_k$, which in turn implies that $(M\_k,\in\_{Ack})$ is *recursively saturated*. Since $M\_k$ is also countable, $(M\_k,\in\_{Ack})$ must be *resplendent* [which means that it has an expansion to every recursive $\Sigma^1\_1$ theory that its elementary diagram is consistent with]. 


Now add a new unary predicate symbol $A$ to the language ${\in}$ of set theory and consider the (recursive) theory $T^A$ consisting of sentences of the form $\phi^A$, where $\phi \in T$, and $\phi^A$ is obtained by relativizing every quantifier of $\phi$ to $A$. It is not hard to show that $T^A$ is consistent with the the elementary diagram of $(M\_k,\in\_{Ack})$, so by replendence the desired $A$ can be produced. 


[I will be glad to add clarifications]


Ressayre's theorem appears in the following paper:


J. P. Ressayre, *Introduction aux modèles récursivement saturés*, Séminaire Général de Logique 1983–1984 (Paris, 1983–1984), 53–72, Publ. Math. Univ. Paris VII, 27, Univ. Paris VII, Paris, 1986. 


