Set Theory inside Arithmetics via the Ackermann Yoga

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 and here 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?

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I'm imagining a non-mathematician reading titles involving the word "yoga" and asking himself in which perverted ways mathematicians use to convey ideas to each other... – Qfwfq Jun 7 '11 at 12:35
LOL!!! well, tell your non-mathematician to check out this MO question: mathoverflow.net/questions/64071/… PS As u certainly know, there are many types of Yoga (Hatha Yoga,the most popular, being only the first level): Raja Yoga, Jnana Yoga, etc. so why not Ackermann 's Yoga? It stretches your mind.... – Mirco A. Mannucci Jun 7 '11 at 12:46

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.

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.

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I don’t see why $T^A$ is consistent with $(M_k,\in_{Ack})$. Doesn’t this require $M_k$ to satisfy all $\Sigma_1$ consequences of $T$, or something like that? – Emil Jeřábek Jun 7 '11 at 17:26
Aha, I’ll answer myself. $M_k$ is not an end-extension of $A$, hence it only needs to satisfy existential consequences of $T$ to make it work. Now, in set theory, there is no MRDP theorem, and existential sentences have very little expressive power: they assert the existence of a finite configuration of points with prescribed elementhood relation on them. It is easy to see that every such sentence is decidable in ZF without infinity (or its negation), hence if $T$ is consistent, its existential consequences are indeed valid in $M_k$. Alright, this is magic. – Emil Jeřábek Jun 7 '11 at 17:54
@Emil: Yes, you have the perfect explanation. – Ali Enayat Jun 7 '11 at 18:17
Indeed this results is a real pearl: totally fascinating! Do you know of any type of classification of nonstandard models of PA on the basis of how strong their internal set theory is? For instance, in certain models an Ackermann "ordinal" could be an inaccessible cardinal.. – Mirco A. Mannucci Jun 8 '11 at 19:10
@Mirco: Ressayre's result applies to all nonstandard model of $PA$, and all consistent extension $T$ of $ZF$, including those that assert the existence of large cardinals; so for better-or-worse, it does not lend itself as a tool of classification. Also, since the model produced in Ressayre's theorem is not "internal" to the ambient model of arithmetic, the theorem does not say much about the "internal set theory" of the model, rather, it shows it is so rich that, externally, we can carve out models of strong set theories from it. – Ali Enayat Jun 8 '11 at 21:07

You might want to look into the notion of the standard system of a nonstandard model $M$ of PA. Any nonstandard member $x$ of $M$, determines, via your favorite coding, a subset $X$ of $M$ that is finite in the internal sense of $M$ but may be infinite when seen from outside $M$. Intersecting this $X$ with the standard part $\mathbb N$ of $M$, we get some subset of $\mathbb N$, and the family of all the intersections obtainable in this way, as $x$ varies over $M$, is called the standard system of $M$. With this definition, it's a collection of subsets of $\mathbb N$, but it corresponds, via Ackermann coding, to a collection of subsets of the set $HF$ of hereditarily finite sets (the standard model of ZF minus infinity).

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