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Peano Arithmetic has an infinite number of axioms because of its induction schema; Likewise $\sf ZFC$ has an infinite number of axioms because of its axiom schema of replacement. $\sf NBG$ however admits a finite axiomatization because of its ontology of classes. Can this class machinery be leveraged to implement a finite axiomatization of $\sf PA$ in first order logic, or does the induction schema require more, like higher order logic?

dezakin
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    In set theory, the Peano's axioms are theorems and nor more axioms. – Mauro ALLEGRANZA Jul 01 '14 at 19:32
  • I'm pretty sure that definitions need to be applied, and they look like axioms, even if from a set theoretic perspective they can all be reduced to primitives. Your point is well taken in that it suggests that $\sf PA$ can be reduced to a finite axiomatization in $\sf NBG$ unless it requires something I've never heard of, like a definition schema. – dezakin Jul 01 '14 at 19:42

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Yes. There is a theory of second-order arithmetic known as $\mathsf{ACA}_0$ which is to PA just as NBG is to ZFC. $\mathsf{ACA}_0$ is a finitely axiomatizable theory in first-order logic. It has two types of objects: natural numbers and sets of natural numbers (much as NBG has sets and classes of sets).

And it is a conservative extension: the sentences in the original language of Peano arithmetic that are provable in $\mathsf{ACA}_0$ are all provable in PA, just as the sentences in the language of set theory that are provable in NBG are all provable in ZFC.

Moreover, the proofs of conservativity are very similar. In each case one extends a model of the one-sorted theory (PA, ZFC) to a model of the two-sorted theory ($\mathsf{ACA}_0$, NBG) by adding only those new objects (sets of numbers, classes of sets) that are definable over the original model.

Carl Mummert
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    Carl, about your last paragraph, you may want to mention that the proof of conservativity you are describing takes place in a theory other than $\mathsf{PA}$. Proving conservativity arguing within $\mathsf{PA}$ is not as straightforward. – Andrés E. Caicedo Jul 02 '14 at 01:43
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    @Andres: that is a very good point. The proof I had in mind is entirely model theoretic and not formalized in PA. I will simply link to two relevant places: http://mathoverflow.net/questions/127080/what-metatheory-proves-mathsfaca-0-conservative-over-pa and http://www.andrew.cmu.edu/user/avigad/Talks/semantic.pdf – Carl Mummert Jul 02 '14 at 01:54
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See Herbert Enderton, Elements of Set Theory (1977), page 70-on : PEANO'S POSTULATES for a proof of Peano's axioms in $\mathsf {ZFC}$.

In first-order languages, individual variables range over "objects" of the domain of interpretation.

In first-order $\mathsf {PA}$ the "objects" are numbers; thus, we need the axiom schema of induction because in this theory it is not possible to quantify over sets of natural numbers.

In set theory The "objects" are sets, i.e. the individual variables range over sets; thus the induction principle is "naturally" expressible and it is provable from $\mathsf {ZFC}$'s axioms.

Mauro ALLEGRANZA
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  • While I can follow the reasoning on why the answer should be 'yes.' this doesn't directly answer it and doesn't reference finitely axiomatizable set theories. I understand that $\sf NBG$ is a conservative extension of $\sf ZFC$ and so every theorem of $\sf NGB$ is also a theorem of $\sf ZFC$ so it would appear that the induction principle should also be present in $\sf NGB$ but this isn't as explicit as it could be. – dezakin Jul 01 '14 at 21:27
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    @dezakin - for a treatment of set-teory according to NBG, see : Elliott Mendelson_Introduction to mathematical logic (4ed - 1997). He prove PROPOSITION 4.13 : TRANSFINITE INDUCTION in several forms, included one called : Induction up to $\omega$ : $\emptyset \in X \land \forall \alpha (\alpha < \omega \land \alpha \in X \Rightarrow \alpha' \in X) \Rightarrow \omega \subseteq X$ – Mauro ALLEGRANZA Jul 02 '14 at 06:25
  • @dezakin - you can see also Patrick Suppes, Axiomatic set theory (1960). He proves the "standard" set-theoretic version of induction axiom [page 139] : Theorem 24. If (i) $\emptyset \in A$, (ii) for every $n$, if $n \in A$ then $n' \in A$, then $\omega \subseteq A$. But he proves also [page 136] the : Theorem Schema 22. If (i) $\varphi(\emptyset)$, (ii) for every $n$, if $\varphi(n)$ then $\varphi(n')$, then for every $n$, $\varphi(n)$. – Mauro ALLEGRANZA Jul 02 '14 at 06:33