The Peano Axioms depend on the concept of sets, i.e., sets need to be defined before the Peano axioms can be used.
Axioms cannot be proven. This means that if I have a system of axioms then no axiom in it can be proven uses any of the other ones.
If I define sets using only the axiom of unrestricted comprehension or naive set theory, then it wouldn't be problematic to then also define the Peano axioms.
However, if I define a set using ZF(C), which is the most common way to define a set nowadays, then I can use those axioms to construct the natural numbers without the need for other axioms as shown here (see pages 1-8) or this StackExchange post here.
Furthermore,
From Wikipedia:
The need to formalize arithmetic was not well appreciated until the work of Hermann Grassmann, who showed in the 1860s that many facts in arithmetic could be derived from more basic facts about the successor operation and induction. In 1881, Charles Sanders Peirce provided an axiomatization of natural-number arithmetic. In 1888, Richard Dedekind proposed another axiomatization of natural-number arithmetic, and in 1889, Peano published a simplified version of them as a collection of axioms in his book, The principles of arithmetic presented by a new method (Latin: Arithmetices principia, nova methodo exposita).
In 1908, Ernst Zermelo proposed the first axiomatic set theory, Zermelo set theory.
The above quotes show that the peano axioms were formed before the modern set axioms, which would explain why the peano axioms construct natural numbers seperately from ZF.
As a result I'd like to ask the following questions:
- are the peano axioms not necessary if ZF(C) is used?
- if so, does that mean that the peano construction of natural numbers is outdated?
- the Paris–Harrington theorem says the strengthened finite Ramsey theorem, is true, but not provable in Peano arithmetic. Is it also not provable for arithmetic built on the set-theoretic construction of numbers.
- and, following from 3, but more generally: how does the fact that the ZF axioms and Peano axioms define natural numbers in a different way affect the way arithmetic works (e.g. what statements can be dis/proven, how certain concepts are defined, etc.)
Best Answer
This is incorrect. There are several ways to deal with the Peano axioms without discussing sets at all.
The only Peano axiom which deals with sets explicitly, in some formulations, is the axiom of induction, which states
$$\forall P \subseteq \mathbb{N} . 0 \in P \land (\forall n \in P . s(n) \in P) \to P = \mathbb{N}$$
However, induction can be rephrased to work with either first or second-order logic.
The second-order logic formulation is a direct rephrasing of the above.
$$\forall P . P(0) \land (\forall n . P(n) \to P(s(n))) \to \forall n . P(n)$$
For first-order logic, we make induction an axiom scheme. The axiom scheme states that for any predicate $\phi(n, x_1, x_2, ..., x_n)$ definable in the language of first-order arithmetic, we have
$$\forall x_1 \forall x_1 ... \forall x_n . \phi(0, x_1, ..., x_n) \land (\forall n . \phi(n, x_1, ..., x_n) \to \phi(s(n), x_1, ..., x_n)) \to \forall n . \phi(n, x_1, ..., x_n)$$
Neither of these formulations explicitly invokes sets.
As for the answers to your questions:
They are no longer axioms, but theorems which hold about a specific set $\mathbb{N}$. They are still extremely useful, as the second-order version of the Peano axioms uniquely characterises the natural numbers up to bijection.
Certainly not.
The Paris-Harrington theorem states that a specific theorem in Ramsey theorem cannot be proved in first-order Peano arithmetic. This version of the Ramsey theorem can be proved in second-order Peano arithmetic, which is part of the set-theoretic logic of natural numbers.
ZF proves that any statements which can be proved in (first-order) Peano arithmetic are actually true about the set of natural numbers. ZF also proves some statements which can be phrased in the language of Peano arithmetic but cannot be proved with first-order Peano arithmetic. The simplest of these statements is that first-order Peano arithmetic is consistent.