4 Natural numbers[1X9]

We want to properly define the set

\[ {\mathbb N}=\{ 0,1,2,\ldots \} \]

of the natural numbers.

A possible model, as shown in Sec. 3.8, is obtained by relying on the theory of Zermelo—Fraenkel.

Here instead we present Peano’s axioms, expressed using the naive version of set theory.

Definition 130 Peano’s axioms

[1XB](Solved on 2022-11-03)

(N1): There is a number \(0∈ ℕ\).
(N2): There is a function \(S:ℕ → ℕ\) (called "successor"), such that ¹
(N3): \(S(x)≠ 0\) for each \(x∈ ℕ\) and
(N4): \(S\) is injective, that is, \(x≠ y\) implies \(S(x)≠ S(y)\).
(N5): If \(U\) is a subset of \(ℕ\) such that: \(0∈ U\) and \(∀ x, x ∈ U⇒ S(x)∈ U\) , then \(U=ℕ\).

We will often write \(Sn\) instead \(S(n)\) to ease notations.

From those two important properties immediately follow. One is the principle of induction, see 133. The other is left for exercise.

Exercise 131

[1YP] Show that every \(n∈ℕ\) with \(n≠ 0\) is successor of another \(k∈ℕ\), proving by induction on \(n\) this proposition

\[ P(n) \, {\stackrel{.}{=}}\, (n=0) ∨ (∃ k ∈ℕ, S(k)=n) \quad . \]

This shows that the successor function

\[ S:ℕ → ℕ⧵\{ 0\} \]

is bijective.

If \(n\neq 0\), we will call \(S^{-1}(n)\) the predecessor of \(n\).

(Part of this result applies more generally, see 1)

The idea is that the successor function encodes the usual numbers according to the scheme

\[ 1=S(0),\quad 2=S(1), \quad 3=S(2)\ldots \]

and (having defined the addition) we will have that \(S(n)=n+1\).

Exercise 132

[1XD]Prerequisites:132.Removing one of the axioms (N1)—(N5), describe a set that satisfies the others but it is not isomorphic to natural numbers.