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The von Neumann naturals form a Peano system

Statement

The triple $(N, 0, σ) = (ω, \emptyset, \cdot^{+})$ satisfies the three Peano axioms (Peano system): (P1) $σ (n) \neq 0$ for all $n$ ; (P2) $σ$ is injective; (P3) induction. Hence a Peano system exists.

Facts & Assumptions

Given: $N = ω$ , $0 = \emptyset$ , $σ (n) = n \cup {n}$ (The natural numbers $N$ (von Neumann)).

[L1]

Induction principle (The principle of mathematical induction).

[L2]

Every natural number is transitive and satisfies $n \notin n$ (Every natural number is a transitive set and is not a member of itself).

Proof

technique · direct

1.1L1

(P3) The induction axiom holds: it is exactly the induction principle.

1.2given

(P1) For every $n$ , $σ (n) = n \cup {n}$ contains $n$ , so $σ (n)$ is nonempty and $σ (n) \neq \emptyset = 0$ .

1.3L2given

(P2) Suppose $σ (m) = σ (n)$ , i.e. $m \cup {m} = n \cup {n}$ ; then $m \in n \cup {n}$ so $m \in n$ or $m = n$ , and symmetrically $n \in m$ or $n = m$ ; if $m \neq n$ then $m \in n$ and $n \in m$ , whence by transitivity [L2] $m \in n$ gives $m \subseteq n$ , and $n \in m \subseteq n$ gives $n \in n$ , contradicting $n \notin n$ [L2]; therefore $m = n$ and $σ$ is injective.

2.1step 1.1step 1.2step 1.3∎

P1, P2, P3 all hold, so $(N, 0, σ)$ is a Peano system; in particular a Peano system exists.

The von Neumann naturals form a Peano system

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources