Minimally Inductive Set forms Peano Structure

Theorem
Let $\omega$ be the minimal infinite successor set.

Then $\omega$ fulfils Peano's axioms, and hence $\omega$ is a Peano structure.

Proof
From the definition:

P1
$\omega \ne \varnothing$ by definition.

P2
Let $s: \omega \to \omega$ be defined as:
 * $\forall X \in \omega: s \left({X}\right) = X^+$

where $X^+$ is the successor of $X$.

P3

 * P3: $\forall m, n \in P: s \left({m}\right) = s \left({n}\right) \implies m = n$

Suppose $m, n \in \N$ and that $s \left({m}\right) = s \left({n}\right)$.

Since $n \in s \left({n}\right)$ it follows that $n \in s \left({m}\right)$.

So either $n \in m$ or $n = m$.

Similarly, either $m \in n$ or $m = n$.

So suppose $n \ne m$.

Then both $n \in m$ and $m \in n$.

But as the Natural Numbers are Transitive Sets, it follows that $n \in n$.

But as $n \subseteq n$, this contradicts Natural Number is Not Subset of Element.

P4
Because of the nature of the empty set, $\varnothing$ is not the successor set of any $n \in S$.

So $s: \omega \to \omega$ is not a surjection.

P5
In context, this reads:


 * $\forall S \subseteq \N: \left({\exists x \in S: \neg \left({\exists y \in \N: x = s \left({y}\right)}\right) \land \left({z \in S \implies s \left({z}\right) \in S}\right)}\right) \implies S = \N$

This is precisely the Principle of Finite Induction that defines a general infinite successor set.

The minimal infinite successor set is an instance of such.

Then $\N$ is an instantiation of the minimal infinite successor set.

All of Peano's axioms are fulfilled.

Hence the result.