Minimal Infinite Successor Set Fulfils Peano Axioms

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.

$\Box$

P2

Let $s: \omega \to \omega$ be defined as:

$\forall X \in \omega: s \left({X}\right) = X^+$

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

$\Box$

P3

We need to show that:

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

Suppose $m, n \in \omega$ 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.

$\Box$

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.

$\Box$

P5

In context, this reads:

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

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

$\Box$

All of Peano's axioms are fulfilled.

Hence the result.

$\blacksquare$

Sources

• Paul R. Halmos: Naive Set Theory (1960)... (previous)... (next): $\S 12$: The Peano Axioms
• Steven G. Krantz: Discrete Mathematics Demystified (2009): $\S 5.2$