Minimally Inductive Set forms Peano Structure

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Let $\omega$ be the minimally inductive set.

Let $\cdot^+: \omega \to \omega$ be the mapping assigning to a set its successor set:

$n^+ := n \cup \set n$

Let $\O \in \omega$ be the empty set.

Then $\struct {\omega, \cdot^+, \O}$ is a Peano structure.


We need to check that all of Peano's axioms hold for $\struct {\omega, \cdot^+, \O}$.

Suppose first that for $m, n \in \omega$, we have $m^+ = n^+$.

Since $n \in n^+$ it follows that $n \in m^+$.

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

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

Now if $n \ne m$, both $m \in n$ and $n \in m$.

By Element of Minimally Inductive Set is Transitive Set, it follows that $n \subseteq m$.

As $m \in n$, this contradicts Finite Ordinal is not Subset of one of its Elements.

Hence it must be that $n = m$, and Peano's Axiom $\text P 3$: $s$ is injective holds.

Next, since $n \in n^+$ for all $n \in \omega$, it follows that $n^+ \ne \O$.

Hence, Peano's Axiom $\text P 4$: $0 \notin \Img s$ holds as well.

Finally, let $S \subseteq \omega$ satisfy:

$\O \in S$
$\forall n \in S: n^+ \in S$

Then by definition, $S$ is an inductive set.

Therefore, by definition of $\omega$ as the minimally inductive set:

$\omega \subseteq S$

Consequently $S = \omega$ by the definition of set equality.

Thus Peano's Axiom $\text P 5$: Principle of Mathematical Induction is seen to hold.

That is, $\struct {\omega, \cdot^+, \O}$ is a Peano structure.