Equivalence of Formulations of Peano's Axioms
Theorem
Let $P$ be a set.
Let $s: P \to P$ be a mapping.
Let $0 \in P$ be a distinguished element.
The following definitions of the concept of Peano's Axioms are equivalent:
Formulation 1
\((\text P 3)\) | $:$ | \(\ds \forall m, n \in P:\) | \(\ds \map s m = \map s n \implies m = n \) | $s$ is injective | |||||
\((\text P 4)\) | $:$ | \(\ds \forall n \in P:\) | \(\ds \map s n \ne 0 \) | $0$ is not in the image of $s$ | |||||
\((\text P 5)\) | $:$ | \(\ds \forall A \subseteq P:\) | \(\ds \paren {0 \in A \land \paren {\forall z \in A: \map s z \in A} } \implies A = P \) | Principle of Mathematical Induction: | |||||
Any subset $A$ of $P$, containing $0$ and | |||||||||
closed under $s$, is equal to $P$ |
Formulation 2
\((\text P 3)\) | $:$ | \(\ds \forall m, n \in P:\) | \(\ds \map s m = \map s n \implies m = n \) | $s$ is injective | |||||
\((\text P 4)\) | $:$ | \(\ds \Img s \ne P \) | $s$ is not surjective | ||||||
\((\text P 5)\) | $:$ | \(\ds \forall A \subseteq P:\) | \(\ds \paren {\paren {\exists x \in A: \neg \exists y \in P: x = \map s y} \land \paren {\forall z \in A: \map s z \in A} } \) | Principle of Mathematical Induction: | |||||
\(\ds \implies A = P \) | Any subset $A$ of $P$, containing an element not | ||||||||
in the image of $s$ and closed under $s$, | |||||||||
is equal to $P$ |
Proof
Formulation 1 implies Formulation 2
For $(\text P 3)$, there is nothing to prove.
Next, axiom $(\text P 4)$ of Formulation 2.
Recall the definition of the image of a mapping as applicable to $s$:
- $\Img s = \set {n \in P: \exists m \in P: \map s m = n}$
From this definition, it follows that $0 \notin \Img s$, and hence:
- $\Img s \ne P$
Lastly, axiom $(\text P 5)$ of Formulation 2.
By Non-Successor Element of Peano Structure is Unique, we know that there is exactly one element of $P$ that is not in the image of $s$.
By axiom $(\text P 3)$ of Formulation 1, we know this element is $0$.
Therefore, the premises:
- $0 \in A$
and:
- $\exists x \in A: \neg \exists y \in P: x = \map s y$
are logically equivalent, and hence so are both forms of axiom $(\text P 5)$.
$\Box$
Formulation 2 implies Formulation 1
By Non-Successor Element of Peano Structure is Unique, we know that there is exactly one element of $P$ that is not in the image of $s$.
We will identify this element with the distinguished element $0$.
This implies that axioms $(\text P 3)$ and $(\text P 4)$ of Formulation 1 are satisfied.
Lastly, axiom $(\text P 5)$ of Formulation 1.
Since $0$ satisfies the premise:
- $\neg \exists y \in P: 0 = \map s y$
we conclude that axiom $(\text P 5)$ of Formulation 1 follows a fortiori from axiom $(\text P 5)$ of Formulation 2.
$\blacksquare$
Sources
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- 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts: Introduction $\S 4$ (implicit)