Non-Successor Element of Peano Axiom Schema is Unique

Theorem

Let $P$ be a set which fulfils the Peano Axiom schema:

• P1: $P \ne \varnothing$

• P2: $\exists s: P \to P$

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

• P4: $\operatorname{Im} \left({s}\right) \ne P$

• P5: $\forall A \subseteq P: \left({\exists x \in A: \neg \left({\exists y \in P: x = s \left({y}\right)}\right) \land \left({z \in A \implies s \left({z}\right) \in A}\right)}\right) \implies A = P$

Then:

$P \setminus s \left({P}\right)$ is a singleton set

where:

• $\setminus$ denotes set difference;
• $s \left({P}\right)$ denotes the image of the mapping $s$.

Proof

Let $T = P \setminus s \left({P}\right)$.

From P4 we know that $T \ne \varnothing$.

Now suppose that $t_1 \in T$ and $t_2 \in T$.

Let us form $A \subseteq P$ such that $t_1 \in A$ and $z \in A \implies s \left({z}\right) \in A$, but such that $t_2 \notin A$.

As $t_1 \in P \setminus s \left({P}\right)$, it follows that $\neg \left({\exists y \in P: t_1 = s \left({y}\right)}\right)$.

Thus $A$ is of the form:

$\left({t_1 \in A: \neg \left({\exists y \in P: t_1 = s \left({y}\right)}\right) \land \left({z \in A \implies s \left({z}\right) \in A}\right)}\right)$

By axiom P5, it follows that $A = P$.

That is, $P \setminus A = \varnothing$.

But $t_2 \notin A$ and so $t_2 \in P \setminus A$.

This contradiction, demonstrates that, given the existence of $t_1 \in P \setminus s \left({P}\right)$, there can be no $t_2 \in P \setminus s \left({P}\right): t_1 \ne t_2$.

Hence $t_1$ is unique, and $P \setminus s \left({P}\right)$ is singleton.

$\blacksquare$

Sources

• Nathan Jacobson: Lectures in Abstract Algebra: I. Basic Concepts (1951): Introduction $\S 4$