Successor Mapping of Peano Structure has no Fixed Point
Theorem
Let $\PP = \struct {P, s, 0}$ be a Peano structure.
Then:
- $\forall n \in P: \map s n \ne n$
That is, the successor mapping has no fixed points.
Proof
Let $T$ be the set:
- $T = \set {n \in P: \map s n \ne n}$
We will use Peano's Axiom $\text P 5$: Principle of Mathematical Induction to prove that $T = P$.
Part $1$: $0 \in T$
From Peano's Axiom $\text P 4$: $0 \notin \Img s$, a fortiori:
- $\map s 0 \ne 0$
Hence $0 \in T$.
$\Box$
Part $2$: $n \in T \implies \map s n \in T$
Let $n \in T$.
Let $m = \map s n$.
Then $n \ne m$, as $\map s n \ne n$.
Aiming for a contradiction, suppose $m \notin T$.
That is:
- $\map s m = m$
Then that would mean:
- $\map s n = \map s m = m$
But from Peano's Axiom $\text P 3$: $s$ is injective, $s$ is injective.
From this contradiction, it follows that $m \in T$.
Hence we have that:
- $n \in T \implies \map s n \in T$
$\Box$
Conclusion
By Peano's Axiom $\text P 5$: Principle of Mathematical Induction, we conclude that $T = P$.
From the definition of $T$, the result follows.
$\blacksquare$
Sources
- 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Introduction $\S 4$: The natural numbers: Exercise $1$
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 12$: The Peano Axioms