Successor Mapping on Natural Numbers has no Fixed Element

Theorem

Let $\N$ denote the set of natural numbers.

Then:

$\forall n \in \N: n + 1 \ne n$

Proof

Consider the set of natural numbers as defined by the von Neumann construction.

From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\N$ is a minimally inductive class under the successor mapping.

Let $s: \N \to \N$ denote this successor mapping:

$\forall x \in \N: \map s x := x + 1$

Aiming for a contradiction, suppose $\exists n \in \N: n = n + 1$

From Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element, $n$ is the greatest element of $\N$.

From Minimally Inductive Class under Progressing Mapping with Fixed Element is Finite it follows that $\N$ is a finite set.

This contradicts the fact that the natural numbers are by definition countably infinite.

$\blacksquare$

Sources

• 1982: Alan G. Hamilton: Numbers, Sets and Axioms ... (previous) ... (next): $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Examples $1.1 \ \text {(d)}$