Successor Mapping on Natural Numbers is Progressing

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Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction.

Let $s: \omega \to \omega$ denote the successor mapping on $\omega$.

Then $s$ is a progressing mapping.

Proof 1

By definition of the von Neumann construction:

$n^+ = n \cup \set n$

from which it follows that:

$n \subseteq n^+$

Hence the result by definition of progressing mapping.


Proof 2

By definition, the successor mapping on $\omega$ is indeed an example of a successor mapping.

The result follows from Successor Mapping is Progressing.