Successor Mapping is Slowly Progressing

Theorem

Let $V$ be a basic universe.

Let $s: V \to V$ denote the successor mapping on $V$:

$\forall x \in V: \map s x := x \cup \set x$

Then $s$ is a slowly progressing mapping.

Proof

From Successor Mapping is Progressing we have that $s$ is a progressing mapping.

Then we have that:

$\set x \notin x$

Thus:

$\card {x \cup \set x} = \card x + 1$

Hence the result.

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 9$ Supplement -- optional