Successor Mapping is 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 progressing mapping.


Proof

Recall

By Set is Subset of Union:

$x \subseteq x \cup \set x$

That is:

$x \subseteq \map s x$

Thus $s$ is by definition a progressing mapping.

$\blacksquare$


Proof

  • 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications
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