Successor Mapping is Progressing
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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
- $x \subseteq x \cup \set x$
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