Natural Number is Union of its Successor

Theorem

Let $n \in \N$ be a natural number as defined by the von Neumann construction.

Then:

$\map \bigcup {n^+} = n$


Proof

\(\ds \) \(\) \(\ds \map \bigcup {n^+}\)
\(\ds \) \(=\) \(\ds \map \bigcup {\set n \cup n}\) Definition of Von Neumann Construction of Natural Numbers
\(\ds \) \(=\) \(\ds \bigcup \set n \cup \bigcup n\) Set Union is Self-Distributive
\(\ds \) \(=\) \(\ds n \cup \bigcup n\) Union of Singleton

From Natural Number is Superset of its Union we have:

$\bigcup n \subseteq n$

Then from Union with Superset is Superset‎:

$\bigcup n \subseteq n \iff \paren {n \cup \bigcup n} = n$

and the result follows.

$\blacksquare$


Sources

  • 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Exercise $3.2 \ \text {(d)}$
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