Natural Number is Union of its Successor
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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 $3.2 \ \text {(d)}$