Natural Number is not Subset of its Union
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Theorem
Let $n \in \N$ be a natural number as defined by the von Neumann construction.
Then, except in the degenerate case where $n = 0$, it is not the case that:
- $n \subseteq \bigcup n$
Proof
First we note that from Union of Empty Set we have:
- $\bigcup \O = \O$
leading to:
- $\O \subseteq \bigcup \O$
thus disposing of the degenerate case.
Let $n \in \N$ such that $n \ne \O$.
By definition of the von Neumann construction:
- $n = \set {0, 1, 2, \ldots, n - 1}$
Thus, by definition, $m \in n$ for $m = 0, 1, 2, \ldots, n - 1$.
From Natural Numbers cannot be Elements of Each Other, it cannot therefore be the case that $n \in m$ for $m \in n$.
By definition of union of class:
- $\bigcup n = \set {x: \exists X \in n: x \in X}$
But it has been established that there is no $X \in n$ such that $n \in X$.
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: $\S 3$ Derivation of the Peano postulates and other results: Exercise $3.2 \ \text {(a)}$