Natural Number is not Subset of its Union

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.