Natural Number is not Subset of its Union

From ProofWiki
Jump to navigation Jump to search

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