Set of Natural Numbers Equals its Union

Theorem
Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction on a Zermelo universe $V$.

Then:
 * $\bigcup \omega = \omega$

Proof
We have that:
 * $\omega \in V$

where $V$ is a Zermelo universe.

By the we have that $\omega$ is a set.

By the, $\omega$ is transitive.

Hence:
 * $\bigcup \omega \subseteq \omega$

Let $n \in \omega$.

Then by definition of the von Neumann construction:
 * $n + 1 = n \cup \set n$

That is:
 * $\exists X \in \omega: n \in X$

where in this case $X = n + 1$.

Thus we have that:
 * $n \in \set {x: \exists X \in \omega: x \in X}$

and so by definition of union of class:
 * $n \in \bigcup \omega$

Thus we have that:
 * $\omega \subseteq \bigcup \omega$

which, together with:
 * $\bigcup \omega \subseteq \omega$

gives us, by set equality:
 * $\bigcup \omega = \omega$