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 by hypothesis that:

$\omega \in V$

where $V$ is a Zermelo universe.

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

By the Axiom of Transitivity, $\omega$ is transitive.

Hence by Class is Transitive iff Union is Subclass:

$\bigcup \omega \subseteq \omega$

$\Box$

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:

$\bigcup \omega = \omega$

$\blacksquare$

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 $3.3 \ \text {(a)}$