Set has Rank

Theorem
If $S$ is a set, then $S$ has a rank.

Proof
Let $U$ be the class of all sets.

Define the mapping $F: \mathbb N \to U$ recursively:


 * $F \left({0}\right) = S$
 * $F \left({n + 1}\right) = \bigcup F \left({n}\right)$

By the axiom of union an finite induction, $F \left({n}\right)$ is a set for each $n \in \mathbb N$.

Let $\displaystyle G = \bigcup_{i \mathop = 0}^\infty F \left({i}\right)$.

By the axiom of union, $G$ is a set.

By the axiom of foundation:


 * $\exists x: x \in G \text{ and } x \cap G = \varnothing$

In particular: