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 function $F \colon \mathbb N \to U$ recursively:


 * $F(0) = S$
 * $F(n+1) = \bigcup F(n)$

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

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

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,