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:


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

Applying the axiom of union inductively, $F \left({n}\right)$ 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 unions, $G$ is a set.

Lemma
$G$ is transitive. That is, if $a \in b$ and $b \in G$, then $a \in G$.

Proof
Suppose $a \in b$ and $b \in G$.

Then by the definition of $G$, there exists an $n\in \mathbb N$, $b \in F(n)$.

By the definition of $F$, $F(n+1) = \bigcup F(n)$. Then by the definition of union, $a \in F(n+1)$.

Thus by the definition of $G$, $a \in G$.

Suppose for the sake of contradiction that for each ordinal $i$ the set $G\setminus V_i$ is non-empty.

Let $i$ be any ordinal.

Then by the axiom of foundation:

$\exists x: x \in G\setminus V_i \text{ and } x \cap (G \setminus V_i)= \varnothing$

Then since $G$ is transitive, $x \subset G$, so $x \subset V_i$, so $x \in V_{i+1}$, so


 * $x \notin G \setminus V_{i+1}$.

Thus $G \setminus V_{i+1} \subsetneq G \setminus V_i$ for each ordinal $i$.

Let $H\colon\operatorname{On}\to P(G)$ be defined by
 * $H(i) = G \setminus V_i$ for each ordinal $i$.

Suppose $p$ and $q$ are ordinals, and $p < q$.

Then $p+1 \le q$, so $H(q) \subseteq H(p+1)$.

By the above, $H(p+1) \subsetneq H(p)$,

so $H(q) \subsetneq H(p)$.

Thus for any two ordinals $i$ and $j$, $H(i) \ne H(j)$,

so $H$ is injective.

But that contradicts the fact that $P(G)$ is a set.