Transitive Set Contained in Von Neumann Hierarchy Level

Theorem
Let $G$ be a transitive set.

Then for some ordinal $i$, $G \subseteq V_i$.

Proof
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:

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

Since $G$ is transitive, $x \subseteq G$.

Since $x \subseteq G$ and $x \cap \left({G \setminus V_i}\right) = \varnothing$:
 * $x \subseteq V_i$

By the definition of $V$, $x \in V_{i+1}$, so:
 * $x \notin G \setminus V_{i+1}$

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

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

Since the class of ordinals is well-ordered, applying Strictly Increasing Mapping on Well-Ordered Class proves that for any ordinals $p$ and $q$, $p < q$ implies $H \left({q}\right) \subsetneq H \left({p}\right)$.

Since the class of ordinals is totally ordered, for any two distinct ordinals $i$ and $j$:
 * $H \left({i}\right) \ne H \left({j}\right)$

so $H$ is injective.

But that contradicts the fact that $P \left({G}\right)$ is a set.