Von Neumann Hierarchy Comparison

Theorem
Let $x$ and $y$ be ordinals such that $x < y$.

Then:


 * $V \left({x}\right) \in V \left({y}\right)$


 * $V \left({x}\right) \subset V\left({y}\right)$

Proof
The proof shall proceed by Transfinite Induction on $y$.

Basis for the Induction
If $y = 0$, then $x \not < y$.

This proves the basis for the induction.

Induction Step
Suppose that $x < y \implies V\left({x}\right) \in V\left({y}\right)$.

Then:

In either case, $V\left({x}\right) \in V\left({y^+}\right)$.

This proves the induction step.

Limit Case
Suppose that $y$ is a limit ordinal and that:


 * $V\left({x}\right) \in V\left({z}\right)$ for all $z \in y$ such that $x < z$.

Since $x < y$, it follows that $x < z$ for some $z \in y$ by Union of Limit Ordinal.

By the inductive hypothesis, $V\left({x}\right) \in V\left({z}\right)$.

But $V\left({z}\right) \subseteq V\left({y}\right)$ by Subset of Union/Family of Sets.

Therefore, $V\left({x}\right) \in V\left({y}\right)$.

This proves the limit case.


 * $V\left({x}\right) \subset V\left({y}\right)$ follows by Von Neumann Hierarchy is Supertransitive.