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
Let $x < y \implies V \left({x}\right) \in V \left({y}\right)$.

Then:

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

This proves the induction step.

Limit Case
Let $y$ be a limit ordinal.

Let:
 * $V \left({x}\right) \in V \left({z}\right)$

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 by Set is Subset of Union: Family of Sets:
 * $V \left({z}\right) \subseteq V \left({y}\right)$

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.