Von Neumann Hierarchy Comparison

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

Then:


 * $\map V x \in \map V y$


 * $\map V x \subset \map V y$

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 $x < y \implies \map V x \in \map V y$.

Then:

In either case:
 * $\map V x \in \map V {y^+}$

This proves the induction step.

Limit Case
Let $y$ be a limit ordinal.

Let:
 * $\map V x \in \map V z$

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:
 * $\map V x \in \map V z$

But by Set is Subset of Union: Family of Sets:
 * $\map V z \subseteq \map V y$

Therefore:
 * $\map V x \in \map V y$

This proves the limit case.


 * $\map V x \subset \map V y$ follows by Von Neumann Hierarchy is Supertransitive.