# Von Neumann Hierarchy Comparison

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## 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

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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.

$\Box$

### Induction Step

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

Then:

 $\displaystyle x$ $<$ $\displaystyle y^+$ $\displaystyle \implies \ \$ $\displaystyle x < y$ $\lor$ $\displaystyle x = y$ Definition of Successor Set $\displaystyle x = y$ $\implies$ $\displaystyle V \left({x}\right) \in \mathcal P \left({ V \left({ x }\right) }\right)$ Definition of Power Set $\displaystyle \implies \ \$ $\displaystyle V \left({x}\right)$ $\in$ $\displaystyle V \left({y^+}\right)$ Definition of Von Neumann Hierarchy $\displaystyle x < y$ $\implies$ $\displaystyle V \left({x}\right) \in V \left({y}\right)$ Inductive Hypothesis $\displaystyle \implies \ \$ $\displaystyle V \left({x}\right)$ $\in$ $\displaystyle V \left({y^+}\right)$ Von Neumann Hierarchy is Supertransitive

In either case:

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

This proves the induction step.

$\Box$

### 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:

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

Therefore:

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

This proves the limit case.

$\Box$

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

$\blacksquare$