# 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

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

Then:

\(\displaystyle x\) | \(<\) | \(\displaystyle y^+\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x < y\) | \(\lor\) | \(\displaystyle x = y\) | Definition of Successor Set | |||||||||

\(\displaystyle x = y\) | \(\implies\) | \(\displaystyle \map V x \in \powerset {\map V x}\) | Definition of Power Set | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map V x\) | \(\in\) | \(\displaystyle \map V {y^+}\) | Definition of Von Neumann Hierarchy | |||||||||

\(\displaystyle x < y\) | \(\implies\) | \(\displaystyle \map V x \in \map V y\) | Inductive Hypothesis | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map V x\) | \(\in\) | \(\displaystyle \map V {y^+}\) | Von Neumann Hierarchy is Supertransitive |

In either case:

- $\map V x \in \map V {y^+}$

This proves the induction step.

$\Box$

### 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.

$\Box$

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

$\blacksquare$

## Sources

- 1971: Gaisi Takeuti and Wilson M. Zaring:
*Introduction to Axiomatic Set Theory*: $\S 9.10$