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:

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

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.

$\Box$


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

$\blacksquare$


Sources