Von Neumann Hierarchy Comparison

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


$\map V x \in \map V y$
$\map V x \subset \map V y$



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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.


Induction Step

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


\(\displaystyle x\) \(<\) \(\displaystyle y^+\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle x < y\) \(\lor\) \(\displaystyle x = y\) 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.


Limit Case

Let $y$ be a limit ordinal.


$\map V x \in \map V z$

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:

$\map V z \subseteq \map V y$


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

This proves the limit case.


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