Von Neumann Hierarchy is Supertransitive
Theorem
Let $V$ denote the Von Neumann Hierarchy.
Let $x$ be an ordinal.
Then $\map V x$ is supertransitive.
Proof
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The proof shall proceed by Transfinite Induction on $x$.
Basis for the Induction
We have that:
- $V_0 = \O$
and $\O$ is supertransitive by the very fact that it has no elements.
This proves the basis for the induction.
$\Box$
Induction Step
Let $\map V x$ be supertransitive.
First, to prove transitivity.
Let $\map V x$ be transitive.
Then:
| \(\ds y\) | \(\in\) | \(\ds \map V {x^+}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(\subseteq\) | \(\ds \map V x\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall z: \, \) | \(\ds z \in y\) | \(\implies\) | \(\ds z \subseteq \map V x\) | Transitivity of $\map V x$ | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall z: \, \) | \(\ds z \in y\) | \(\implies\) | \(\ds z \in \map V {x^+}\) | Definition of Von Neumann Hierarchy | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(\subseteq\) | \(\ds \map V {x^+}\) | Definition of Subset |
Next, to prove supertransitivity:
| \(\ds y\) | \(\in\) | \(\ds \powerset {\map V x}\) | by hypothesis | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(\subseteq\) | \(\ds \map V x\) | Definition of Power Set | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \powerset y\) | \(\subseteq\) | \(\ds \powerset {\map V x}\) | Power Set Preserved Under Subset | ||||||||||
| \(\ds \) | \(=\) | \(\ds \map V {x^+}\) | Definition of Von Neumann Hierarchy |
This proves the induction step.
$\Box$
Limit Case
Let $x$ be a limit ordinal.
Furthermore, let $\map V y$ be transitive for all $y \in x$.
Then:
| \(\ds z\) | \(\in\) | \(\ds \map V x\) | by hypothesis | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists y \in x: \, \) | \(\ds z\) | \(\in\) | \(\ds \map V y\) | Definition of Von Neumann Hierarchy | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists y \in x: \, \) | \(\ds z\) | \(\subseteq\) | \(\ds \map V y\) | Transitivity of $V_y$ | |||||||||
| \(\ds \) | \(\subseteq\) | \(\ds \map V x\) | Set is Subset of Union of Family |
This proves transitivity.
Now, to prove supertransitivity:
| \(\ds z\) | \(\in\) | \(\ds \map V x\) | by hypothesis | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists y \in x: \, \) | \(\ds z\) | \(\in\) | \(\ds \map V y\) | Definition of Von Neumann Hierarchy | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists y \in x: \, \) | \(\ds z\) | \(\subseteq\) | \(\ds \map V y\) | Transitivity of $V_y$ | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists y \in x: \, \) | \(\ds \powerset z\) | \(\subseteq\) | \(\ds \map V {y + 1}\) | Power Set Preserved Under Subset | |||||||||
| \(\ds \) | \(\subseteq\) | \(\ds \map V x\) | Set is Subset of Union of Family |
This proves the limit case.
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 9.10$