Set has Rank
Theorem
Let $S$ be a set.
Then $S$ has a rank.
Proof 1
The proof shall proceed by Epsilon Induction on $S$.
Suppose that all the elements $a \in S$ have a rank.
That is, $a \in \map V x$ for some $x$.
Let:
- $\ds \map F a = \inf \set {x \in \On : a \in \map V x}$
be the rank of $a$.
Let:
- $\ds y = \sup \set {\map F a : a \in S}$
be the least level of the Von Neumann Hierarchy containing all elements of $S$.
Then, for any $a \in S$:
\(\ds a\) | \(\in\) | \(\ds \map V {\map F a}\) | Definition of $F$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds a\) | \(\in\) | \(\ds \map V y\) | Definition of $y$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds S\) | \(\subseteq\) | \(\ds \map V y\) | Definition of Subset | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds S\) | \(\in\) | \(\ds \powerset {\map V y}\) | Definition of Power Set | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds S\) | \(\in\) | \(\ds \map V {y + 1}\) | Definition of Von Neumann Hierarchy |
Therefore $S \in \map V z$ for some ordinal $z = y + 1$.
Thus by Epsilon Induction every set has a rank.
$\blacksquare$
Proof 2
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Let $G$ be the smallest transitive set containing $S$ as a subset.
By Set Contained in Smallest Transitive Set, $G$ must exist.
By Transitive Set Contained in Von Neumann Hierarchy Level, $G \subseteq V_i$ for some ordinal $i$.
Therefore:
- $G \in V_{i + 1}$
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We have that Ordinals are Well-Ordered
From Proper Well-Ordering determines Smallest Elements, there exists a smallest ordinal $k$ such that $G \in V_{k + 1}$.
Hence, by definition, $G$ has rank $k$.
$\blacksquare$