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:

$G \in V_{i + 1}$

This article, or a section of it, needs explaining. In particular: Proper Well-Ordering determines Smallest Elements only applies once it's been shown that initial segments are sets You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

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 $k$.

$\blacksquare$