Set has Rank/Proof 1

Theorem

Let $S$ be a set.

Then $S$ has a rank.

Proof

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.

$\blacksquare$

Sources

• 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 9.13$