Maximal Injective Mapping from Ordinals to a Set

Theorem
Let $F$ be a mapping satisfying the following properties:


 * The domain of $F$ is $\On$, the class of all ordinals
 * For all ordinals $x$, $\map F x = \map G {F \restriction x}$.
 * For all ordinals $x$, if $A \setminus \Img x \ne \O$, then $\map G {F \restriction x} \in A \setminus \Img x$ where $\Img x$ is the image of the subset $x$ under $F$.
 * $A$ is a set.

Then there exists an ordinal $y$ satisfying the following properties:


 * $\forall x \in y: A \setminus \Img x \ne \O$
 * $\Img y = A$
 * $F \restriction y$ is an injective mapping.

Note that the first third and fourth properties of $F$ are the most important. For any mapping $G$, a mapping $F$ can be constructed satisfying the first two properties using transfinite recursion.

Proof
Set $B$ equal to the class of all ordinals $x$ such that $A \setminus \Img x \ne \O$.

Assume $B = \On$.

Then:

By Condition for Injective Mapping on Ordinals, $A$ is a proper class.

This contradicts the fact that $A$ is a set.

Therefore $B \subsetneq \On$.

Because $B$ is bounded above, $\bigcup B \in \On$.

By Union of Ordinals is Least Upper Bound, the union of ordinals is the least upper bound of $B$.

Setting $\bigcup B = x$:
 * $(1): \quad A \setminus \Img x = \O \land \forall y \in x: A \setminus \Img y \ne \O$

The first condition is satisfied.

In addition:
 * $(2): \quad A \subseteq \Img x$

Take any $y \in \Img x$.

Then:

This means that:
 * $\Img x \subseteq A$

Combining with $(2)$:
 * $\Img x = A$

$F$ is a mapping, so $F \restriction x$ is a mapping.

Take any $y, z \in x$ such that $y$ and $z$ are distinct.

, allow $y \in z$ (justified by Ordinal Membership Trichotomy).

From this, we may conclude that $F$ is injective.

Also see

 * Condition for Injective Mapping on Ordinals
 * Transfinite Recursion
 * Order Isomorphism between Ordinals and Proper Class