Maximal Injective Mapping from Ordinals to a Set

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


 * The domain of $F$ is $\operatorname{On}$, the ordinal class
 * For all ordinals $x$, $F(x) = G(F \restriction x)$.
 * For all ordinals $x$, if $(A \setminus \operatorname{Im}(x) ) \ne \varnothing$, then $G(F \restriction x) \in (A \setminus \operatorname{Im}(x))$ where $\operatorname{Im}(x)$ is the image of $x$ under $F$.
 * $A$ is a set.

Then, there is an ordinal $y$ satisfying the following properties:


 * $\forall x \in y: (A \setminus \operatorname{Im}(x) ) \ne \varnothing$
 * $\operatorname{Im}(y) = A$
 * $F \restriction y$ is an injective mapping.

Proof
Set $B$ equal to the class of all ordinals $x$ such that $( A \setminus \operatorname{Im}(x) ) \ne \varnothing$.

Assume $B = \operatorname{On}$.

By Condition for Injective Mapping on Ordinals, $A$ is a proper class, which contradicts the fact that it is a set.

Therefore $B \subsetneq \operatorname{On}$.

Because $B$ is bounded above, $\bigcup B \in \operatorname{On}$. The union of ordinals is the least upper bound of $B$. Setting $\bigcup B = x$:


 * $\displaystyle ( A \setminus \operatorname{Im}(x) ) = \varnothing \land \forall y \in x: ( A \setminus \operatorname{Im}(y) ) \ne \varnothing$ (1)

The first condition is satisfied. In addition $A \subseteq \operatorname{Im}(x)$ (2).

Take any $y \in \operatorname{Im}(x)$.

This means that $\operatorname{Im}(x) \subseteq A$ and, combining with (2), we have that $\operatorname{Im}(x) = A$.

$F$ is a function, so $(F \restriction x)$ is a function. Take any $y,z \in x$ such that $y$ and $z$ are distinct. Without loss of generality, 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