Condition for Injective Mapping on Ordinals

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


 * ($1$) The domain of $F$ is $\operatorname{On}$, the ordinal class
 * ($2$) For all ordinals $x$, $F(x) = G(F \restriction x)$.
 * ($3$) For all ordinals $x$, $G(F \restriction x) \in (A \setminus \operatorname{Im}(x))$ where $\operatorname{Im}(x)$ is the image of $x$ under $F$.

Let $\operatorname{Im}(F)$ denote the image of $F$. Then, the following properties hold:


 * $\operatorname{Im}(F) \subseteq A$
 * $F$ is injective
 * $A$ is a proper class

Note that only the third property of $F$ is the most important. For any function $G$, a function $F$ can be constructed satisfying the first two using transfinite recursion.

Proof
Let $x$ be an ordinal. Then, $F(x) = G(F \restriction x)$ and $G(F \restriction x) \in A$ by hypothesis. Therefore, $F(x) \in A$. This satisfies the first statement.

Take two distinct ordinals $x$ and $y$. Without loss of generality, assume $x \in y$ (we are justified in this by Ordinal Membership is Trichotomy).

For distinct ordinals $x$ and $y$, $F(x) \not = F(y)$. Therefore, $F$ is injective.

$F$ is injective and $F : \operatorname{On} \to A$. Therefore, if $A$ is a set, then $\operatorname{On}$ is a set. But by the Burali-Forti Paradox, this is impossible, so $A$ is not a set. Therefore, $A$ is a proper class.

Also see

 * Maximal Injective Mapping from Ordinals to a Set
 * Transfinite Recursion