Condition for Injective Mapping on Ordinals

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


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

Let $\operatorname{Im} \left({F}\right)$ denote the image of $F$.

Then the following properties hold:


 * $(1): \quad \operatorname{Im} \left({F}\right) \subseteq A$
 * $(2): \quad F$ is injective
 * $(3): \quad A$ 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 $F \left({x}\right) = G \left({F \restriction x}\right)$ and $G \left({F \restriction x}\right) \in A$ by hypothesis.

Therefore, $F \left({x}\right) \in A$.

This satisfies the first statement.

Take two distinct ordinals $x$ and $y$.

WLOG assume $x \in y$ (we are justified in this by Ordinal Membership is Trichotomy).

Then:

Thus for distinct ordinals $x$ and $y$, $F \left({x}\right) \ne F \left({y}\right)$.

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