Condition for Injective Mapping on Ordinals

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

$(1): \quad$ The domain of $F$ is $\On$, the class of all ordinals
$(2): \quad$ For all ordinals $x$, $\map F x = \map G {F \restriction x}$
$(3): \quad$ For all ordinals $x$, $\map G {F \restriction x} \in \paren {A \setminus \Img x}$ where $\Img x$ is the image of $x$ under $F$.

Let $\Img F$ denote the image of $F$.

Then the following properties hold:

$(1): \quad \Img F \subseteq A$
$(2): \quad F$ 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 the First Principle of Transfinite Recursion.



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Let $x$ be an ordinal.

Then $\map F x = \map G {F \restriction x}$ and $\map G {F \restriction x} \in A$ by hypothesis.

Therefore, $\map F x \in A$.

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).


\(\ds x \in y\) \(\leadsto\) \(\ds \map F x \in \Img y\)
\(\ds \) \(\leadsto\) \(\ds \map F x \in \Img y \land \map F y \notin \Img y\) by hypothesis
\(\ds \) \(\leadsto\) \(\ds \map F x \ne \map F y\)

Thus for distinct ordinals $x$ and $y$, $\map F x \ne \map F y$.

Therefore, $F$ is injective.

$F$ is injective and $F: \On \to A$.

Therefore, if $A$ is a set, then $\On$ 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