Condition for Injective Mapping on Ordinals

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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

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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:

\(\displaystyle x \in y\) \(\implies\) \(\displaystyle F \left({x}\right) \in \operatorname{Im} \left({y}\right)\)
\(\displaystyle \) \(\implies\) \(\displaystyle F \left({x}\right) \in \operatorname{Im} \left({y}\right) \land F \left({y}\right) \notin \operatorname{Im} \left({y}\right)\) by hypothesis
\(\displaystyle \) \(\implies\) \(\displaystyle F \left({x}\right) \ne F \left({y}\right)\)

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.

$\blacksquare$


Also see


Sources