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According to this old answer, global choice implies its own version of the well-ordering theorem. Namely that

Every proper class can be put in bijection with the class of ordinals.

There are two things about this that are shocking to me. First, the statement itself, that there is only one cardinality for classes strikes me as a surprising fact, although I have no intuition to believe otherwise. Second and more shocking (to me at least) is that this class well-ordering theorem seems to make no reference to ordering whatsoever. So my question is two-fold: how do we prove this claim, and why is it parallel to the well-ordering theorem?

tox123
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  • The last paragraph of the cited answer says that global choice is effectively stronger than the axiom of choice (for sets). – Berci Jul 26 '21 at 15:10

1 Answers1

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If $X$ is a class, then $X=\bigcup_{\alpha\in\rm Ord}X\cap V_\alpha$. Use global choice to choose a well ordering for each $X\cap V_\alpha$, and combine them to create a bijection with $\rm Ord$. Of course, it is enough to consider $X=V$, so $X\cap V_\alpha=V_\alpha$.

Now, if there is a well-ordering of $V$, then for every non-empty set $X$, simply choose the least element in that set relative to the well-ordering.

Exactly like how you'd prove the axiom of choice from "every set can be well-ordered".

Asaf Karagila
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