# On models of ZFC, does there exist a bijection between Von Neumann universe and the ordinal class?

Von Neumann universe $$V:=\bigcup_{\alpha \in Ord}V_\alpha$$, in which $$V_{\alpha+1}=\wp(V_\alpha)$$ and $$V_\alpha=\bigcup_{\xi<\alpha}V_\alpha$$ for limit ordinals.

My question: Is $V$ equipotent to the ordinal class Ord? Note that they are both proper class, hence seems do not have cardinality, but it seems bijections(seems proper class too) can still be built...

## 2 Answers

This is the axiom of global choice. Suppose that there were a formula $\varphi(x,y)$ defining a class bijection from $\mathbf{ON}$ to $V$; then $V$ would have a global well-ordering and would satisfy a strong form of the axiom of choice. Thus, such a formula is not provable in $\mathsf{ZF}$, or even in $\mathsf{ZFC}$.

The axiom of global choice does follow from $V=L$ .

• V=L seems so strong, many axioms(like GCH, also CH) follow it. – Popopo Mar 21 '13 at 16:08
• @Popopo: Yes, it’s a very strong axiom. – Brian M. Scott Mar 21 '13 at 16:13

The assertion "$V$ is in bijection with $\sf Ord$" is equivalent to the axiom of global choice, namely there is a choice function on every class of non-empty sets.

This is consistent with ZFC, for example in a model of the axiom $V=L$ (the constructible universe) this axiom is true, but it is consistent that the axiom fails. The construction (that I have in mind) of a counterexample can be a bit technical, as it involves class forcing.

But the point is that this is consistent with ZFC, but it is not provable from ZFC.

• Asaf just out of curiosity, how is the PhD coming along? You must be nearing completion by now. – goblin Mar 28 '14 at 9:17
• Near completion of my research proposal, but there is still a lot if work in that direction. – Asaf Karagila Mar 28 '14 at 11:23
• Good luck. $\!\;$ – goblin Mar 28 '14 at 11:27