Assume ZFC (and AC in particular) as the background theory.

If $(M,\in^M)$ is a model of ZFC (not necessarily transitive or standard), must there exist a bijection between $M$ and $$\{x \in M \mid (M,\in^M) \models x \mbox{ is an ordinal number}\}?$$

I am also interested in the cases where $M$ is assumed to be a model of ZFC2.

Remark. I think this is different from what was asked here. Please comment if you believe otherwise; I am happy to discuss.

  • $\begingroup$ You can easily prove that from $\sf ZFC+\lnot\operatorname{Con}(ZFC)$! ;-) $\endgroup$ – Asaf Karagila Apr 3 '14 at 18:01

The answer is yes. However, the answer is no if we require the model itself to know the bijection. More specifically, the existence of a class bijection between $V$ and its ordinals is equivalent to the axiom of global choice, and it is consistent that $\mathsf{ZFC}$ holds but global choice fails.

Now, given any (set) model $M$ of $\mathsf{ZFC}$, for each $\alpha$ ordinal of $M$ there is (in $M$) a bijection $f$ between $V_\alpha$ and some ordinal $\beta$ of $M$. Any such $f$ gives us a true bijection between $\hat\beta=\{a\in M\mid M\models a<\beta\}$ and $\hat V_\alpha=\{b\in M\mid M\models b\in V_\alpha\}$, simply by setting $$\hat f=\{(a,b)\in M\times M \mid M \models a<\beta,b\in V_\alpha,\& \,b=f(a)\}.$$ This proves that $$|M|=\left|\bigl(\bigcup_{a\in\mathsf{ORD}^M}V_a\bigr)^M\right|\le|\mathsf{ORD}^M|\sup_{a\in\mathsf{ORD}^M}|\hat V_a|\le |\mathsf{ORD}^M|\sup_{b\in\mathsf{ORD}^M}|\hat b|=|\mathsf{ORD}^M|.$$

The answer is no if we replace $\mathsf{ZFC}$ with $\mathsf{ZF}$. Indeed, we can have transitive models of $\mathsf{ZF}$ that are uncountable and have countably many ordinals, see for instance here.


Modulo possible subtleties about non transitive models, I understand that is the same question because being an ordinal is absolute for transitive models. About (2), infinite cardinals (so all the cardinals) and ordinals are in bijective correspondence with ordinals via the $\aleph$ function.


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