# Is there a “natural” bijection between models of $ZFC$ and $ZF\neg C$?

I'm quite confident that there is the same number of models of $ZFC$ and of $ZF\neg C$, which means that there would exist a bijection between the former and the latter, by definition. However, I was wondering - can we somehow employ a forcing techniques (or some other tools) to find an explicit bijection between them? Far more relaxed version of this question is: is there always a definable bijection between these?

Thanks in advance.

## 1 Answer

Well, the simple answer is that if there is a single model, there is a proper class of models, of each cardinality possible.

But let's instead restrict to transitive models. Then the answer is considerably different. It is consistent that all the transitive models of $\sf ZFC$ have the same height, say $\alpha$. Of course, in this case $\alpha<\omega_1$ and we have that every transitive model is countable.

This means that there are at most $2^{\aleph_0}$ models, and in fact there are at least $2^{\aleph_0}$ models as well, since there is a perfect set of reals which are Cohen generic for the model.

But the work of Harvey Friedman showed that there are models of $\sf ZF$ of height $\alpha$, of cardinality $\beth_\alpha$. This means that there are plenty of intermediate models in between, and that there are many many more transitive models of $\sf ZF$ than transitive models of $\sf ZFC$.

Harvey Friedman, Large models of countable height, Trans. Amer. Math. Soc. 201 (1975), 227--239.