It is well-known that in $\mathsf{ZF}$ there cannot exist a set containing all of the ordinals (transitive sets of transitive sets).
Is the same true for cardinals? (ordinals satisfying $(\forall \alpha\in\kappa)[\alpha\lnsim\kappa]$)
Of course the answer is "yes" assuming the axiom of choice because we can define successor cardinals, but what about a setting without choice?
Best Answer
Here's a simple proof that the class of cardinals $(\dagger)$ (in $\mathrm{ZF}$) is a proper class:
We already now that the class of ordinals $\mathrm{Ord}$ is a proper class and as such unbounded in the von Neumann hierarchy. Now just note that for every $\alpha$ there is a cardinal $\alpha^+ > \alpha$ (this doesn't use any form of choice -- see Hartogs number). Hence the class of cardinals is unbounded in the von Neumann hierarchy (as it is $\in$-cofinal in $\mathrm{Ord}$) and as such it must be a proper class.
$(\dagger)$ This shows a bit more: The class of wellordered cardinals (or alephs) is a proper class.