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?

• If $A$ is a set, the cardinality of $\mathcal P(\bigcup A)$ is strictly larger than the cardinality of any set in $A$. – Andrés E. Caicedo Nov 17 '18 at 16:17

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.
• Hartogs number! I notice that the axiom of replacement is crucial here because $V_{\omega+\omega}$ has very few cardinals. – Alberto Takase Nov 17 '18 at 10:07
• @AlbertoTakase Replacement certainly plays a role here but notice that the powerset axiom is also important: Let $\kappa$ be in infinite cardinal. Then (assuming $\mathrm{ZFC}$ in our background universe) $(H_{\kappa+}; \in) \models \mathrm{ZFC}^-$, that is $\mathrm{ZFC}$ without the powerset axiom and $(H_{\kappa^+}; \in) \models \forall x \exists f \colon \kappa \to x \text{ surjective}$. In particular $(H_{\kappa^+}; \in) \models \kappa \text{ is the largest cardinal}$. – Stefan Mesken Nov 17 '18 at 10:16