TITLE: Do there Exist Proper Classes that aren't "Too Big"
QUESTION [27 upvotes]: Some proper classes are "too big" to be a set in the sense that they have a subclass that can be put in bijection with $\alpha$ for every cardinal $\alpha$. It is implied in this post that every proper class is "too big" to be a set in this sense, however I have been unable to prove it. It's true if every two proper classes are in bijection, but it's consistent with ZFC for there to be a pair of non-bijective classes.
So, is the following true in ZFC:

For all proper classes, $C$, and $\alpha\in\mathbf{Card}$, $\exists S\subset C$ such that $|S|=\alpha$?

If not, is there something reasonable similar that preserves the intuition about classes that are "too big to be sets"?

REPLY [14 votes]: Here is an alternative proof to that of Eric.
Since $C$ is a proper class, there is no $\alpha$ such that $C\subseteq V_\alpha$. Consider the class $\{C\cap V_\alpha\mid\alpha\in\mathrm{Ord}\}$, if there is an upper bound on cardinality to this class, some $\kappa$, then there cannot be more than $\kappa^+$ different sets in that class, which would make $C$ a set.
Therefore there are arbitrarily large $C\cap V_\alpha$'s, and therefore there is one larger than your given cardinal.

Interestingly, without choice, it is consistent that there is a proper class which does not have any countably infinite subsets. Although it is true in that every proper class can be mapped onto the class of ordinals.