Resolving Russell's Paradox in NBG 
Looking back at the form of the paradox we see that we now have a way out. In order to derive $ R \in R$ we would need the extra assumption that $R$ is a set... contin
Topoi, a categorical analysis of logic

To be clear, is the element relation undefined for proper classes?
 A: The way the proper class/set divide works is that sets can be elements of classes, but proper classes can't. A comprehension term $\{x: \varphi(x)\}$ denotes the class of all sets that satisfy $\varphi(x).$
Russel's paradox tells us that $R:=\{x: x\notin x\},$ the class of all sets that do not have themselves as members, cannot be a set. But there's no contradiction there since NBG does not promise us every comprehension term denotes a set. It just means $R$ must be a proper class, and that's fine.
(In ZFC the situation is a bit more nuanced, since we are promised every comprehension term denotes a set, but on the other hand, we are only allowed bounded comprehensions of the form $\{x\in y: \varphi(x)\}$... so what Russel's argument tells us is that there is no set $y$ containing all sets $x$ such that $x\notin x.$ Again, fine.)
As for whether the membership relation is defined when the LHS is a proper class, that's an implementation detail. One can formulate NBG in a two-sorted way, where classes and sets are different kinds of object (and saying a given class is proper just means there is no set with the same elements as it). In this case it is most natural to say $A\in B$ is syntactically illegal when $A$ is a class, so it is undefined in that sense. Or one can formulate it in a single-sorted way where all objects are classes and a set is a special type of class, in which case $A\in B$ is a well-defined, but false statement whenever $A$ is a proper class. Moreover, one can define what a set is this way: if a class is an member of some class, it is a set, otherwise it is a proper class.
