Why the Russell class can't be a member of itself?

Question

From my understanding, the Russell class can't be a member of other classes because by definition it contains all the members of other classes and hence any class is contained in the Russell class. But what is confusing me is that why the Russell class can't be a member of itself? What contradiction would that lead to? Thank you!

Answer 1

I think you've completely misunderstood its definition. The Russell class $R$ satisfies this: for any $x$, $x\in R$ iff $x$ is a set and $x\notin x$. To avoid a paradox, $R$ cannot be a set, since then we'd have $R\in R\iff R\notin R$. Since $R$ isn't a set, $R\notin R$.

Answer 2

If you're talking NBG, where the domain of discourse is classes and any formula specifies a class, then $R=\{x:x\notin x\}$ is a perfectly valid class, but one has to be careful: in the class builder operator $x$ only refers to sets. In NBG a class $a$ is a set if and only if there exists a class $b$ such that $a\in b$. If we denote by $M$ the unary predicate “being a set”, the class $R$ is more precisely defined as $$ R=\{x:M(x)\land x\notin x\} $$ We can have $R\in R$ only if $M(R)$. However, this leads to a contradiction:

If $R\in R$, then $R$ is a set and satisfies $R\notin R$.

Thus we conclude that $R$ is not a set (hence a proper class). In particular $R\notin R$, because otherwise $R$ would be a set.