$\{x\in A: x\not\in x\}$ is not a member of $A$

Question

For any set $A$, the set $r(A):=\{x\in A: x\not\in x\}$ is not a member of $A$. It follows that the collection of all sets in not a set.

The proof goes like this. By the separation axiom, $r(A)$ is a set. If $r(A)\in A$ then $r(A)\in r(A) \iff r(A)\not\in r(A)$, which is absurd.

Why do we need to know that $r(A)$ is a set? And how does it follow that the collection of all sets is not a set?

Answer 1

The reason you want to know it is a set is, at least in part, so that you can prove that there is no set of all sets. If all you knew was “Given a set $A$, there is an object $X_A$ (which depends on $A$, which may or may not be a set, and that is not an element of $A$”, you would not be able to do the following argument.

(Also, in some theories, such as ZF or ZFC, the theory can only talk about sets; you cannot speak about other types of objects; this is not the case in other theories, such as GBN, where the primitive objects are classes, and sets are special types of classes; or set theories with ur-elements, where there are some objects which can be elements of sets, but are not themselves sets).

Now, why does this argument show that there is no “set of all sets”? Look carefully at what you’ve proven: you’ve proven that for every set $A$, there exists a set $X_A$, which depends on $A$, such that $X_A\notin A$. Formally, $$\forall A\bigl(A\text{ is a set}\rightarrow \exists X_A(X_A\text{ is a set, and }X_A\notin A)\bigr)\tag{$\star$}$$

What would be a “set of all sets”? If would be a set $C$ such that for every $X$, if $X$ is a set then $X\in C$. That is: $$\exists C\Bigl(C\text{ is a set}\text{ and }\bigl(\forall X(X\text{ is a set}\rightarrow X\in C)\bigr)\Bigr).$$ This is precisely the negation of $(\star)$. Hence, since $(\star)$ has been proven, this statement is false. Thus, there is no “set of all sets”.

Answer 2

Well, in order to talk about an object set-theoretically, we have to know it is a set, don't we?

This Claim is essentially a set-up for Russell's Paradox. If there were a set of all sets, say $U,$ then by Separation, we would have that $r(U)$ is a set, but by the Claim, this would lead to a contradiction. Thus, there is no set of all sets.