In Daniel J. Velleman's How To Prove It there is an exercise which asks us to consider the existence of the set of all sets and its implications:
Suppose U were the collection of all sets. Note that in particular U is
a set, so we would have U ∈ U. This is not yet a contradiction; although
most sets are not elements of themselves, perhaps some sets are elements
of themselves. But it suggests that the sets in the universe U could be
split into two categories: the unusual sets that, like U itself, are elements
of themselves, and the more typical sets that are not. Let R be the set of
sets in the second category. In other words, R = {A ∈ U | A ∈/ A}. This
means that for any set A in the universe U, A will be an element of R iff
A ∈/ A. In other words, we have ∀A ∈ U(A ∈ R ↔ A ∈/ A)(a) Show that applying this last fact to the set R itself (in other words,
plugging in R for A) leads to a contradiction.
Since $R \in U$, we have that $R \in R ↔ R \notin R$, a contradiction, Russel's famous paradox. We are then asked to reflect on its implications. I used to think that Russel's paradox refuted the existence of a set which contains itself, but it seems to me now that it just refutes the idea that any definable collection is a set, or that one can subdivide any set into any two subsets. In this question I found out that sets that contain themselves can exist even in the context of ZFC.
I also can't see how this paradox proves that there can't be a set of all sets. I've seen proofs of it based on Cantor's Theorem, but once again the paradox seems to me not to prove the impossibility of the existence of the set of all sets, but the impossibility of the construction of a certain subset of it.
Concluding, is Russel's paradox just an argument against the idea than any definable collection is a set, or is there more profound implications of it I am not aware of?
Best Answer
It does prove that there is no set of sets in ZFC, where we have as an axiom (separation or comprehension) that if $y$ is a set and $\phi(x)$ is some predicate with free variable $x$ (ignoring parameters for now) then $\{x \in y\mid \phi(x)\}$ is also a set. So if $U$ (the set of all sets) existed, so would $\{x \in U\mid x \notin x\}$ and this leads to a contradiction. So $U$ does not exist in a system obeying this axiom (like ZFC, which we do assume in almost all of maths).