In Russell's famous paradox ("Does the set of all sets which do not contain themselves contain itself?") he obviously makes the assumption that a set can contain itself. I do not understand how this should be possible and therefore my answer to Russell's question would simply be "No, because a set cannot contain itself in the first place."
How can a set be exactly the same set as the one that contains it? To me it seems unavoidable that the containing set will always have one more additional level of depth compared to all the sets which it contains, just like those russian matryoshka-dolls where every doll contains at least one more doll than all the dolls inside it.
Of course one can define something like "the set of all sets with at least one element" which of course would include a lot of sets and therefore by definition should also include itself, but does it necessarily need to include itself just because its definition demands so? To me this only seems to prove that it's possible to define something that cannot exist beyond its pure definition.
Best Answer
Yes, this is an issue.
Naively, this issue cannot be dealt with, and we'll get to that in a moment. But in 1917 mathematicians already noticed that "normal sets" do not contain themselves, and in fact have an even stronger property. Namely, there are no infinite decreasing chains in $\in$, so not only that $a\notin a$ it is also true that $a\notin b$ whenever $b\in a$, and that $a\notin c$ whenever for some $b\in a$ we have $c\in b$; and more generally there is no sequence $x_n$ such that $x_{n+1}\in x_n$ for all $n$.
This is exactly what the axiom of regularity came to formalize. It says that the membership relation is well-founded, which assuming the axiom of choice, is equivalent to saying that there are no decreasing chains. In particular $A\notin A$, for any set $A$.
But we know, nowadays, that it is consistent relative to the other axioms of modern set theory (read: $\sf ZFC$) that there are sets which include themselves, namely $x\in x$. We can even go as far as having $x=\{x\}$. You can even arrange for infinitely many sets of the form $x=\{x\}$.
This shows that naively we cannot prove nor disprove that sets which contain themselves exist. Because naive set theory has no formal axioms, and is usually taken as a subset of axioms which include very little from $\sf ZFC$ in terms of axioms, and certainly it does not include the axiom of regularity.
But it also tells us that we cannot point out at a set which includes itself, if we do not assume the axiom of regularity. Since these sets cannot be defined in a nontrivial way. They may exist and may not exist, depending on the universe of sets we are in. But we do know that in order to do naive set theory and even more, we can safely assume that this situation never occurs.