About Russell's paradox and Russell's anti-sets

Question

It is written on Wikipedia:

According to naive set theory, any definable collection is a set. Let R be the set of all sets that are not members of themselves. If R is not a member of itself, then its definition dictates that it must contain itself, and if it contains itself, then it contradicts its own definition as the set of all sets that are not members of themselves. This contradiction is Russell's paradox. Symbolically:

${{\text{Let }}R=\{x\mid x\not \in x\}{\text{, then }}R\in R\iff R\not \in R}$

I would like to define Russell's anti-set as:

A set $R$ is a Russell's anti-set if and only if it contains some sets that are members of themselves.

If we allow, in some set-theory, an existence of the (or a?) set of all sets and suppose that set of all sets is also a set, then, since set of all sets contains all sets and since set of all sets is a set, then it contains itself, so is a Russell's anti-set.

My question is:

Is set of all sets the only Russell's anti-set?

Also, is this a legal question in at least one reasonable set-theory (except in naive set-theory?)?

Answer 1

In any reasonable sense, no.

Either your theory does not permit any set to contain itself (e.g. $\sf ZF$), in which case the question is trivially answered.

Or there is a set $x$ that contains itself, but then $\{x\}$ contains a set which contains itself. So $\{x\}$ is an anti-Russell set. As is $\{x,\varnothing\}$ and $\{x\}\cup a$ for any set $a$.

Indeed, the existence of a universal set is irrelevant here. This holds in such theories, e.g. $\sf NF$, as well as in theories like $\sf ZF-Reg+\exists x(x\in x)$ which have no universal set.

The only thing you need is that every set is a member of a different set. For example singletons. If a theory cannot prove that, it's not much of a set theory.