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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?)?

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Grešnik
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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.

Asaf Karagila
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  • Is the existence of set of all sets in contradiction with AC? Could we define also a choice function on the set of all sets (that is, on the set of all sets without empty set)? – Grešnik Jun 21 '19 at 08:04
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    Well, NF disproves AC. But NFU doesn't, and it does admit a set of all sets. – Asaf Karagila Jun 21 '19 at 08:07
  • I know nothing about NFU, but, can the concept of cardinality be defined in that NFU set-theory? If it can, what is the cardinality of the set of all sets? Is it some limit of cardinalities of smaller sets? – Grešnik Jun 21 '19 at 08:10
  • I also know nothing about NFU except that it is NF+Urelements, and that it is consistent with AC. – Asaf Karagila Jun 21 '19 at 08:13
  • Haha :) I already like that NFU if we have inside both AC and the set of all sets. I am also currently thinking about finding very weak axiom W so that ZFC+W proves CH. – Grešnik Jun 21 '19 at 08:16
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    CH itself is that axiom. – Asaf Karagila Jun 21 '19 at 08:17
  • Yes of course, but I want to find W which is not equivalent to CH, and, obviously, CH is equivalent to CH: – Grešnik Jun 21 '19 at 08:18
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    Well then, CH+SH. – Asaf Karagila Jun 21 '19 at 08:19
  • Those are two axioms, not one? I want to find one that implies CH, but such that CH does not imply it. Although there is some CA already in some set-theories (Axiom of Constructibility). – Grešnik Jun 21 '19 at 08:25
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    So... you disallow conjunctions and disjunctions in your logic? – Asaf Karagila Jun 21 '19 at 08:40
  • Not necessarily, but truth tables are not at all important to me. The methods of proof are more important and all mathematical sentences seem to be of the form "if...then" or "if and only if" or $A=B$ or $A \leq B$ or $A<B$ , more generally, they are either equality statements, or inequality statements, or "if...then" statements, or "iff statements", where inequality can be thought of, more generally, as some order on some set. It seems that all sentences are either conjuctions or disjunctions of those elementary sentences. And we should not forget quantifiers. – Grešnik Jun 21 '19 at 08:49
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    You're missing my point. CH+SH is a conjunction of two statements. But it is one statement, on its own. – Asaf Karagila Jun 21 '19 at 09:07
  • It is written "Please avoid extended discussions in comments. " If you want, we can chat in some chat room, so there we can share ideas. I do not know how to make one. – Grešnik Jun 21 '19 at 09:19