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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    $\begingroup$ Related post : How can a set contain itself ? $\endgroup$ Jun 21, 2019 at 8:00
  • $\begingroup$ If $U$ is the set of all sets (and if it exists) then the single element set $\{U\}$ contains a set which is a member of itself. So too does the two-element set $\{U,\emptyset\}$, etc. and thus these are different examples $\endgroup$
    – Henry
    Jun 21, 2019 at 8:12

1 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.

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    $\begingroup$ Well, NF disproves AC. But NFU doesn't, and it does admit a set of all sets. $\endgroup$
    – Asaf Karagila
    Jun 21, 2019 at 8:07
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    $\begingroup$ CH itself is that axiom. $\endgroup$
    – Asaf Karagila
    Jun 21, 2019 at 8:17
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    $\begingroup$ Well then, CH+SH. $\endgroup$
    – Asaf Karagila
    Jun 21, 2019 at 8:19
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    $\begingroup$ So... you disallow conjunctions and disjunctions in your logic? $\endgroup$
    – Asaf Karagila
    Jun 21, 2019 at 8:40
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    $\begingroup$ You're missing my point. CH+SH is a conjunction of two statements. But it is one statement, on its own. $\endgroup$
    – Asaf Karagila
    Jun 21, 2019 at 9:07

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