# Positive set theory, antifoundation, and the “co-Russell set”

Tl;dr version: are there "reasonable" theories which prove/disprove "the set of all sets containing themselves, contains itself"?

Inspired by this question, I'd like to ask a question which has been vaguely on my mind for a while but which I've never looked into.

Working naively for a moment, let $S=\{x: x\in x\}$ be the "dual" to Russell's paradoxical set $R$. There does not seem to be an immediate argument showing that $S$ is or is not an element of itself, nicely paralleling the fact that there are of course arguments for $R$ both containing and not containing itself (that's exactly what the paradox is, of course).

However, it's a bit premature to leap to the conclusion that there actually are no such arguments. Namely, if we look at the Godel situation, we see something quite different: while the Godel sentence "I am unprovable (in $T$)" is neither provable nor disprovable (in $T$), the sentence "I am provable (in $T$)" is provable (in $T$) (as long as we express "is provable" in a reasonable way)! So a certain intuitive symmetry is broken. So this raises the possibility that the question

$$\mbox{Does S contain itself?}$$

could actually be answered, at least from "reasonable" axioms.

Now ZFC does answer it, albeit in a trivial way: in ZFC we have $S=\emptyset$. So ideally we're looking for a set theory which allows sets containing themselves, so that $S$ is nontrivial. Also, to keep the parallel with Russell's paradox, a set theory more closely resembling naive comprehension is a reasonable thing to desire.

All of this suggests looking at some positive set theory - which proves that $S$ exists, since "$x\in x$" is a positive formula, but is not susceptible to Russell's paradox since "$x\not\in x$" is not a positive formula - possibly augmented by some kind of antifoundation axiom.

To be specific:

Is there a "natural" positive set theory (e.g. $GPK_\infty^+$), or extension of such by a "natural" antifoundation axiom (e.g. Boffa's), which decides whether $S\in S$?

In general, I'm interested in the status of "$S\in S$" in positive set theories. I'm especially excited by those which prove $S\in S$; note that these would have to prove the existence of sets containing themselves, since otherwise $S=\emptyset\not\in S$.

• @egreg I'm aware. Note that positive set theory does prove that $S$ is a set, as I said in the paragraph before "to be specific." – Noah Schweber Sep 15 '17 at 23:15
• @egreg Actually, ZFC does let us define $S$ (note that it can also define the class of sets not containing themselves) - even better it proves that $S$ is a set, since it proves that $S=\emptyset$. – Noah Schweber Sep 15 '17 at 23:16
• Well, $x\in x$ is not stratified, so NF is not gonna prove it is a set. – Asaf Karagila Sep 15 '17 at 23:17
• @AsafKaragila NF is not positive set theory. I didn't mention NF. – Noah Schweber Sep 15 '17 at 23:17
• I know I know. I'm just making conversation. I should hit the hay anyway. – Asaf Karagila Sep 15 '17 at 23:17