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A more focused version of this question has now been asked at MO.

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 $S=\emptyset\not\in S$.

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  • $\begingroup$ @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." $\endgroup$ Sep 15, 2017 at 23:15
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    $\begingroup$ @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 $S=\emptyset$. $\endgroup$ Sep 15, 2017 at 23:16

2 Answers 2

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I found a paper of Cantini's that contains an argument that can be used to establish that $S \in S$ under fairly weak assumptions (on both the amount of comprehension and the underlying logic). Ultimately the proof is a fixed-point argument in the vein of Löb's theorem. This argument is strong enough to establish that $S \in S$ in $\mathsf{GPK}$. While Cantini is concerned with a contractionless logic, I would like to avoid writing out sequent calculus in this answer, so I will state and prove a weaker result in classical first-order logic.

EDIT: I recently found out that adding an abstraction operator (i.e., set builder notation) is far less innocuous than I had realized. (This is discussed by Forti and Hinnion in the introduction of this paper. My understanding of the issue is that it allows you to code negation with equality.) I suspect that the old version of my answer was only vacuously correct in that the resulting theory is inconsistnt, so I have fixed it, although I have specialized the argument to the particular case at hand. I've also cleaned up the argument a bit, mostly to make sure I actually understood it.

We need to assume that our theory $T$ has enough machinery for the following:

  • $T$ entails extensionality.
  • There is definable pairing function $(x,y) \mapsto \langle x,y\rangle$.
  • For any $a$ and $f$, there is a set $f[a]$ such that $x \in f[a]$ if and only if $\langle a,x\rangle \in f$. Note that $(f,a) \mapsto f[a]$ is a definable function by extensionality.
  • There is a set $D$ such that every element of $D$ is an ordered pair $\langle x,y\rangle$ and $\langle x,y\rangle \in D$ if and only if either $y \in y$ or $y = x[x]$.

(It is easy to check that $\mathsf{GPK}$ satisfies all of these.)

Now let $I = D[D]$. Unpacking, we have that $x \in I$ if and only if $\langle D,x\rangle \in D$ if and only if either $x \in x$ or $x = D[D] = I$. Therefore $I$ contains precisely the elements of the co-Russell class $S$ and $I$ itself, but since $I \in I$, $I \in S$ and so $I = S$, whence $S \in S$.

(Incidentally, a similar argument also resolves a question in my earlier answer to your related question. In particular, $\mathsf{GPK}$ does entail the existence of a Quine atom by the above argument if we just say that $\langle x,y\rangle \in D$ if and only if $y = x[x]$.)

In light of this, I wonder whether there even is a 'reasonable' set theory in which $S$ is non-trivial and $S \notin S$ is consistent.

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This is not a direct answer to your specific question, but it might shed an idea on a possible solution within the arena of $\mathsf{GPK}_\infty^+$ in which your question is decidable and to the positive!

Around three months ago I've asked Olivier Esser if whether adding the following condition is consistent with $\mathsf{GPK}_\infty^+$:

$``$ if $\phi$ is purely positive without free variables other than $y,A$, and without using the false formula, then: $$\exists A \forall y (y \in A \iff \phi)"$$ By this principle we can construct Quine atoms and alike sets, which are not constructible in merely $\mathsf{GPK}_\infty^+$

However Olivier Esser see that it's unclear whether such addition is consistent or not? So this principle is itself debatable?

The idea is that everything depends on what "reasonable" is? If the above principle is considered as more or less reasonable and if found consistent, then an answer is there! However, we are not there yet!

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