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Assuming the set theory we're working with allows self-containment, as well as arbitrary set building of the form $\{x:\Phi(x)\}$, if we define $S=\{x:x\in x\}$, is $S\in S$ knowable?

As we see from assuming self-containment of $S$ or not, it seems to not give any new information:

$$S\in S \implies S\in S$$

$$S\notin S \implies S\notin S$$

However, unlike Russel's Paradox, there does not seem to be a contradiction for this set's existence (at least naively).

Edit: For clarity, I'm asking if the provability (or rather unprovability of) $S\in S$, is itself, provable.

Andrés E. Caicedo
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Graviton
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    As you say, the formula $S \in S$ does not produce Russell's paradox. But, if we do not restrict the comprehension axiom: for every $\Phi$, the set $A = { x \mid \Phi(x) }$ exists, we can reproduce Russell's paradox: $R \in R \leftrightarrow R \notin R$. This is a contradicition and thus, using the Principle of Explosion we can prove every formula in the language of the system. – Mauro ALLEGRANZA May 26 '20 at 09:23
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    This means that we can prove $S \in S$ as well as $S \notin S$. – Mauro ALLEGRANZA May 26 '20 at 09:23
  • @MauroALLEGRANZA I should have expected that. On the grand assumption that this hypothetical set theory had other axioms that forbid Russel's paradox or other contradictions that would lead to the principle of explosion, then $S\in S$ would be unprovable? – Graviton May 26 '20 at 09:31

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Let me interpret this as asking the question in $\sf ZF-Foundation$ as our set theory. Can it decide this statement?

The answer is no. Since $\sf Foundation$ is consistent with our set theory, it is consistent that $\{x\mid x\in x\}=\varnothing$, and of course, $\varnothing\notin\varnothing$. So any chance for deciding anything must be for the negative.

Alas, we can arrange for a situation where the only set that contains itself is a unique Quine atom, that is $a=\{a\}=\{x\mid x\in x\}$, and in that case the set of all sets that contain themselves contains itself.

But we're not over yet. It is consistent also to have a proper class of Quine atoms, which means that the class $\{x\mid x\in x\}$ is not even a set to begin with. So we can't even prove that this collection is a set.

Asaf Karagila
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