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This mathoverflow question makes the following 2 points:

  • The Gödel sentence, "this sentence is not provable", which indeed is not provable in whichever base theory that was used when formulating it and hence is true yet unprovable.
  • The Löb sentence, "this sentence is provable," is indeed provable, remarkably, by Löb's theorem.

In the context of naive set theory, let us momentarily consider what one might name the Löb set:

$$L \equiv \left\{x\, |\, x\in x\right\}$$

Of course in $ZF$ set theory such a construction is not allowed, as there is no universal set over which a predicate in $x$ can be considered. But at the surface level, unlike the Russell set in which $R\in R\Leftrightarrow R\not\in R$, there is no immediate contradiction evident when either statement $L\in L$, or $L\not\in L$, are considered.

Questions

Given the well-known correspondence between Russell's Paradox and Gödel's second incompleteness theorem, one might consider the correspondence between $L$ and the Löb sentence. Moreover since Löb's theorem shows that the Löb sentence is provable, is there a corresponding theorem in set theory that proves $L\in L$?

Is there a way to even ask this question by improving the definition of $L$ within the context of axiomatic set theory, or is the notion of $L$ dead-on-arrival with no way to survive outside of naive set theory?

Matt
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    This is a very natural question, but it has been asked before. Briefly: under mild hypotheses, we do indeed get the parallel situation, where the set of all sets that do contain themselves contains itself. (We have to leave $\mathsf{ZFC}$ to make this occur, though.) – Noah Schweber Sep 11 '22 at 16:02
  • I'm not sure why this got a downvote - it is a very good question! – Noah Schweber Sep 11 '22 at 17:10
  • Interesting, thank you for linking to your prior exploration of this question. A similar positive result appears to exist for the question Will this Turing machine find a proof of its halting? – Matt Sep 11 '22 at 18:34
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    Also, note that in fact in $\mathsf{ZF}$ the class ${x: x\in x}$ is a set - it's $\emptyset$. Meanwhile, the alternative set theory $\mathsf{NFU}$ does have self-containing sets but proves that ${x:x\in x}$ is not a set (since every set has a complement). The interesting theories are things like $\mathsf{GPK_\infty^+}$ which prove that ${x:x\in x}$ is a nonempty set. – Noah Schweber Sep 13 '22 at 01:29

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