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Russell's paradox deals with the question: "Does the set of all sets that do not contain themselves, contain itself?"

What about the question: "Does the set of all sets that contain themselves, contain itself?"

I find the latter an equally interesting and contradictory question/statement.

Doesn't the set of all sets that contain themselves imply that for every such set $P$ you have to create an additional set $Q$ containing $P$ and all the sets in $P$, this process going on infinitely?

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  • $\begingroup$ Naively it seems like it could not be well-founded, but non-well-founded set theories exist and do not lead to contradictions. $\endgroup$ – Ian May 23 '16 at 0:02
  • $\begingroup$ Minor correction to my previous comment: naively it seems like it would have to either be empty (if the notion of "set" somehow excludes the possibility of self-containment by itself) or non-well-founded. But if it is not well founded then there is still no automatic contradiction. $\endgroup$ – Ian May 23 '16 at 0:11
  • $\begingroup$ See also math.stackexchange.com/questions/2431146/…. $\endgroup$ – Noah Schweber Aug 15 '18 at 18:56
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There's no paradox here.

If there are no sets that contain themselves (such as in modern ZFC set theory), then $\{x\mid x\in x\}$ is simply the empty set -- and we know the empty set exists!

If there are sets that contain themselves (but not too many of them), then we can still assert that there is a set that consists of all of them. That set may or may not contain itself; neither choice leads to a contradiction.

On the other hand if you have too many self-containing sets -- such that if we're working with an anti-foundation axiom that guarantees that for every $x$ there is an $y$ such that $y=\{x,y\}$ -- then we can't have a set of all of them, at least not in a ZF background, because its union would be a set of all sets, which leads to the ordinary Russell and Cantor paradoxes.

Doesn't the set of all sets that contain themselves imply that for every such set $P$ you have to create an additional set $Q$ containing $P$ and all the sets in $P$, this process going on infinitely?

Well, suppose $P=\{x\mid x\in x\}$ exists. Either $P\in P$ or $P\notin P$.

If $P\in P$, then $Q=P\cup\{P\}$ is just $P$ itself, so nothing new has been produced.

If $P\notin P$, then $Q=P\cup\{P\}$ certainly ought to exist, but it is not a set of all sets that contain themselves, because one of its elements is $P$ and $P$ doesn't contain itself. ($Q$ may or may not contain itself, but that's still not a problem -- it will if and only if it does).

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  • $\begingroup$ More accurately: They don't lead to a contradiction that we are aware of and even if they lead to a contradiction, their negation does too. That is to say: These notions are consistent relative to ZF. $\endgroup$ – Stefan Mesken May 23 '16 at 10:35
  • $\begingroup$ @Stefan: Yes, I considered writing something like that -- but it's really not clear that ZF (with no Foundation, one presumes) would be the background theory in which the OP thinks that $\{x\mid x\in x\}$ would be paradoxical. $\endgroup$ – Henning Makholm May 23 '16 at 10:38
  • $\begingroup$ I agree and to be honest. My comment was initially motivated by @Ian 's comment that ill-founded set theories don't lead to contradictions. Since this was implicitly mentioned in your answer as well, I decided to post it here instead. $\endgroup$ – Stefan Mesken May 23 '16 at 10:41

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