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I know pretty much nothing about set theory beyond first year undergraduate maths, so apologies if this is a stupid question.

The axiom of regularity in ZFC as I have understood it would forbid the existence of the following sets:

$\{a,\{a\},\{\{a\}\},\{\{\{a\}\}\},\ldots\}$

$\{a,\{a\},\{a,\{a\}\},\{a,\{a\},\{\{a\}\}\},\ldots\}$

Why should such sets not exist? Can it be shown that their existence leads to a contradiction? I know about Russell's paradox but isn't forbidding sets like the above a bit overkill, as they seem not to lead to paradoxes like Russell's set does.

Also is it possible to create a set of axioms that allow the maximum number of sets to exist such that they do not lead to a contradiction?

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    $\begingroup$ Why apologies? No need to do so; as long as a question is "readable", it is a good question. No stipulation is made such that every question should be a great conjecture (which is impossible in nature); let some neurotics that you possibly met here leave questions that interest not them alone. :) $\endgroup$ – Megadeth Dec 9 '15 at 5:00
  • $\begingroup$ Interestingly you're thinking about exactly the right questions in set theory. The goal in defining the axioms is basically to have axioms that permit a lot, including large sets and even huge sets, but not sets so big as to invoke Russell's paradox. $\endgroup$ – djechlin Dec 9 '15 at 5:27
  • $\begingroup$ Btw I think this is a duplicate but heck if I know how to find it. $\endgroup$ – djechlin Dec 9 '15 at 5:27
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    $\begingroup$ No contradiction at all, they exist (if $a$ does). The standard definition of the integers in set theory is almost like that, and either of these can serve as a proxy for $\Bbb N$. $\endgroup$ – BrianO Dec 9 '15 at 6:52
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You have it backwards, those sets are perfectly fine. In fact, the ordinal numbers are a hugely important class of sets of the form of the second example.

The axiom of regularity says that there are no infinite descending chains under inclusion, i.e. that there do not exist infinitely many $S_i$ such that $\cdots \in S_{i+2}\in S_{i+1}\in S_i\in\cdots\in S_1$. It is a consequence of the axiom of regularity that no set is an element of itself.

The axiom of regularity effective prevents infinitely nested sets like $\cdots\{\{a\}\}\cdots$. If we look at the example in the title, the issue at stake here is the existence of the limit of the sequence defined by the elements of the sets. That limit does not correspond to a set

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  • $\begingroup$ ah yes, I see why those sets are fine now. I misunderstood the axiom. $\endgroup$ – user85798 Dec 9 '15 at 5:50
  • $\begingroup$ You want to say that no set is an element of itself. $\endgroup$ – Stefan Mesken Dec 9 '15 at 8:53
  • $\begingroup$ Ah yes. Thanks. $\endgroup$ – Stella Biderman Dec 9 '15 at 20:05

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