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Essentially, I want to prove that $\forall x(x \notin x)$. I'm not sure where to begin with this. I have tried manipulating the axiom of specification to achieve this, as my guess is this is the axiom that enforces this (given it's purpose of eliminating Russell's paradox and the set of all sets, etc.).

In fact, I am trying to prove this statement as part of proving that the set of all sets doesn't exist. Here's my planned method:

  1. Assume that the set of all sets exist. From this I can derive $\exists x(x \in x)$
  2. Prove $\forall x(x \notin x)$, from which it follows that $\neg \exists x(x \in x)$
  3. A contradiction arises, disproving the assumption in 1.

I am working through a set theory book, and trying to write proofs to satisfy myself that various statements in the book are true. For all I know, this might be proven further in the book, and may require further axioms that I have not yet encountered. At this point the only axioms I have are extensionality and the axiom schema of specification. Perhaps I need other axioms not yet encountered?

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    The axiom of regularity is necessarily needed to prove it. – Hanul Jeon Jul 17 '18 at 04:59
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    Perhaps surprisingly, separation is not what enforces $\forall x(x\notin x)$. It is foundation (and it is not provable without foundation). An easier way to show that a set of all sets doesn't exist is just from Russell's paradox alone. If it did, we could use separation to show the existence of Russell's paradoxical set. (Note that foundation is not necessary here.) – spaceisdarkgreen Jul 17 '18 at 04:59
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    Also, on a side note, it's probably a bad habit to think of specification/separation as 'eliminating Russell's paradox' because adding axioms can never do anything but increase your chances of being inconsistent (the upside of course is that you can prove more things). What avoids Russell is that specification is a weakened form of the unrestricted comprehension axiom that was part of naive set theory. – spaceisdarkgreen Jul 17 '18 at 05:42
  • Some authors use the word Regularity; others say Foundation. – DanielWainfleet Jul 24 '18 at 18:41
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    Possible duplicate of Set that has itself as an element? – Git Gud Sep 16 '18 at 09:35

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You need the axiom of regularity to prove your statement. The axiom of regularity states every non-empty set has a $\in$-minimal element in the following sense: if $x\in A$ is a $\in$-minimal element of $A$ then no $y\in x$ satisfy $y\in A$.

Now you can prove your statement as follows: assume that there is $x$ such that $x\in x$. Take $A=\{x\}$, then $A$ must have a $\in$-minimal element. As $A$ is singleton, such $\in$-minimal element must be $x$. However it is impossible, since $x\in x$ satisfies $x\in A$.

(In fact, you can prove stronger theorem: the axiom of regularity proves there is no $\in$-descending chain. That is, there is no sequence $\langle x_n \mid n=0,1,2\cdots\rangle$ such that $x_0\ni x_1\ni x_2\ni\cdots$.)

Axiom of regulariry is necessary to prove the statement. Assuming there is a set $q$ such that $q=\{q\}$ does not derive any contradiction under ZFC without regularity. (A set $q$ such that $q=\{q\}$ is known as Quine atom.)

Hanul Jeon
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    Kunen's textbook (Chapter IV, Exercises 18 & 19) shows how to obtain, from ZFC, a class model $\Bbb M=(M,E)$ that satisfies all the axioms of ZFC except Regularity (Foundation) but $\Bbb M$ satisfies $ \exists x; (x={x}).$ So if the rest of the axioms are consistent then they cannot disprove the existence of a Quine atom. – DanielWainfleet Jul 24 '18 at 19:00
  • @DanielWainfleet Thank you for your informative comment. – Hanul Jeon Jul 25 '18 at 10:12