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I have seen all the available answers to this question here and I am still in doubt. I know that any set x containing a single element, ie itself, can easily shown to be non-existent.

But, how do we go about proving that the following is not possible? -

x = {y, x} - were y is another entity completely disjoint from x. Say, y = 1.

TheSilverBullet
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    Note that in those duplicates nobody is assuming that $x={x}$. Just that $x\in x$. – Asaf Karagila Sep 11 '20 at 20:09
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    The axiom of regularity alone does not prove it. You have to also use the axiom of pairing: https://en.wikipedia.org/wiki/Axiom_of_pairing#:~:text=Axiom%20of%20pairing.%20In%20axiomatic%20set%20theory%20and,case%20of%20his%20axiom%20of%20elementary%20sets%20. – user247327 Sep 11 '20 at 20:40

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If $t\in t$, then $\{t\}\ne \emptyset$ and $t\in \{t\}\cap t$, therefore $\{t\} $ fails the axiom of regularity. Specialise this argument to the case $t=\{x,y\}$, and you obtain that $x=\{x,y\}$ implies that $\{x\}$ (which is also equal to $\{\{x,y\}\}$) fails regularity.

  • So, you are saying that, we repeatedly apply foundation axiom on the elements - in this case? If x = {x,1}, we can argue that {x} fails regularity. Am I correct? – TheSilverBullet Sep 11 '20 at 19:56
  • Why repeatedly? Just once: the one and only element of ${{x,y}}$ is ${x,y}$, and ${{x,y}}\cap {x,y}={x}\cap {x,y}$ by hypothesis on $x$. Therefore ${{x,y}}\cap {x,y}\ne \emptyset$, therefore ${{x,y}}\cap u\ne\emptyset$ for all $v\in {{x,y}}$. –  Sep 11 '20 at 20:04
  • Last was supposed to be $u\in{{x,y}}$, not $v$. –  Sep 12 '20 at 06:05