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I want to prove for each set $S$ that $S\notin S$.

I know that I can prove this with the Regularity Axiom which says: $\exists x(x\in S) \implies (\exists y\in S)(\forall z \in S)(z \notin y)$

However if I take a simple example, let's say S = {a,{a},S} Then I can pick $y = a$ so that $a \notin a$ and {a}$ \notin a$ and $S \notin a$.

Where is my logic wrong?

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marked as duplicate by Asaf Karagila set-theory Sep 23 '17 at 17:32

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  • $\begingroup$ Could someone explain to me where my example I gave is wrong? $\endgroup$ – macco Sep 23 '17 at 17:37
  • $\begingroup$ You didn't go wrong with your logic, you just applied the axiom to the wrong set. Apply it to $\{S\}$ instead of $S$. $\endgroup$ – Bob Jones Sep 23 '17 at 17:50
  • $\begingroup$ @BobJones But even then, {S} is not element of a. $\endgroup$ – macco Sep 23 '17 at 17:52
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    $\begingroup$ No, what I mean is to let $T$ be a new set $\{S\}$, and then apply the axiom to $T$. $T$ is nonempty, so the axiom says that there is an element of $T$ that doesn't contain any element of $T$; this is false because there's only one element of $T$, namely $S$, which does contain itself. $\endgroup$ – Bob Jones Sep 23 '17 at 17:59

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