# Prove that for each set $S$ that $S\notin S$ with regularity axiom. [duplicate]

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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?

## marked as duplicate by Asaf Karagila♦ set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 23 '17 at 17:32

• You didn't go wrong with your logic, you just applied the axiom to the wrong set. Apply it to $\{S\}$ instead of $S$. – Bob Jones Sep 23 '17 at 17:50
• 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. – Bob Jones Sep 23 '17 at 17:59