Why does $x = \{\{\emptyset\}\}$ not contradict the axiom of regularity in set theory? The axiom of regularity: $\forall x, x=\emptyset \lor\exists y\in x $ $\forall z\in y$ , $z\notin x $. 
But what about the set $x = \{\{\emptyset\}\}$? From my understanding of the axiom, it states that there exists an element $y$ in $x$ such that $y$ and $x$ share no elements. In this case, it seems $y$ would have to be $\{\emptyset\}$, which has one element, namely, $\emptyset$, but that is an element as well is it not? Since the empty set is an element of every set? Thus $x$ is actually equal to $x = \{\emptyset ,\{ \emptyset\}\}$? 
I am assuming here that $y$ cannot be equal to the empty set, as that has no members, so I cannot make sense of the idea that the elements of $y = \emptyset$ are not elements of $x$. 
Where has my reasoning gone wrong? 
 A: The emptyset is not an element of every set - it's a subset of every set, but that's not the same thing. In particular, $\emptyset\not\in\{\{\emptyset\}\}$: the only element of $\{\{\emptyset\}\}$ is $\{\emptyset\}$, and that's not the emptyset ($\emptyset$ has no elements but $\{\emptyset\}$ has one element, namely $\emptyset$).
Similarly, it is not the case that $\{\{\emptyset\}\}=\{\emptyset,\{\emptyset\}\}$.
A: 
the empty set is an element of every set

Wrong.
A: First of all, if we were talking about $\{\emptyset, \{\emptyset\}\}$, then this would absolutely satisfy Regularity. In general, never assume that the empty set is excluded from a "there exists" statement unless explicitly stated - $y$ is absolutely permitted to be $\emptyset$. The axiom states that, provided $x$ is nonempty, there exists a $y \in x$ so that $y$ does not have any members in common with $x$; if $y$ is the empty set, then it does not have any members at all, so of course it does not have any members in common with $x$!
Second, never assume something is a member of a set unless explicitly stated. The notation $x = \{\{\emptyset\}\}$ literally means "$x$ is that set which includes $\{\emptyset\}$ and absolutely nothing else". The empty set is only a member of those sets which $\emptyset$ is a member of - it sounds tautological, but it's important. What you're misremembering is something that causes a lot of students confusion - the empty set is a subset of every set, not an element. Recall that $a$ is a subset of $b$ if every member of $a$ is a member of $b$; another way of saying it is that $a$ does not have any members that aren't members of $b$. Since $\emptyset$ has no members at all, $\emptyset$ does not have any members that aren't members of $b$ no matter what $b$ is, so $\emptyset$ is automatically a subset of $b$.
So the $y$ required by Regularity in this case is $\{\emptyset\}$; $\{\emptyset\}$'s only element is $\emptyset$, which is not an element of $x$.
