$A=\{A,\emptyset\}$ and axiom of regularity The axiom of regularity says:
(R) $\forall x[x\not=\emptyset\to\exists y(y\in x\land x\cap y=\emptyset)]$.
From (R) it follows that there is no infinite membership chain (imc). 
Consider this set: $A=\{A,\emptyset\}$. 
I am confused because it seems to me that A violates and does not violate (R). 
It seems to me that A does not violate (R) because $\emptyset\in A$ and $\emptyset\cap A=\emptyset$. 
It seem to me that A violates (R) because we can define the imc $A\in A\in A\in ...$
I checked some sources, but I am still confused. Can anyone help me? Thanks. 

The set $A=\{A, \emptyset\}$ itself does not violate Regularity, since indeed there exists an element inside $A$ whose intersection with $A$ is empty.
$$\emptyset\cap A = \emptyset$$
(Incidentally this property holds true for any set $A$.)
However the set $\{A\}$ does violate Regularity: every element (there is only one) inside $\{A\}$ intersects $\{A\}$ in a nonempty way.
$$A \cap \{A\} = A \ne \emptyset$$
because $A \in \{A\}$ and $A \in A$.
Therefore although it may seem that $A$ doesn’t violate regularity, it has a consequence that does.
