$A=\{A,\emptyset\}$ 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. 

Suppose there were a set $A$ with the property $A=\{A,\emptyset\}$. Clearly $A \not = \emptyset$ since $A$ would have at least one element and possibly two.
Now let $x=\{A\}$. Clearly $x \not = \emptyset$ since $x$ would have exactly one element.   So the axiom of regularity would imply $\exists y\,(y\in x\land x\cap y=\emptyset)$. 
Since $x=\{A\}$ would only have one element, this would mean $y\in x$ would imply $y=A$ and then $x\cap y = \{A\} \cap \{A,\emptyset\} = \{A\} = x \not = \emptyset$.
So the property $A=\{A,\emptyset\}$ is inconsistent with the axiom of regularity for any set $A$.  
