$A=\{A,\emptyset\}$ 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 other answers and the various comments clarified this important point. There are two senses of the expression 
(a) 'A violates (R)'. 
First, (a) may mean: if you let x=A in (R), then you get a false sentence. This is not true: A does not violate (R) in this sense. 
Second, (a) may mean: if you assume that A exists (plus the axioms of ZFC minus (R)), then the denial of (R) holds. This is true (as people showed). A violates (R) in this sense. Many thanks.
