TITLE: If $A$ is a non-empty set, then $A \notin A$
QUESTION [7 upvotes]: Exercise 3.2.2
Use the axiom of regularity and the singleton set axiom to show that if $A$ is a set then $A \notin A$.
Axiom of Regularity:
If $A$ is a non-empty set, then there is at least one element x of A which is either not a set, or is disjoint from A.
Singleton Set Axiom:
If $a$ is an object, then there exists a set $\{a\}$ whose only element is $a$, we refer to $\{a\}$ as the singleton set whose element is $a$
Furthermore, if $a$ and $b$ are objects, then there exists a set $\{a,b\}$ whose only elements are $a$ and $b$

I think I can give the proof when A is the non-empty set which contains finite elements. However, can anyone give any hint for the general case?
Suppose $A \in A$ 
Then eixsts $x$ such that either $x$ is not a set or $x \cap A$=$\emptyset$
Then it is clear thar $x \ne A$ (if $x = A$, then $x \cap A \ne \emptyset$)
We observe that as an element, $A$ $\ne$ $x$, $A$ $\in$ $A$
Then $A$ $\in$ $A-x$ 
Define the new set $A-x$ as $A_1$
then exists $x_1$ such that $A \in A_1-x_1$
After finite steps
we find that $A \in \emptyset $
which leads to contradiction.

REPLY [4 votes]: Let $A$ a set. By Singleton axiom, $\{A\}$ is a set, which contradicts the axioms of regularity.
EXTRA. Let $A$ and $B$ sets. By pair axiom, $\{A,B\}$ is a set, a contradiction of axiom of regularity. Hence $A\notin B$ or $B\notin A$.