# If $A$ is a non-empty set, then $A \notin A$ [duplicate]

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 $A \in \emptyset$

## marked as duplicate by Git Gud, Davide Giraudo, Andrés E. Caicedo, user91500, user147263 Dec 28 '14 at 17:55

HINT: If $A\in A$, then $X=\{A\}$ is a counterexample to the axiom of regularity.
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 $B\notin A$.