# Axiom of Regularity and the set that contains itself

I'm trying to verify that my interpretation of the Axiom of Regularity and its logical consequence that no set is an element of itself is correct:

$A\cap\{A\}=\emptyset$ but $\{A\}\cap\{A\}=A$

In other words, if $A$ does not equal $\{A\}$ then $A\in\{A\}$ and $A\cap\{A\}=\emptyset$, but when I assume that $A=\{A\}$ such that $A\in A$ or $\{A\}\in\{A\}$ then $\{A\}\cap\{A\}$ is no longer disjoint because they both share the common element, $A$.

Thanks for your time and assistance!

• $\{A\}\cap\{A\}=\{A\}$. This is not the same as $A$, unless $A=\{A\}$, which cannot happen under regularity. – Andrés E. Caicedo Jul 8 '13 at 18:07
• Okay, so if A={A} then A∩{A} is equivalent to {A}∩{A} which is NOT disjoint. By regularity, A does not equal {A} for this reason. Is this line of reasoning correct (or almost correct)? – user84815 Jul 8 '13 at 18:14
• Yes, that's correct. (I would say it is "non-empty" rather than "not disjoint".) – Andrés E. Caicedo Jul 8 '13 at 18:29