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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!

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    $\begingroup$ $\{A\}\cap\{A\}=\{A\}$. This is not the same as $A$, unless $A=\{A\}$, which cannot happen under regularity. $\endgroup$ – Andrés E. Caicedo Jul 8 '13 at 18:07
  • $\begingroup$ 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)? $\endgroup$ – user84815 Jul 8 '13 at 18:14
  • $\begingroup$ Yes, that's correct. (I would say it is "non-empty" rather than "not disjoint".) $\endgroup$ – Andrés E. Caicedo Jul 8 '13 at 18:29

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