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So we have axiom of regularity which says that any set (let's call it $A$) has a subset $B$ such that $A\cap B = \emptyset$. But thinking about such sets as $C = \{1, 2\}$ and $D = \{3, 4\}$ we still have that $C\cap D = \emptyset$. So can it be that $\emptyset$ actually sometimes represents $2$ sets' common element (set) and sometimes represents that these $2$ sets are just disjoint? Or everything is ok with this symbolism?

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    $\begingroup$ To start with, you've got the axiom of regularity wrong. It says that any nonempty set $A$ (not just any set, it has to be nonempty) has an element (not subset) $B$ suth that $A\cap B=\emptyset$. $\endgroup$ – bof Jul 24 '14 at 9:22
  • $\begingroup$ Regularity states that $A$ (if nonempty) has a member $B$ with $A\cap B=\emptyset$. That there is a subset of $A$ with this property doesn't require regularity. In any case, $\emptyset$ means only one thing: the set that has no members. $\endgroup$ – Malice Vidrine Jul 24 '14 at 9:24
  • $\begingroup$ bof, yes you right. I just wanted to write a question shorter. ;) $\endgroup$ – Martynas Riauka Jul 24 '14 at 9:25
  • $\begingroup$ My point of question was just about symbolism. I thought that if we do not knew anything about sets A and B (or element), but we knew that A∩B=∅ then what's happening there... Are they disjoint with all they subsets or disjoint just in one element. Whatever, stupid yes.... $\endgroup$ – Martynas Riauka Jul 24 '14 at 9:41
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    $\begingroup$ That A∩B is empty. $\endgroup$ – Martynas Riauka Jul 24 '14 at 9:53
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$\emptyset$ is not a member of $A\cap B$. It is $A\cap B$. So $\emptyset$ represents the set of common elements of $A$ and $B$ but is not a common element itself. Also you got the axiom wrong as explained in comments.

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