Exercise 3.2.2 (Axiom of regularity)

To show $A\notin A$, consider the singleton set $\{A\}$ (by Axiom 3.3, such a set exists). Axiom 3.9 implies that $A\cap \{A\}=\varnothing$. If $A\in A$ then $A\cap \{A\}=\{A\}$, leading to a contradiction.

To prove the latter claim, suppose $A\in B$ and $B\in A$. Consider the pair set $\{A,B\}$ (which, again, by Axiom 3.3, exists). Then, $A\cap \{A,B\}=\{B\}$ and $B\cap \{A,B\}=\{A\}$ which contradicts Axiom 3.9.

P1: Assume we have a set $A$ that $A\in A$, then let $B=\{A\}$. According to the Axiom of Regularity, B as a set should at least contain an element which is a) either not a set, b)or disjoint from itself $B$. Apparently a) is not valid cause the only element of $B$ is a set; thus b) should be valid. So the only element of $B$ $A$ should be disjoint from $B$ which is $\{A\}$, but $A\cap \{A\}=A$. (This statement is a little bit confusing but just try to understand it.) Thus both a) and b) are invalid, and so come to a paradox. So we can’t have a set $A$ that $A\in A$.

P2: Assume we have two sets $A$ and $B$ which contains each other. Then we can have a set $\{A,B\}$. For each element in $\{A,B\}$, say $A$, we have $A$ is a set and $A\cap \{A,B\}=A$, and so is B. This contradicts the Axiom of Regularity. So we can’t have such $A$ and $B$.