Definition 3.1.4 (Equality of sets). Two sets \(A\) and \(B\) are \(equal\), \(A=B\), iff every element of \(A\) is an element of \(B\) and vice versa. To put it another way, \(A=B\) if and only if every element \(x\) of \(A\) belongs also to \(B\), and every element \(y\) of \(B\) belongs also to \(A\).
$$ (\forall x, x \in A \implies x \in B) \land (\forall x, x \in B \implies x \in A) $$
\(Proof~by~Tao\). We prove by contradiction. Suppose there does not exist any object \(x\) such that \(x\in A\). Then for all objects \(x\), we have \(x\notin A\). Also, by Axiom 3.2 we have \(x\notin\emptyset\). Thus \(x\in A\iff x\in\emptyset\) (both statements are equally false), and so \(A=\emptyset\) by Definition 3.1.4, a contradiction.
\(Proof~by~Tao\). We prove just the associativity identity \((A\cup B)\cup C=A\cup (B\cup C)\), and leave the remaining claims to Exercise 3.1.3. By Definition 3.1.4, we need to show that every element \(x\) of \((A\cup B)\cup C\) is an element of \(A\cup(B\cup C)\), and vice versa. So suppose first that \(x\) is an element of \((A\cup B)\cup C\). By Axiom 3.4, this means that at least one of \(x\in A\cup B\) or \(x\in C\) is true. We now divide into two cases. If \(x\in C\), then by Axiom 3.4 again \(x\in B\cup C\), and so by Axiom 3.4 again we have \(x\in A\cup(B\cup C)\). Now suppose instead \(x\in A\cup B\), then by Axiom 3.4 again \(x\in A\) or \(x\in B\). If \(x\in A\) then \(x\in A\cup(B\cup C)\) by Axiom 3.4, while if \(x\in B\) then by consecutive applications of Axiom 3.4 we have \(x\in B\cup C\) and hence \(x\in A\cup(B\cup C)\). Thus in all cases we see that every element of \((A\cup B)\cup C\) lies in \(A\cup(B\cup C)\). A similar argument shows that every element of \(A\cup(B\cup C)\) lies in \((A\cup B)\cup C\), and so \((A\cup B)\cup C=A\cup(B\cup C)\) as desired. \(\square\)
Exercise 3.1.2 Using only Definition 3.1.4, Axiom 3.1, Axiom 3.2, and Axiom 3.3, prove that the sets \(\emptyset, \{\emptyset\}, \{\{\emptyset\}\}\), and \(\{\emptyset, \{\emptyset\}\}\) are all distinct (i.e, no two of them are equal to each other).