*Lemma 3.1.13 If \(a\) and \(b\) are objects, then \(\{a,b\}=\{a\}\cup\{b\}\). If \(A,B,C\) are sets, then the union operation is commutative (i.e., \(A\cup B= B\cup A\)) and associative (i.e., \((A\cup B)\cup C=A\cup(B\cup C))\). Also, we have \(A\cup A=A\cup\emptyset=\emptyset\cup A=A\).
\(Proof~by~Tao\). 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. \(\square\)