(a) For a nonempty subset $B$ , if there is $a\in B\cap A_{1}$ then the smallest element of $B$ is the $<_{1}$ -smallest element of $B\cap A_{1}$ , otherwise $B\subset A_{2}$ , and the smallest element of $B$ is the $<_{2}$ -smallest element of $B\subset A_{2}$ .

(b) Let $\{A_{\alpha}\}_{\alpha\in J}$ be a family of disjoint well-ordered (by $<_{\alpha}$ , respectively) sets where the index set $J$ is well-ordered by $<_{J}$ . Define an order relation on $A=\cup_{\alpha\in J}A_{\alpha}$ such that $a<b$ iff either $a\in A_{\alpha}$ , $b\in A_{\beta}$ and $\alpha<_{J}\beta$ , or $a,b\in A_{\alpha}$ and $a<_{\alpha}b$ . Then, $<$ is a well-ordering on $A$ . Indeed, for a nonempty set $B\subset A$ , the set of indexes $I=\{\alpha|B\cap A_{\alpha}\neq\emptyset\}$ is not empty, hence, there is the $<_{J}$ -smallest element $\gamma$ of $I$ . Further, the set $C=B\cap A_{\gamma}$ is not empty, and, hence, there is the $<_{\gamma}$ -smallest element $c$ of $C$ . For every $b\in B$ , $b\in A_{\alpha}$ for some $\alpha$ , and either $\gamma<_{J}\alpha$ , or $\gamma=\alpha$ and $c\le_{\gamma}b$ . In either case, we conclude that $c\le b$ , and $c$ is the smallest element of $B$ .