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Next, we show a general fact about the dual space $B^*$ :

Proof. Clearly, every element of $\mathcal{B}$ is clopen, by definition. Conversely, suppose $U$ is clopen. Then $U=\bigcup \lbrace M(a_i)\mid i\in I\rbrace$ for some index set $I$ , since $U$ is open. But $U$ is closed, so $B^*-U=\bigcup \lbrace M(a_j) \mid j\in J\rbrace$ for some index set $J$ . Hence $B^*=\bigcup \lbrace M(a_k) \mid k\in I\cup J \rbrace$ . Since $B^*$ is compact, there is a finite subset $K$ of $I\cup J$ such that $B^* =\bigcup \lbrace M(a_k)\mid k\in K\rbrace$ . Let $V= \bigcup \lbrace M(a_i)\mid i\in K\cap I\rbrace$ . Then $V\subseteq U$ . But $B^*-V \subseteq B^*-U$ also. So $U=V$ . Let $y=\bigvee \lbrace a_i \mid i\in K\cap I\rbrace$ , which exists because $K\cap I$ is finite. As a result, $$U=V= \bigcup \lbrace M(a_i)\mid i\in K\cap I\rbrace = M(\bigvee \lbrace a_i\mid i\in K\cap I\rbrace)=M(y)\in \mathcal{B}.$$

Finally, based on the result of this entry, $B$ is isomorphic to the field of sets $$F:=\lbrace F(a)\mid a\in B\rbrace,$$ where $F(a)=\lbrace P\mid P\mbox{ prime in }B,\mbox{ and }a\notin P\rbrace$ . Realizing that prime ideals and maximal ideals coincide in any Boolean algebra, the set $F$ is precisely $\mathcal{B}$ .