Exercise 1.4.3 (Atomic algebras are determined up to relabeling)

Recall the Example 1.4.7.

• Suppose that

  $$\left\{\bigcup _{\alpha \in K}{A}_{\alpha }:K\subseteq I\right\}=\left\{\bigcup _{\beta \in M}{B}_{\beta }:M\subseteq J\right\}.$$

  Pick an $A_{\alpha }$ for some arbitrary $\alpha \in I$. Then $A_{\alpha }={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in \{\alpha \}}{A}_i$ with $\{\alpha \}\subseteq I$, and is thus contained in the right-hand side. By the assumtion $A_{\alpha }$ is also contained in the right-hand side; thus, $A_{\alpha }={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{\beta \in M}{B}_{\beta }$ for some index $M\subseteq J$. But then $M$ must consist of only one element $\{\beta \}$. Suppose for the sake of contradiction that $M=\{\beta ,{\beta }^{\prime }\}$. Then $A_{\alpha }=B_{\beta }\cup B_{{\beta }^{\prime }}$. Since $B_{\beta }={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in \{\beta \}}{B}_i$ and $B_{{\beta }^{\prime }}={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in \{{\beta }^{\prime }\}}{B}_i$ are also subsets of the right-hand side by themselves, and are thus contained in the left-hand side. But $B_{\beta },B_{{\beta }^{\prime }}$ are not generated by any of the ${(A_{\alpha })}_{\alpha \in I}$ since its elements are disjoint, and only a larger set $A_{\alpha }$ from the beginning contains them. Thus, we arrive at contradiction, and $A_{\alpha }=B_{\beta }$ for some $\beta \in J$. Since the elements are disjoint, we also have the uniqueness. A similar argument gives us the opposite direction. In other words, for each $\alpha \in I$ there is a unique $\beta \in J$ such that $A_{\alpha }=B_{\beta }$, and vice versa. This shows the existence of a bijection.

• Pick an arbitrary element from the first atomic algebra, i.e., $A={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in K}{A}_{\alpha }$ for some $K\subseteq I$. Define $M:=\{\varphi (\alpha ):\alpha \in K\}$. Using the surjectivity assumption we then have

  $$B:=\bigcup _{\beta \in M}{B}_{\beta }=\bigcup _{\alpha \in K}{B}_{\varphi (\alpha )}=\bigcup _{\alpha \in K}{A}_{\alpha }$$

  and this element is contained in the $B$-atomic algebra. The converse statement follows similarly by taking the inverses ${\varphi }^{-1}(\beta )$ for $\beta \in M$ for some $M\subseteq J$.