Let $Ω$ be a set and let $E$ $=$ {$A$, $B$} where $A$, $B$ ∈ $Ω$ and none of $A$, $B$, $A$ $∩$ $B$, $A$ \ $B$ are empty. By constructing a partition of $Ω$ based on ${A, B}$ prove that $σ(E)$has 16 members.
Not really sure how to approach this. I've attempted constructing a partition but I'm having trouble tying this to the requirement for the sigma algebra of of $E$ having 16 members. Any help is very appreciated.
Let $P=\{A\cap B,A^c\cap B,A\cap B^c,A^c\cap B^c \}$. Clearly, $P$ is a partition of $\Omega$.
1) $\sigma(P)=\sigma(E)$: since every element of $P$ is in $\sigma(E)$, then the $\sigma$-algebra generated by $P$ is contained on $\sigma(E)$. That is $\sigma(P)\subseteq\sigma(E)$. On the other hand, note that $A=(A\cap B)\cup(A\cap B^c)$. Then $A\in\sigma(P)$. On the same way $B\in\sigma(P)$. Then, since $\sigma(P)$ is a $\sigma$-algebra containing $A$ and $B$, we have $\sigma(A,B)\subseteq\sigma(P)$. This proves $\sigma(P)=\sigma(E)$.
2) Rewrite $P$ as $P=\{E_i\}_{i=1}^4$. Now, by hypothesis, none element of $P$ is empty, and since $P$ is partition, $\sigma(P)$ can be characterized as $\sigma(P)=\{\cup_{i\in I} E_i|I\subseteq\{1,2,3,4\}\}$ (we convent that $\cup_\emptyset E_i=\emptyset$). Since the number of subsets of $\{1,2,3,4\}$ is $2^4=16$, then $|\sigma(P)|=16$.
Indeed there is a general result: if $A_1,...,A_n\subseteq\Omega$, then $\sigma(A_1,...,A_n)$ has at most $2^{2^n}$ elements. The proof is basically the same that is above.