Let $\Omega = \{ 1,2,3,4,5,6\}$. Determine the smallest $\sigma$ algebra Let $\Omega = \{ 1,2,3,4,5,6\}$. Determine the smallest $\sigma$- algebra that contains sets $A_{1} = \{1,3,5\},A_{2} = \{2,4,6\},A_{3} = \{1,3\},A_{4} = \{2,4\}$.
I was thinking of putting $A_{i}^{'} = \{i\}, i=1,...,6$ and now, $\sigma (A_{1},...,A_{4}) =\sigma (A_{1}^{'},...,A_{6}^{'})= P(\Omega)$. Is that good thinking?
 A: As you need each of $\emptyset,\{1,2,3,4,5,6\},\{1,3,5\},\{2,4,6\},\{1,3\},\{2,4\}$ and any combination of unions, intersections, complements, and differences that can be made from these, it helps to try to spot what the essential "atoms" are in this sigma algebra.
You should be able to see that the list of atoms are $\{1,3\},\{2,4\},\{5\},\{6\}$
The sigma algebra then contains all sets from $\mathcal{P}(\{1,2,3,4,5,6\})$ with the exception of those where $1$ appears without $3$, or where $2$ appears without $4$ or vice versa.  Another way to visualize the set is to consider $\mathcal{P}(\{a,b,5,6\}$ and anywhere you see an $a$ replace it with "$1,3$" (including the comma) and anywhere you see a $b$ replace it with "$2,4$."
A: Let's use an exhaust technique. The number of possible subsets of $ \Omega$ is $2^{|\Omega|}=2^6=64$. Let $ \mathcal{A}_1$ be the collection of the $n_1= 8 $ sets below.
$$
A_1, \quad A_2,\quad A_3,\quad A_4,\quad
A_{5}=A_1^c, \quad A_{6}=A_2^c,\quad A_{7}=A_3^c,\quad A_{8}=A_4^c,
$$
If  $\mathcal{A}_{1}$ is a sigma algebra and the work is finished.
Otherwise, let $\mathcal{A}_2$ be the collection of the $8\cdot 8=64$  sets below.
$$
\begin{array}
\,A_{i_1i_2}=A_{i_1}\cup A_{i_2} & \mbox{ for }\quad i_1>i_2 \mbox{ and } A_{i_1},A_{i_2}\in \mathcal{A}_1
\\
\,A_{i_1i_2}=A_{i_1}\cup A_{i_2} & \mbox{ for }\quad i_1=i_2 \mbox{ and } A_{i_1},A_{i_2}\in \mathcal{A}_1
\\
\,A_{i_1i_2}=(A_{i_1}\cup A_{i_2})^c & \mbox{ for }\quad i_1<i_2 \mbox{ and } A_{i_1},A_{i_2}\in \mathcal{A}_1
\\ 
\end{array}
$$
Notice that $\mathcal{A}_1\subset \mathcal{A}_2$, $\mathcal{A}_2\subset 2^\Omega$ and $|2^\Omega|=64$ then some of those sets above are eventually repeated. Let $ n_2 $ be the number of subsets different from each other of $ \mathcal{A}_2 $. If  $\mathcal{A}_{2}$ is a sigma algebra and the work is finished. Otherwise, let $\mathcal{A}_3$ the colection of $n_2\cdot n_2$ sets below.
$$
\begin{array}
\,A_{i_1i_2}=A_{i_1}\cup A_{i_2} & \mbox{ for }\quad i_1>i_2 \mbox{ and } A_{i_1},A_{i_2}\in \mathcal{A}_2
\\
\,A_{i_1i_2}=A_{i_1}\cup A_{i_2} & \mbox{ for }\quad i_1=i_2 \mbox{ and } A_{i_1},A_{i_2}\in \mathcal{A}_2
\\
\,A_{i_1i_2}=(A_{i_1}\cup A_{i_2})^c & \mbox{ for }\quad i_1<i_2 \mbox{ and } A_{i_1},A_{i_2}\in \mathcal{A}_2
\\ 
\end{array} 
$$
Notice that $\mathcal{A}_2\subset \mathcal{A}_3$, $\mathcal{A}_3\subset 2^\Omega$ and $|2^\Omega|=64$ then some of those sets above are eventually repeated. Let $ n_3 $ be the number of subsets different from each other of $ \mathcal{A}_3 $. If  $\mathcal{A}_{3}$ is a sigma algebra and the work is finished. Otherwise, let $\mathcal{A}_4$ the colection of $n_3\cdot n_3$ sets below.
$$
\begin{array}
\,A_{i_1i_2}=A_{i_1}\cup A_{i_2} & \mbox{ for }\quad i_1>i_2 \mbox{ and } A_{i_1},A_{i_2}\in \mathcal{A}_3
\\
\,A_{i_1i_2}=A_{i_1}\cup A_{i_2} & \mbox{ for }\quad i_1=i_2 \mbox{ and } A_{i_1},A_{i_2}\in \mathcal{A}_3
\\
\,A_{i_1i_2}=(A_{i_1}\cup A_{i_2})^c & \mbox{ for }\quad i_1<i_2 \mbox{ and } A_{i_1},A_{i_2}\in \mathcal{A}_3
\\ 
\end{array} 
$$
This process ends after a finite number $ k $ of interactions meaning that $\mathcal{A}_k =\mathcal{A}_{k+1}=\mathcal{A}_{k+2}=\ldots $. Thus, $ \mathcal{A}_k $ will be the smallest sigma algebra containing the sets $A_1$,$A_2$,$A_3$ and $A_4$.
