sigma algebra generated by a set example I wanted to check my understanding of this concept. The sigma algebra $M(\psi)$ generated by a set $\psi$ is the intersection of all sigma-algebras that contain $\psi$.
As an example if $X=\{1,2,\dots,6\}, \psi=\{\{2,4\},\{6\}\}$ then
$M(\psi)$ would be the sigma algebra that is the intersection of all sigma algebras that contain $\psi$. For a sigma algebra to contain $\psi$ it must (if I've calculated correctly) contain $\{\psi, \emptyset, \{2,4,6\},\}$ and then also the complements of these and the complements of $\{2,4\}$ and $\{6\}$.
Other than direct computation is there a faster way to calculate this?
 A: Here is a method to compute the $\sigma$-algebra generated by two non-empty disjoint sets say $S_1$ and $S_2$ on a set $S$.

*

*Let $S_3:=S\setminus (S_1\cup S_2)$. Then $\{S_1,S_2,S_3\}$ is a partition of $S$.

*Let  $E$ a subset of $S$. Define $E^0=\emptyset$ and $E^1=E$. Let
$$A_{i_1,i_2,i_3}=S_1^{i_1}\cup S_2^{i_2}\cup S_3^{i_3}, i_1,i_2,i_3\in \{0,1\}.
$$
Then $\sigma(S_1,S_2)=\{A_{i_1,i_2,i_3},i_1,i_2,i_3\in \{0,1\}\}$.

This extends readily to the $\sigma$-algebra generated by a finite collection of non-empty pairwise disjoint subsets $S_1,\dots,S_n$: let $S_{n+1}:= S\setminus \bigcup_{i=1}^nS_i$: then
$$
\sigma(S_i,1\leqslant i\leqslant n)=\left\{  \bigcup_{q=1}^{n+1} S_q^{i_q},i_q\in \{0,1\}    \right\}.
$$
A: Actually $\psi$ must be a subset (I think this is what you meant) since both are by definitions subsets of $2^X$. My approach would be the following:
I know that $\emptyset, X, \lbrace 2,4\rbrace $ and $\lbrace 6 \rbrace $ must be contained. You must add $\lbrace 2,4,6 \rbrace$, otherwise not every union would be contained. Do we now have a $\sigma$-algebra? No, because we need the complements. So we have to add the complement of each set to obtain $$\lbrace \emptyset, \lbrace 2,4 \rbrace, \lbrace 6\rbrace,\lbrace 2,4,6\rbrace,\lbrace 1,3,5,6 \rbrace, \lbrace 1,2,3,4,5\rbrace ,\lbrace 1,3,5\rbrace, X \rbrace.$$
This set is constructed such that it must be contained in every $\sigma$-algebra which contains $\psi$, so all you have to do now ist to verify that this is a $\sigma$-algebra. Then, it would also be the smallest.
A: At first hand - if $\psi=\{A_1,\cdots,A_n\}$ (so is finite) - go for sets of the form $E_1\cap\cdots\cap E_n$ where $E_i\in \{A_i,A_i^{\complement}\}$ for $i\in\{1,\dots,n\}$.
In your case that leads to the sets:


*

*$\{2,4\}\cap\{6\}=\varnothing$

*$\{2,4\}\cap\{6\}^{\complement}=\{2,4\}$

*$\{2,4\}^{\complement}\cap\{6\}=\{6\}$

*$\{2,4\}^{\complement}\cap\{6\}^{\complement}=\{1,3,5\}$
At second hand go for the sets that can be written as a union of these sets.
They together form the $\sigma$-algebra generated by $\psi$. 
(Actually the algebra but in finite case it coincides with the  $\sigma$-algebra.)
There are $2^3=8$ of such unions in your case.
