I am stuck with this problem. I want to prove this equality but I don't know how to do it. First of all I am going to define what is a $\sigma$-algebra is:
Let $X$ be a set. Then a sigma-algebra $F$ is a nonempty collection of subsets of $X$ such that the following hold:
$X$ is in $F$.
If $A$ is in $F$, then so is $(X\setminus A)$.
If $A_n$ is a sequence of elements of $F$, then the union of the $A_n$ is in $F$.
If $S$ is any collection of subsets of $X$, then we can always find a sigma-algebra containing $S$, namely the power set of $X$. By taking the intersection of all sigma-algebras containing $S$, we obtain the smallest such sigma-algebra. We call the smallest sigma-algebra containing $S$ the sigma-algebra generated by $S$.
Being $\Omega$ a set and $\{A_1, A_2,...,A_n\}$ a partition of it. I have to prove that:
$$\sigma (\{A_1, A_2,...,A_n\})=\left\{\bigcup_{i \in I} A_i : I \subseteq \{1,...,n\}\right\}$$
Can somebody help me?
You want to show $\sigma\{A_1, \ldots, A_n\} = \left\{\bigcup_{i \in I} A_i : I \subset \{1, \ldots, n\}\right\}$.
To prove the inclusion $\subseteq$, just show the right-hand side is a $\sigma$-algebra.
To prove the inclusion $\supseteq$, show that any $\sigma$-algebra that contains the left-hand side must also contain the right-hand side.
Edit to clarify my last sentence:
I should have written "any $\sigma$-algebra that contains $\{A_1, \ldots, A_n\}$ must also contain the right-hand side."
First show that if $\mathcal{F}$ is a $\sigma$-algebra that contains $\{A_1, \ldots, A_n\}$, then $\mathcal{F}$ must contain $\{\bigcup_{i \in I} A_i : I \subset \{1, \ldots, n\}\}$.
Then note that the intersection of all such $\sigma$-algebras $\mathcal{F}$ containing $\{A_1, \ldots, A_n\}$ must also contain $\{\bigcup_{i \in I} A_i : I \subset \{1, \ldots, n\}\}$.
Finally, note that $\sigma\{A_1, \ldots, A_n\}$ is the intersection of all such $\mathcal{F}$.