Letting $\Omega =\{1,2,3,4,5\}$ find the $\sigma$-field for $\Omega$
generated by $\mathcal{B}=\{\{4,5\},\{2,4,5\}\}$.
I'm new to measure theory and just wanted to make sure I'm understanding this correctly. Just going off the definition:
Let $X$ be a set and $\mathcal B$ be a non-empty collection of subsets
of $X$. The smallest $\sigma$–algebra containing all the sets of
$\mathcal B$ is denoted by $\sigma(\mathcal B)$ and is called the
sigma-algebra generated by the collection $\mathcal B$.
Would I just include the necessary elements to $\mathcal B$ which make it a $\sigma$-algebra (contains the sample space, the empty set, and is closed under complementation)? That would give
$$\sigma(\mathcal{B})=\{\Omega,\emptyset, \{4,5\},\{2,4,5\}, \{1,2,3\}, \{1,3\}\}$$
Is my reasoning correct or are there other elements that need to be added to $\sigma(\mathcal B)$?
Best Answer
Your reasoning is correct but incomplete. The algebra must be closed for countable unions and (relative)complements, so must also include $\{2,4,5\}^{\small\complement}\cup\{4,5\}$ and $(\{2,4,5\}^{\small\complement}\cup\{4,5\})^{\small\complement}$.
That is: $\{1,3,4,5\}$ and $\{2\}$.