# $\sigma$-algebra generated by a subset $\mathcal{B}$

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 $$\sigma(\mathcal B)$$?

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 $$\{2\}$$.
• Just to clarify, it should also contain $\{2,4,5\}\cup\{4,5\}^{\small\complement}=\Omega$, $(\{2,4,5\}\cup\{4,5\}^{\small\complement})^{\small\complement}=\emptyset$, $\{2,4,5\}^{\small\complement}\cap\{4,5\}^{\small\complement}=\{1,3\}$, and $\{2,4,5\}\cap\{4,5\}=\{4,5\}$ but you omitted these because they were already accounted for?