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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 $\sigma(\mathcal B)$?

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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 $\{2\}$.

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  • $\begingroup$ 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? $\endgroup$
    – Remy
    Commented Sep 9, 2020 at 4:05
  • $\begingroup$ Yes, "also" means "in addition to". $\endgroup$
    – Graham Kemp
    Commented Sep 9, 2020 at 4:08

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