I understand that an algebra $F \subset 2^\Omega$ is called a $\sigma$-algebra if it additionaly satisfies:
$(A_i)_{i \in \mathbb{N}}$ with $A_i \in F$ pairwise disjoint, then also $\cup_{i \in \mathbb{N}} A_i \in F$
, i.e., each finite union of pairwise disjoint subsets must again be in $F$.
However, I have difficulties imagining an example for an algebra which is not $\sigma$-algebra.
Can there be an algebra which is no $\sigma$-algebra e.g. on $\Omega = \{1,2,3,4,5,6\}$ ?
Best Answer
Consider:
$$\mathcal{C}=\{A \subset \mathbb{Z} : A \text{ is finite or } \mathbb{Z}\backslash A \text{ is finite} \}$$
Then it's algebra: it's closed under finite sum and complement $\emptyset, \mathbb{Z} \in \mathcal{C}$, but not $\sigma$-algebra, for example $\{2n\}\in\mathcal{C}$ for all $n \in \mathbb{Z}$, but:
$$\cup_{n \in \mathbb{Z}}\{2n\} \not\in \mathcal{C}$$