For finite $\sigma$-algebras, to check closure of countable unions, does it suffice to show that the union of two events is also an event?
I have a finite, supposedly $\sigma$-algebra $\Sigma$ and I want to show if it is closed under countable unions. My question is if it is true that if $A,B\in\Sigma$ and $A\cup B\in\Sigma$ it means that $\Sigma$ is closed under countable unions.
I think it is because in the context of finite $\sigma$-algebras i think closure of countable unions is equivalent to the closure of finite unions, and finite unions are basically unions of two sets, repeatedly.
But I don't know how to prove it, so I cannot believe myself yet.
Yes, it is enough.
Let $A_1,A_2,\dots\in\mathcal A$.
Based on $A,B\in\mathcal A\implies A\cup B\in\mathcal A$ with induction it can be shown that $\bigcup_{k=1}^nA_k\in\mathcal A$ for every $n$.
So for $n$ large enough - because $\mathcal A$ is finite - we will have: $$\bigcup_{k=1}^{\infty}A_k=\bigcup_{k=1}^{n}A_k\in\mathcal A$$
(If no such $n$ exists then it can be proved that $\{A_k\mid k\in\mathbb N\}$ is infinite, contradicting that $\mathcal A$ is finite).