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?

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.

• Hint: $A \cup B \cup C = (A \cup B) \cup C$. – user295959 Dec 26 '18 at 14:33

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).

• Another explanation: You want to show that $\mathcal{B}\subseteq\mathcal{A}$ for $\mathcal{B}$ countable implies $\bigcup\mathcal{B}\in\mathcal{A}$. If $\mathcal{B}$ is finite, then we're done by induction. And $\mathcal{B}$ cannot be countably infinite, else by assumption, $\mathcal{B}\subseteq\mathcal{A}$, i.e. a countably infinite set is a subset of a finite set, contradiction. So there's no need to check the case for $\mathcal{B}$ countably infinite. – user524154 Dec 26 '18 at 15:13
• I knew induction plays a role here! Thanks. – Garmekain Dec 26 '18 at 15:48
• You are welcome. – drhab Dec 26 '18 at 15:49