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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.

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

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  • $\begingroup$ 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. $\endgroup$ – user524154 Dec 26 '18 at 15:13
  • $\begingroup$ I knew induction plays a role here! Thanks. $\endgroup$ – Garmekain Dec 26 '18 at 15:48
  • $\begingroup$ You are welcome. $\endgroup$ – drhab Dec 26 '18 at 15:49

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