# Proof that $\sigma$-algebra is closed under intersection

I want to show in this proof that a $\sigma$-algebra is closed under intersection. But I am not sure if my second to last implication is true?

Let $\mathcal{F}$ be a subset of the sample space $\Omega$ with the three properties of a $\sigma$-algebra. Prove that an event $A_k\in\mathcal{F}$ is also closed under countable intersection for all $k\in\mathbb{N}$.

Properties
1) $\emptyset\in\mathcal{F}$
2) $A\in\mathcal{F}\implies A^C\in\mathcal{F} \quad \text{(closed under complementation)}$
3) $A_k\in\mathcal{F}\implies \bigcup_{k\in I} A_k\in\mathcal{F}$ for all $k\in I \quad\text{(closed under countable union)}$

My Solution
If $\mathcal{F}$ is modeled as a $\sigma$-algebra, using the second property and third property yields

\begin{align} A_k\in\mathcal{F}&\implies\bigcup_{k\in\mathbb{N}}A_k\in\mathcal{F} &&\text{(closed under union)}\\ &\implies\left(\bigcup_{k\in\mathbb{N}}A_k\right)^C\in\mathcal{F} && \text{(closed under complementation)}\\ &\implies\bigcap_{k\in\mathbb{N}}A_k^C\in\mathcal{F} && \text{(de Morgan's Law)}\\ &\implies\bigcap_{k\in\mathbb{N}}(A_k^C)^C\in\mathcal{F} && (A_k^C\in\mathcal{F} \text{ closed under complementation) }\\ &\implies \bigcap_{k\in\mathbb{N}}A_k\in\mathcal{F}\\ &\implies \mathcal{F}\;\textit{is closed under intersection.} \end{align}

New Solution
Taking @AnyAD's advice into consideration: \begin{align} A_k\in\mathcal{F}&\implies A_k^C\in\mathcal{F} && \text{(closed under complementation)}\\ &\implies\left(\bigcup_{k\in I} A_k^C\right)\in\mathcal{F} && \text{(closed under union)}\\ &\implies\left(\bigcup_{k\in I} A_k^C\right)^C\in\mathcal{F} && \text{(closed under complementation)}\\ &\implies\left(\bigcap_{k\in I} (A_k^C)^C\right)\in\mathcal{F} && \text{(de Morgan's Law)}\\ &\implies\bigcap_{k\in I}A_k\in\mathcal{F}\\ &\implies \mathcal{F}\;\textit{is closed under intersection.} \end{align}

• It would probably be helpful if you can edit the post to indicate what the three $\sigma$-algebra properties you're referring to are. – Aaron Montgomery Jun 29 '18 at 1:28
• @AaronMontgomery Sure, I'm on it! – 冬海愛衣 Jun 29 '18 at 1:29
• How do you justify (properly) the 4th step? Maybe start with $A_k^C$ instead. – AnyAD Jun 29 '18 at 1:30

## 1 Answer

I think your proof as written is missing a logical link that lets you assert the step that I think concerns you -- specifically, it's not clear that $\bigcap_{k \in \mathbb N} A_k^C \in \mathcal F \implies \bigcap_{k \in \mathbb N} A_k \in \mathcal F$. Instead, try taking complements first: $A_k \in \mathcal F \implies A_k^C \in \mathcal F \implies \bigcup_{k \in \mathbb N} A_k^C \in \mathcal F \implies \dots$

• Yes, I hope I did it right this time! – 冬海愛衣 Jun 29 '18 at 1:43
• Looks good!${}$ – Aaron Montgomery Jun 29 '18 at 1:51
• If I want to show (using property 1 and 2, same setup) that $\Omega\in\mathcal{F}$, could I argue like this? $\emptyset\in\mathcal{F}\implies\emptyset^C\in\mathcal{F}$ which in turn implies $\Omega\setminus\emptyset\in\mathcal{F}$, thus $\Omega\in\mathcal{F}$? – 冬海愛衣 Jun 29 '18 at 2:11
• Yep, that works! – Aaron Montgomery Jun 29 '18 at 3:14