- Prove that if $F\in\mathcal{A}$, then we have
$$\bigcap\mathcal{A}\subseteq F \subseteq \bigcup\mathcal{A}$$.
Note: $\bigcap\mathcal{A}=\bigcap_{U\in\mathcal{A}}U$ and $\bigcup\mathcal{A}=\bigcup_{U\in\mathcal{A}}U$.
My proof. Let $F\in\mathcal{A}.$
$(\subseteq)$. Let $x\in\bigcap\mathcal{A}=\bigcap_{U\in\mathcal{A}}U$. It means that $x\in U$ for all $U\in\mathcal{A}$, so $x\in F$ because $F\in\mathcal{A}$. Hence
$$\bigcap\mathcal{A}\subseteq F.$$
$(\supseteq).$ $x\in\bigcup U$ iff there is a some $U\in \mathcal{A}$ such that $x\in A$. But $F$ is an exactly such an $U$. Can you check my proof? Thankss...