Monotone class that is closed under unions Let $X$ be a set, $E\subset X$ a set-algebra and $m$ the smallest monotone class containing $E$.
Let $Y=\{A\in m:A^\complement\in m\}$.

I want to show that $Y$ is closed under countable unions

My idea was that for a sequence $(A_n)_{n}$ we can define $U_k=\bigcup_{i=1}^k A_i.$ Then $(U_k)_k$ is an increasing sequence. However, for that to hold, $Y$ should be closed under finite unions, which beforehand is not necessary.
How do I solve this problem?
 A: Lemma 1 For all $A \in m, B \in E$ we have $A \cap B \in m$.
Proof. Let 
$$\mathcal{A} = \{ A \in m : (\forall B \in E) \, A \cap B \in m \}.$$
Since $E$ is an algebra of sets, $E \subseteq \mathcal{A}$. Moreover $\mathcal{A}$ is a monotone class: for instance let $A_1 \subseteq A_2 \subseteq A_3 \subseteq \ldots$ be a sequence of sets in $\mathcal{A}$ and let $B \in E$. From the assumption each $A_n \cap B$ is in $m$ and $A_1 \cap B \subseteq A_2 \cap B \subseteq A_3 \cap B \subseteq \ldots$ is an ascending sequence, so the set
$$\bigcup_{n=1}^{\infty} A_n \cap B = \bigcup_{n=1}^{\infty} (A_n \cap B)$$
is in $m$, hence $\displaystyle \bigcup_{n=1}^{\infty} A_n \in m$. It's similar to prove that $\mathcal{A}$ is closed under descending sequences, using
$$\bigcap_{n=1}^{\infty} A_n \cap B = \bigcap_{n=1}^{\infty} (A_n \cap B).$$
Hence $\mathcal{A}$ is a monotone class containing $E$, therefore $m \subseteq \mathcal{A}$ which implies our claim. $\ \square$
Lemma 2 For all $A, B \in m$ we have $A \cap B \in m$. 
Proof. Fix $A \in m$ and let
$$\mathcal{B} = \{ B \in m : A \cap B \in m \}.$$
From lemma 1 it follows that $E \subseteq \mathcal{B}$. Moreover $\mathcal{B}$ is a monotone class: for instance let $B_1 \subseteq B_2 \subseteq B_3 \subseteq \ldots$ be a sequence of sets in $\mathcal{B}$. From the assumption $A \cap B_1 \subseteq A \cap B_2 \subseteq A \cap B_3 \subseteq \ldots$ is an ascending sequence of sets in $m$, so the set 
$$A \cap \bigcup_{n=1}^{\infty} B_n = \bigcup_{n=1}^{\infty} (A \cap B_n)$$
is in $m$, hence $\displaystyle \bigcup_{n=1}^{\infty} B_n \in \mathcal{B}$. It's similar to prove that $\mathcal{B}$ is closed under descending sequences, using
$$A \cap \bigcap_{n=1}^{\infty} B_n = \bigcap_{n=1}^{\infty} (A \cap B_n).$$
Hence $\mathcal{B}$ is a monotone class containing $E$, therefore $m \subseteq \mathcal{B}$ which implies our claim. $\ \square$
Now that we know that $m$ is closed under finite intersections, it follows immediately that $Y$ is closed under finite unions, which is what you were missing.
