Problem: Let $(X, \mathscr{A})$ be a measurable space and $(\mathscr{A}_n)_{n\in\mathbb{N}}$ be a strictly increasing sequence of $\sigma$-algebras. Show that
$$ \mathscr{A}_{\infty} := \bigcup_{n\in\mathbb{N}} \mathscr{A}_n $$
is never a $\sigma$-algebra.
This is a problem of a book of Rene Schilling. He shows an answer of this problem on his homepage. However I don't understand the last part, Step 5. I don't know there always exists the smallest set $B_n$.
Here is his solutions. This problem is Problem 3.8. [pp.24-26]
The problem reduces to the following.
Indeed, define $$ I:=\left\{i\in\mathbb N\mid \exists A_i\in\mathcal A,k\in A_i,i\notin A_i\right\}. $$ For each $i\in I$, choose $A_i\in\mathcal A$ such that $k\in A_i$ and $i\notin A_i$. Define $$ E_k:= \bigcap_{i\in I}A_i. $$ Then $E_k\in\mathcal A$ and $k\in E_k$. Let $B\in\mathcal A$ be such that $k\in B$. We have to check that $E_k\subset B$. Observe that $i\notin A_i$ hence $$ E_k= \bigcap_{i\in I}A_i\setminus \{i\}=\bigcap_{i\in I}A_i\cap \left(\mathbb N\setminus \{i\}\right)=E_k\setminus I. $$ Let $j\in E_k=E_k\setminus I$. We know that for all $S\in\mathcal A$, we either have $k\notin S$ or $j\in S$. Apply this to $B$ to get that $k\notin B$ or $j\in B$. Since $k\in B$, the only possibility is that $j\in B$, which proves that $E_k\subset B$.