Show that there exists the smallest algebra containing $\mathcal{C}$

Question

Suppose that $\mathcal{A}$ is a collection of subsets of $X$ and $(A_i)$ a sequence of sets in $\mathcal{A}$. Show that there exists a sequence disjoint sets $(B_n)_{n \in \mathbb{N}}$ in $\mathcal{A}$ such that $$\bigcup_{n=1}^{\infty}{B_n} = \bigcup_{n=1}^{\infty}{A_n}$$

I manage to show the existence part, i.e. for all $n \geq 2$, $B_n=A_n\setminus \bigcup_{1=i}^{n-1}{A_n}$. I don't know how to show the equality $$\bigcup_{n=1}^{\infty}{B_n} = \bigcup_{n=1}^{\infty}{A_n}.$$

By using the definition of $B_n$, I can show that $\bigcup_{n=1}^{\infty}{B_n} \subset \bigcup_{n=1}^{\infty}{A_n}$, as for each $n$, $B_n \subset A_n$. But I have no avail on the other inclusion.

Answer 1

What you want to do is define $B_n$ as $A_n\setminus \bigcup\limits_{i=1}^{n-1} A_i$.

If $x$ appears in at least one $A_n$ you can let $A_k$ be the first one, and you have $x\in B_k$. On the other hand it is clear that $B_k\subseteq A_k$, so $\bigcup \limits_{n=1}^\infty B_n=\bigcup\limits_{n=1}^\infty A_n$.

To prove that $B_k$ and $B_j$ with $k