Finite and infinite unions

Question

Let $\mathcal{F}\subseteq\mathcal{P}(X)$ an algebra. Then for all sequence $\{E_k\}\subseteq\mathcal{F}$ exists a disjoint sequence $\{F_k\}\subseteq\mathcal{F}$ such that

1) $F_k\subseteq E_k$ for all $k\in \mathbb{N}$

2) $\bigcup_{k=1}^{+\infty} E_k=\bigcup_{k=1}^{+\infty} F_k$.

It is sufficient to consider the sequence \begin{align} F_1& =E_1 \\ F_n & =E_n\setminus \bigcup_{k=1}^{n-1} E_k. \end{align} Now, we observe that \begin{equation} \bigcup_{k=1}^{n} E_k=\bigcup_{k=1}^{n} F_k. \end{equation} At this point how can I conclude 2) ?

Thanks!

Answer 1

I don't think the equality of the finite unions is actually a useful step along the way.

Instead, prove $\bigcup E_k\subseteq\bigcup F_k$ and $\bigcup F_k\subseteq\bigcup E_k$ separately by considering an arbitrary element of the left side and showing it must be on the right side too.

There's no need to make it an indirect proof.

Answer 2

Hint: Suppose not. Let $n$ be minimal such that $$ \bigcup_{k =1}^{n} E_k \neq \bigcup_{k=1}^{n} F_k. $$ Since $F_k \subseteq E_k$ for all $k$ there is thus some $x \in E_n \setminus F_n$ (by the minimality of $n$). But $F_n = \ldots$ which leads to the desired contradiction.