Proof of $ \bigcup\limits_{i=1}^n F_i = \bigcup\limits_{i=1}^n E_i $ Prove the following relations: For any sequence of events $E_1,E_2,\dots$, define a new sequence $F_1,F_2,\dots,$ of disjoint events (that is, events such that $F_i\cap F_j=\emptyset$ whenever $i\neq j$) such that for all $n\geq 1$,
$$ \bigcup\limits_{i=1}^n F_i = \bigcup\limits_{i=1}^n E_i $$
Proof: Define $F_i=E_i \backslash \left(\bigcup\limits_{j=1}^{i-1} E_j \right)$ for all $i$. Then
\begin{equation*}
\begin{aligned}
x\in \bigcup\limits_{i=1}^n F_i 
& \iff x\in F_i  \text{ for all } i=1,\dots,n \\
& \iff x\in E_i \backslash \left(\bigcup\limits_{j=1}^{i-1} E_j \right) \text{ for all } i=1,\dots,n \\
& \iff x\in E_i 
\text{ and } x\not\in \bigcup\limits_{j=1}^{i-1} E_j \text{ for all } i=1,\dots,n \\
& \iff x\in E_i 
\text{ and } x\not\in E_j \text{ for all } i=1,\dots,n \text{ and } j=1,\dots, i-1\\
\end{aligned}
\end{equation*}
Note that last statement implies disjointness. So an element of $E_1$ only appears in $E_1$ .
Can I just state that the last statement is $ \iff x\in \bigcup\limits_{i=1}^n E_i$? Or do I have to mention something more?
 A: Let $E_1,E_2,\dots$ be a sequence of events. Now define a new sequence of events $F_1, F_2,\dots$ of disjoint events in the following manner:
\begin{equation*}
\begin{aligned}
F_1 & = E_1 \\
F_2 & = E_2 - E_1 = E_2 \backslash E_1 \\
F_3 & = E_3 - E_1 - E_2 = E_3 \backslash (E_1 \cup E_2) \\
\vdots \\
F_i & = E_i \left\backslash \left(\bigcup\limits_{j=1}^{i-1} E_j \right)\right. \text{ for all } i
\end{aligned}
\end{equation*}
By construction of $F_i$,
$$ \bigcup\limits_{i=1}^n F_i \subset \bigcup\limits_{i=1}^n E_i $$
What is left to show is the other direction.
Let $x\in\bigcup\limits_{i=1}^n E_i \iff x\in E_j$ for some $j$. By the well ordering principle of the natural numbers, we can say that there exists $i_0$ such that $x\in E_{i_0}$ and $x\not\in E_k$ for $k<i_0$. Hence $x\in F_{i_0}$ and thus $x\in\bigcup_{i=0}^n F_j$.
A: $\space\space\space$ As has been mentioned in the comments, this proof doesn't really work. We do have a relation between the sets $F_i$ and $E_i$ but it is not what you have stated. Using your definition of $F_i$, we have;
$$x\in \bigcup_{i=1}^{n} F_i \iff x\in F_i \text{ for some }i\in\{1,\cdots ,n\} $$
$$\iff x\in E_i \setminus \left(\bigcup_{j=1}^{i-1}E_j\right) \text{ for some }i\in\{1,\cdots ,n\}$$
$$\iff x\in E_i \space \land \space x\ \not\in \bigcup_{j=1}^{i-1}E_j \text{ for some }i\in\{1,\cdots ,n\}$$
$$\iff x\in E_i \space \land \space x\ \not\in E_j \text{ for some }i\in\{1,\cdots ,n\} \text { and for all } j \in {\{1,\cdots ,i-1\}} $$
So, every $x$ in some $F_i$ does, indeed, appear in only one $E_i$. However, not every $x$ in some $E_i$ is covered by the $F_i$ sets. Namely, the union of the $F_i$ sets is missing, precisely, the intersection of the $E_i$ sets. So the actual relation between the union of the $F_i$ sets and the $E_i$ sets would be;
$$\bigcup_{i=1}^{n}F_i=\bigcup_{i=1}^{n}E_i \setminus \left(\bigcap_{i=1}^{n}E_i \right)$$
A: Using the given definition of $F_i = F_i=E_i \backslash\left(\bigcup\limits_{j=1}^{i-1} E_j \right)$ we get :-
$$x\in \bigcup_{i=1}^{n} F_i \iff x\in F_i \text{ for some }i\in\{1,\cdots ,n\}$$
$$\iff x\in E_i \setminus \left(\bigcup_{j=1}^{i-1}E_j\right) \text{ for some }i\in\{1,\cdots ,n\}$$
$$\iff x\in E_i \space \land \space x\ \not\in \bigcup_{j=1}^{i-1}E_j \text{ for some }i\in\{1,\cdots ,n\}$$
Now this as mentioned implies disjointness. Also $F_i\subseteq E_i$ for $i \in\{1,\cdots ,n\}$ and also $F_i\not\in \bigcup_{j=1}^{i-1}E_j$
Now let $$y \in E_i \text{ for some }i\in\{1,\cdots ,n\}$$
$$\iff y\in E_i \setminus \left(\bigcup_{j=1}^{i-1}E_j\right) \text{ for some }i\in\{1,\cdots ,n\} \hspace 1cm \text{ [This is as all }E_i\text{ are disjoint] }$$
$$\iff y\in F_i \text{ for some }i\in\{1,\cdots,n\}$$
Thus $E_i\subseteq F_i$ for $i \in\{1,\cdots ,n\}$
Thus it follows that $E_i = F_i$
So we conclude:-
$$\bigcup\limits_{i=1}^n F_i = \bigcup\limits_{i=1}^n E_i$$
