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$\{ E_j \}_{j=1}^{\infty}$ : sequence of sets

We define $\{ F_j \}_{j=1}^{\infty}$ with \begin{align} &F_1=E_1 \\ &F_j=E_j-(E_1 \cup \cdots \cup E_{j-1}) \ (j\geqq 2) \end{align}

Then, prove that

$ F_j \cap F_k= \varnothing$ for $j\neq k$.

$\bigcup_{j=1}^{\infty} E_j =\bigcup_{j=1}^{\infty} F_j $.

I could prove $ F_j \cap F_k=\varnothing$ for $j\neq k$ but I cannot prove $\bigcup_{j=1}^{\infty} E_j =\bigcup_{j=1}^{\infty} F_j $.

Because $F_j=E_j-(E_1 \cup \cdots \cup E_{j-1})\subset E_j$ for all $j$, $\bigcup_{j=1}^{\infty} F_j \subset\bigcup_{j=1}^{\infty} E_j $ holds.

I have no idea to prove $\bigcup_{j=1}^{\infty} F_j \supset\bigcup_{j=1}^{\infty} E_j .$

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Consider some $x\in \bigcup \limits_{j=1}^\infty E_j$. By definition $x\in E_j$ for some $j$ and we would like to say that $x\in F_j$. But what if eg. $x\in E_{j-1}$ as well? We need to make sure that the definition of $F_j$ doesn’t kill our element $x$. We can do this using the wellordering principle of the natural numbers: There is a minimal $j_0$ st $x\in E_{j_0}$ and $x\notin E_k$ for $k<j_0$. But then $x \in F_{j_0}$ and thus $x \in \bigcup_{j=0}^\infty F_j$.

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