# Given a sequence of sets $\{E_j\}$, define $F_1=E_1$ and $F_j=E_j-(E_1\cup\dots\cup E_{j-1})$ for $j>1$. Why does $\bigcup E_j=\bigcup F_j$?

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

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 $$k. But then $$x \in F_{j_0}$$ and thus $$x \in \bigcup_{j=0}^\infty F_j$$.