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In many texts, the construction of a countable disjoint union of sets from a sequence of sets, $E_1, E_2,E_3,\ldots$ follows from:

Let $F_1 = E_1, F_2 = E_2\setminus E_1,F_3 = E_3\setminus (E_1\cup E_2),\ldots,F_n=E_n \setminus \bigcup\limits_{k=1}^{n-1} E_k$, etc.

I'm wondering how to show that $\bigcup\limits_{n=1}^{\infty}F_n = \bigcup\limits_{k=1}^\infty E_k$. I can visualize why this is true, but analytically, I find it boggling.

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Every $F_k$ is by construction a subset of $E_k$, for each $k$, so one inclusion is clear. On the other hand, if $x \in \bigcup_{k=1}^{\infty} E_k$, then let $m$ be the smallest index $i$ such that $x \in E_i$ (there is at least one such index, as $x$ is in the union, and non-empty subsets of $\mathbb{N}$ have a minimum).

Then by minimality, $x \notin \bigcup_{k=1}^{m-1} E_k$, but it is in $E_m$, so $x \in F_m$. So $x \in \bigcup_{k=1}^{\infty} F_k$.

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  • $\begingroup$ Thank you. Does the indexing not matter? Is $F_{n}$ a subset of $E_{k}$? $\endgroup$ – guest Oct 17 '14 at 17:31
  • $\begingroup$ No $F_k \subset E_k$ and $F_n \subset E_n$. $\endgroup$ – Henno Brandsma Oct 17 '14 at 17:32
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Note $\bigcup\limits_{k=1}^{n}E_{k}=\bigcup\limits_{k=1}^{n}F_{k} \quad \forall n$.

$x\in \bigcup\limits_{k=1}^{\infty}F_{k}$ if $\exists n, x\in \bigcup\limits_{k=1}^{n}F_{k}=\bigcup\limits_{k=1}^{n}E_{k}\subset \bigcup\limits_{k=1}^{\infty}E_{k}$. So $\bigcup\limits_{n=1}^{\infty}F_{n} \subset \bigcup\limits_{k=1}^{\infty}E_{k}$

Similarly for the other side.

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Clearly, $F_1=E_1$ and $F_n=E_n\setminus E_{n-1}$, for $n>1$, implies that $$ \bigcup_{n\in\mathbb N} F_n\subset\bigcup_{n\in\mathbb N} E_n. \tag{1} $$ To show the opposite direction, let $x\in \bigcup_{n\in\mathbb N} E_n$. Then $x\in E_n$, for some $n\in\mathbb N$. Find all such $n$'s for which $x\in E_n$, and pick the least one $n_0$. This means $$ x\in E_{n_0} $$ but $x\ne E_n$, for all $n<n_0$. This implies that $x\not\in E_{n_0-1}$ and hence $$ x\in E_{n_0}\setminus E_{n_0-1}=F_{n_0}. $$ Thus $x\in F_{n_0}$, and finally $x\bigcup_{n\in\mathbb N} F_n$, and hence $$ \bigcup_{n\in\mathbb N} E_n\subset \bigcup_{n\in\mathbb N} F_n. \tag{2} $$ Now $(1)$ and $(2)$ imply that $$ \bigcup_{n\in\mathbb N} E_n= \bigcup_{n\in\mathbb N} F_n. $$

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