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Let $\{ E_n \}_{n=1}^{\infty} $be a collection of countable sets and let

$$ F_k = E_k \setminus ( \bigcup_{j=1}^{k-1} E_j ) $$

Then $F_k$ are pairwise disjoint and $\bigcup^{\infty} F_k = \bigcup^{\infty} E_k $

Attempt

Write

$$ F_k = E_k \cap ( \bigcup_{j=1}^{k-1} E_j )^c $$

say $l \neq k$ we show $F_k \cap F_l = \varnothing $. WLOG we can assume $l < k $ and write

$$ F_k \cap F_l = E_k \cap ( \bigcup_{j=1}^{k-1} E_j )^c \cap E_l \cap ( \bigcup_{j=1}^{l-1} E_j )^c$$

By Demorgans we get all intersections:

$$ F_k \cap F_l = E_k \cap ( E_1^c\cap ... \cap E_{k-1}^c) \cap E_l \cap ( E_1^c \cap ... \cap E_{l-1}^c)$$

Since $l < k$, then $E_k^c $ must be somewhere in $( E_1^c \cap ... \cap E_{l-1}^c)$ and since $E_k \cap E_k^c = \varnothing $ we have

$$ F_k \cap F_l = \varnothing $$ as desired.

As for the other part of the problem, If $x \in \bigcup^{\infty} F_k$, then $x \in F_{k_0}$ for some $k_0$. But $F_{k_0} \subset E_{k_0} \subset \bigcup^{\infty} E_k $. So,

$$ \bigcup F_k \subset \bigcup E_k $$

Next, notice

$$ F_k = E_k \cap ( \bigcup_{j=1}^{k-1} E_j )^c \implies E_k = F_k \cap ( \bigcup_{j=1}^{k-1} E_k )$$

and so by the same argument as above we get the equality of the unions.

Is this a correct approach? or is it unnecessary long? I know it is a trivial result, but I would like to get feedback. Thanks

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  • $\begingroup$ Related. $\endgroup$ – Git Gud Nov 25 '14 at 8:16
  • $\begingroup$ I think the first part would be shorter with a proof by contradiction. $\endgroup$ – H_B Nov 25 '14 at 8:23
  • $\begingroup$ Suppose there exists $x\in F_k, F_l$ wlog $k<l$ then $x\in F_k\subseteq ( \bigcup_{j=1}^{l-1} E_j )$ contradicting $x\in F_l$ $\endgroup$ – H_B Nov 25 '14 at 8:26
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Your answer to the "disjoint" part is fine. You could perhaps make it just slightly quicker:

Assume $l<k$ and $x\in F_l$. Then $x\in E_l\subseteq\bigcup_{j=1}^{k-1}E_j$, so $x\notin F_k$. Hence $F_l\cap F_k=\emptyset$.

To prove the unions are equal, you would do exactly as you have done to show $\bigcup F_j\subseteq \bigcup E_j$, but unfortunately it does not work the same way for the reverse inclusion (indeed if your argument were correct, we would have $E_k=F_k$ for all $k$). To get you started:

Let $x\in\bigcup_{k=1}^\infty E_k$. Then $\{k\in\mathbb{N}\,|\,x\in E_k\}$ is a non-empty subset of $\mathbb{N}$, so it has a minimal element, say $m_0$. See if you can finish it off from here.

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