Union of Indexed Family of Sets Equal to Union of Disjoint Sets

Theorem
Let $\left\{ {E_n}\right\}_{n \mathop \in \N}$ be a countable indexed family of sets where at least two $E_n$ are distinct.

Then there exists a countable indexed family of disjoint sets $\left\{ {F_n}\right\}_{n \mathop \in \N}$ defined by:


 * $\displaystyle F_k = E_k \setminus \left({ \bigcup_{j \mathop = 0}^{k \mathop - 1} E_j }\right)$

satisfying:


 * $\displaystyle \bigsqcup_{n \mathop \in \N} F_n = \bigcup_{n \mathop \in \N} E_n$

where $\bigsqcup$ denotes disjoint union.

Corollary
The countable family $\left\{ { F_k }\right \}_{k \mathop \in \N}$ can equivalently be constructed by:


 * $\displaystyle F_k = \bigcap_{j \mathop = 0}^{k \mathop - 1} \left({E_k \setminus E_j}\right)$

Proof
Denote:


 * $\displaystyle E = \bigcup_{k \mathop \in \N} E_k$


 * $\displaystyle F = \bigcup_{k \mathop \in \N} F_k$

where:


 * $\displaystyle F_k = E_k \setminus \left({\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j }\right)$

We first show that $E = F$.

That $x \in E \implies x \in F$ follows from the construction of $F$ from subsets of $E$.

Thus $E \subseteq F$.

For the other implication:

so $F \subseteq E$.

Thus $E = F$ from the definition of set equality.

To show that the sets in $F$ are disjoint, consider an arbitrary $x \in F$.

Then $x \in F_k$ for some $F_k$.

By the well-ordering principle, there is a smallest such $k$.

Then for any $j < k$, $x \notin F_j$

Choose any distinct $\ell, m \in \N$.

If $m > \ell$, then:


 * $x \in F_\ell \implies x \in E_\ell$


 * $x \in F_m \implies x \notin E_m$

If $m < \ell$, then:


 * $x \in F_m \implies x \in E_m$


 * $x \in F_\ell \implies x \notin E_\ell$

So the sets $F_\ell, F_m$ are disjoint.

Thus $F$ is the disjoint union of sets equal to $E$:


 * $\displaystyle \bigcup_{k \mathop \in \N} E_k = \bigsqcup_{k \mathop \in \N} F_k$

Proof of Corollary
This is a direct application of De Morgan:


 * $\displaystyle E_k \setminus \left({ \bigcup_{j \mathop = 0}^{k \mathop - 1} E_j }\right) = \bigcap_{j \mathop = 0}^{k \mathop - 1} \left({E_k \setminus E_j}\right)$