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

Theorem
Let $\family {E_n}_{n \mathop \in \N}$ 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 $\family {F_n}_{n \mathop \in \N}$ defined by:


 * $\ds F_k = E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j}$

satisfying:


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

where $\bigsqcup$ denotes disjoint union.

Proof
Denote:

where:


 * $\ds F_k = E_k \setminus \paren {\bigcup_{j \mathop = 0}^{k \mathop - 1} E_j}$

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 $E \subseteq F$.

Then:

so $F \subseteq E$.

Thus $E = F$ by definition of set equality.

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

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

By the Well-Ordering Principle, there exists a smallest such $k$.

Then:
 * $\forall j < k: x \notin F_j$

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

We have:

If $m > \ell$, then:

If $m < \ell$, then:

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

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


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