Union of Indexed Family of Sets Equal to Union of Disjoint Sets/General Result

Theorem
Let $I$ be a set which can be well-ordered by a well-ordering $\preccurlyeq$.

Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be a countable indexed family of sets indexed by $I$ where at least two $E_\alpha$ are distinct.

Then there exists a countable indexed family of disjoint sets $\family {F_\alpha}_{\alpha \mathop \in I}$ defined by:


 * $\ds F_\beta = E_\beta \setminus \paren {\bigcup_{\alpha \mathop \prec \beta} E_\alpha}$

satisfying:


 * $\ds \bigsqcup_{\alpha \mathop \in I} F_n = \bigcup_{\alpha \mathop \in I} E_n$

where:
 * $\bigsqcup$ denotes disjoint union.
 * $\alpha \prec \beta$ denotes that $\alpha \preccurlyeq \beta$ and $\alpha \ne \beta$.

Proof
Denote:

where:


 * $\ds F_\beta = E_\beta \setminus \paren {\bigcup_{\alpha \mathop \prec \beta} E_\alpha}$

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 $x \in F_\beta$ for some $F_\beta$.

By the Well-Ordering Principle, there is a smallest such $\beta$ with respect to $\preccurlyeq$.

Then:
 * $\forall \gamma \prec \beta: x \notin F_\gamma$

Choose any distinct $\eta, \zeta \in I$.

We have:

If $\eta \prec \zeta$, then:

If $\zeta < \eta$, then:

So the sets $F_\eta, F_\zeta$ are disjoint.

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


 * $\ds \bigcup_{\alpha \mathop \in I} E_\alpha = \bigsqcup_{\alpha \mathop \in I} F_\alpha$