# 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:

 $\ds E$ $=$ $\ds \bigcup_{\beta \mathop \in I} E_\beta$ $\ds F$ $=$ $\ds \bigcup_{\beta \mathop \in I} F_\beta$

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:

 $\ds x$ $\in$ $\ds \bigcup_{\beta \mathop \in I} F_\beta$ $\ds \leadsto \ \$ $\ds \exists \beta \in I: \,$ $\ds x$ $\in$ $\ds F_\beta$ $\ds \leadsto \ \$ $\ds \exists \beta \in I: \,$ $\ds x$ $\in$ $\ds E_\beta$ $\, \ds \land \,$ $\ds x$ $\notin$ $\ds \paren {\bigcup_{\gamma \mathop \prec \beta} E_\gamma}$ $\ds \leadsto \ \$ $\ds \exists \beta \in I: \,$ $\ds x$ $\in$ $\ds E_\beta$ Rule of Simplification

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:

 $\ds x \in F_\eta$ $\implies$ $\ds x \in E_\eta$ $\ds x \in F_\zeta$ $\implies$ $\ds x \notin E_\zeta$

If $\zeta < \eta$, then:

 $\ds x \in F_\zeta$ $\implies$ $\ds x \in E_\zeta$ $\ds x \in F_\eta$ $\implies$ $\ds x \notin E_\eta$

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

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

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

$\blacksquare$