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 $E$:
- $\ds \bigcup_{\alpha \mathop \in I} E_\alpha = \bigsqcup_{\alpha \mathop \in I} F_\alpha$
$\blacksquare$
Sources
- 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $13$