# 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.

### Corollary

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

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

### General Result

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:

 $\ds E$ $=$ $\ds \bigcup_{k \mathop \in \N} E_k$ $\ds F$ $=$ $\ds \bigcup_{k \mathop \in \N} F_k$

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:

 $\ds x$ $\in$ $\ds \bigcup_{k \mathop \in \N} F_k$ $\ds \leadsto \ \$ $\ds \exists k \in \N: \,$ $\ds x$ $\in$ $\ds F_k$ $\ds \leadsto \ \$ $\ds \exists k \in \N: \,$ $\ds x$ $\in$ $\ds E_k$ $\, \ds \land \,$ $\ds x$ $\notin$ $\ds \paren {E_0 \cup E_1 \cup E_2 \cup \cdots \cup E_{k - 1} }$ $\ds \leadsto \ \$ $\ds \exists k \in \N: \,$ $\ds x$ $\in$ $\ds E_k$ 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 $x \in F_k$ 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:

 $\ds x \in F_\ell$ $\implies$ $\ds x \in E_\ell$ $\ds x \in F_m$ $\implies$ $\ds x \notin E_m$

If $m < \ell$, then:

 $\ds x \in F_m$ $\implies$ $\ds x \in E_m$ $\ds x \in F_\ell$ $\implies$ $\ds x \notin E_\ell$

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

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

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

$\blacksquare$