# Union is Commutative/Family of Sets

## Theorem

Let $\family {S_i}_{i \mathop \in I}$ be an indexed family of sets.

Let $\ds I = \bigcup_{i \mathop \in I} S_i$ denote the union of $\family {S_i}_{i \mathop \in I}$.

Let $J \subseteq I$ be a subset of $I$.

Then:

$\ds \bigcup_{i \mathop \in I} S_i = \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k = \bigcup_{k \mathop \in \relcomp I J} S_k \cup \bigcup_{j \mathop \in J} S_j$

where $\relcomp I J$ denotes the complement of $J$ relative to $I$.

## Proof

We have that both $\ds \bigcup_{j \mathop \in J} S_j$ and $\ds \bigcup_{k \mathop \in \relcomp I J} S_k$ are sets.

Hence by Union is Commutative we have:

$\bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k = \bigcup_{k \mathop \in \relcomp I J} S_k \cup \bigcup_{j \mathop \in J} S_j$

It remains to be demonstrated that $\ds \bigcup_{i \mathop \in I} S_i = \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k$.

So:

 $\ds x$ $\in$ $\ds \bigcup_{i \mathop \in I} S_i$ $\ds \leadstoandfrom \ \$ $\ds \exists i \in I: x$ $\in$ $\ds S_i$ Definition of Union of Family $\ds \leadstoandfrom \ \$ $\ds \exists j \in J: x$ $\in$ $\ds S_j$ Definition of Relative Complement $\, \ds \lor \,$ $\ds \exists k \in \relcomp I J: x$ $\in$ $\ds S_k$ $\ds \leadstoandfrom \ \$ $\ds x$ $\in$ $\ds \bigcup_{j \mathop \in J} S_j$ Definition of Union of Family $\, \ds \lor \,$ $\ds x$ $\in$ $\ds \bigcup_{k \mathop \in \relcomp I J} S_k$ Definition of Union of Family $\ds \leadstoandfrom \ \$ $\ds x$ $\in$ $\ds \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k$ Definition of Set Union

That is:

$\ds x \in \bigcup_{i \mathop \in I} S_i \iff x \in \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k$

The result follows by definition of set equality.

$\blacksquare$