# Union is Associative

## Theorem

$A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$

### Family of Sets

Let $\family {S_i}_{i \mathop \in I}$ and $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$ be indexed families of sets.

Let $\displaystyle I = \bigcup_{\lambda \mathop \in \Lambda} I_\lambda$ denote the union of $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$.

Then:

$\displaystyle \bigcup_{i \mathop \in I} S_i = \bigcup_{\lambda \mathop \in \Lambda} \paren {\bigcup_{i \mathop \in I_\lambda} S_i}$

## Proof

 $\displaystyle$  $\displaystyle x \in A \cup \paren {B \cup C}$ Definition of Set Union $\displaystyle$ $\leadstoandfrom$ $\displaystyle x \in A \lor \paren {x \in B \lor x \in C}$ Definition of Set Union $\displaystyle$ $\leadstoandfrom$ $\displaystyle \paren {x \in A \lor x \in B} \lor x \in C$ Rule of Association: Disjunction $\displaystyle$ $\leadstoandfrom$ $\displaystyle x \in \paren {A \cup B} \cup C$ Definition of Set Union

Therefore:

$x \in A \cup \paren {B \cup C}$ if and only if $x \in \paren {A \cup B} \cup C$

Thus it has been shown that:

$A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$

$\blacksquare$