Union is Associative

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Theorem

Set union is associative:

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


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


Then:

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


Proof

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

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

$\blacksquare$


Also see


Sources

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