Intersection is Associative
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Theorem
Set intersection is associative:
- $A \cap \paren {B \cap C} = \paren {A \cap B} \cap C$
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 $\ds I = \bigcap_{\lambda \mathop \in \Lambda} I_\lambda$.
Then:
- $\ds \bigcap_{i \mathop \in I} S_i = \bigcap_{\lambda \mathop \in \Lambda} \paren {\bigcap_{i \mathop \in I_\lambda} S_i}$
Proof
\(\ds \) | \(\) | \(\ds x \in A \cap \paren {B \cap C}\) | ||||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds x \in A \land \paren {x \in B \land x \in C}\) | Definition of Set Intersection | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds \paren {x \in A \land x \in B} \land x \in C\) | Rule of Association: Conjunction | |||||||||||
\(\ds \) | \(\leadstoandfrom\) | \(\ds x \in \paren {A \cap B} \cap C\) | Definition of Set Intersection |
Therefore:
- $x \in A \cap \paren {B \cap C}$ if and only if $x \in \paren {A \cap B} \cap C$
Thus it has been shown that:
- $A \cap \paren {B \cap C}\ = \paren {A \cap B} \cap C$
$\blacksquare$
Also see
Sources
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