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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