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 $\left \langle{S_i}\right \rangle_{i \in I}$ and $\left \langle{I_\lambda}\right \rangle_{\lambda \in \Lambda}$ be indexed families of sets.

Let $\displaystyle I = \bigcup_{\lambda \mathop \in \Lambda} I_\lambda$.


Then:

$\displaystyle \bigcap_{i \mathop \in I} S_i = \bigcap_{\lambda \mathop \in \Lambda} \left({\bigcap_{i \mathop \in I_\lambda} S_i}\right)$


Proof

\(\displaystyle \) \(\) \(\displaystyle x \in A \cap \paren {B \cap C}\) $\quad$ $\quad$
\(\displaystyle \) \(\leadstoandfrom\) \(\displaystyle x \in A \land \paren {x \in B \land x \in C}\) $\quad$ Definition of Set Intersection $\quad$
\(\displaystyle \) \(\leadstoandfrom\) \(\displaystyle \paren {x \in A \land x \in B} \land x \in C\) $\quad$ Rule of Association: Conjunction $\quad$
\(\displaystyle \) \(\leadstoandfrom\) \(\displaystyle x \in \paren {A \cap B} \cap C\) $\quad$ Definition of Set Intersection $\quad$


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