# Intersection is Commutative

## Theorem

$S \cap T = T \cap S$

### Family of Sets

Let $\family {S_i}_{i \mathop \in I}$ be an indexed family of sets.

Let $\ds I = \bigcap_{i \mathop \in I} S_i$ denote the intersection of $\family {S_i}_{i \mathop \in I}$.

Let $J \subseteq I$ be a subset of $I$.

Then:

$\ds \bigcap_{i \mathop \in I} S_i = \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k = \bigcap_{k \mathop \in \relcomp I J} S_k \cap \bigcap_{j \mathop \in J} S_j$

where $\relcomp I J$ denotes the complement of $J$ relative to $I$.

## Proof

 $\ds x$ $\in$ $\ds \paren {S \cap T}$ $\ds \leadstoandfrom \ \$ $\ds x \in S$ $\land$ $\ds x \in T$ Definition of Set Intersection $\ds \leadstoandfrom \ \$ $\ds x \in T$ $\land$ $\ds x \in S$ Conjunction is Commutative $\ds \leadstoandfrom \ \$ $\ds x$ $\in$ $\ds \paren {T \cap S}$ Definition of Set Intersection

$\blacksquare$