Intersection is Commutative

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Theorem

Set intersection is commutative:

$S \cap T = T \cap S$


Family of Sets

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

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

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


Then:

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

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


Proof

\(\displaystyle x\) \(\in\) \(\displaystyle \paren {S \cap T}\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x \in S\) \(\land\) \(\displaystyle x \in T\) Definition of Set Intersection
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x \in T\) \(\land\) \(\displaystyle x \in S\) Conjunction is Commutative
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x\) \(\in\) \(\displaystyle \paren {T \cap S}\) Definition of Set Intersection

$\blacksquare$


Also see


Sources