Intersection is Commutative

From ProofWiki
Jump to navigation Jump to search

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 $\ds I = \bigcap_{i \mathop \in I} S_i$ 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

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Intersection_is_Commutative&oldid=569200"