Intersection is Commutative

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

$\blacksquare$

Also see

• Union is Commutative
• Set Difference is Anticommutative
• Symmetric Difference is Commutative

Sources

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• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): intersection: $\text {(ii)}$
• 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): algebra of sets: $\text {(ii)}$