Union is Commutative

Theorem

Set union is commutative:

$S \cup T = T \cup S$

Family of Sets

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

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

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

Then:

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

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

Proof

$\blacksquare$

Also see

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