Union is Commutative

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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

\(\ds x\) \(\in\) \(\ds \paren {S \cup T}\)
\(\ds \leadstoandfrom \ \ \) \(\ds x \in S\) \(\lor\) \(\ds x \in T\) Definition of Set Union
\(\ds \leadstoandfrom \ \ \) \(\ds x \in T\) \(\lor\) \(\ds x \in S\) Disjunction is Commutative
\(\ds \leadstoandfrom \ \ \) \(\ds x\) \(\in\) \(\ds \paren {T \cup S}\) Definition of Set Union

$\blacksquare$


Also see


Sources

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