Union is Commutative/Family of Sets

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Theorem

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

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

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


Then:

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

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


Proof

We have that both $\displaystyle \bigcup_{j \mathop \in J} S_j$ and $\displaystyle \bigcup_{k \mathop \in \relcomp I J} S_k$ are sets.

Hence by Union is Commutative we have:

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


It remains to be demonstrated that $\displaystyle \bigcup_{i \mathop \in I} S_i = \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k$.

So:

\(\displaystyle x\) \(\in\) \(\displaystyle \bigcup_{i \mathop \in I} S_i\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \exists i \in I: x\) \(\in\) \(\displaystyle S_i\) Definition of Union of Family
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \exists j \in J: x\) \(\in\) \(\displaystyle S_j\) Definition of Relative Complement
\(\, \displaystyle \lor \, \) \(\displaystyle \exists k \in \relcomp I J: x\) \(\in\) \(\displaystyle S_k\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x\) \(\in\) \(\displaystyle \bigcup_{j \mathop \in J} S_j\) Definition of Union of Family
\(\, \displaystyle \lor \, \) \(\displaystyle x\) \(\in\) \(\displaystyle \bigcup_{k \mathop \in \relcomp I J} S_k\) Definition of Union of Family
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x\) \(\in\) \(\displaystyle \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k\) Definition of Set Union

That is:

$\displaystyle x \in \bigcup_{i \mathop \in I} S_i \iff x \in \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k$

The result follows by definition of set equality.

$\blacksquare$


Also see


Sources