Union is Commutative/Family of Sets

Theorem
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
We have that both $\ds \bigcup_{j \mathop \in J} S_j$ and $\ds \bigcup_{k \mathop \in \relcomp I J} S_k$ 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 $\ds \bigcup_{i \mathop \in I} S_i = \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k$.

So:

That is:
 * $\ds x \in \bigcup_{i \mathop \in I} S_i \iff x \in \bigcup_{j \mathop \in J} S_j \cup \bigcup_{k \mathop \in \relcomp I J} S_k$

The result follows by definition of set equality.

Also see

 * General Commutativity of Set Intersection