Set Union is Self-Distributive/General Result
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Theorem
Let $\family {\mathbb S_i}_{i \mathop \in I}$ be an $I$-indexed family of sets of sets.
Then:
- $\ds \bigcup_{i \mathop \in I} \bigcup \mathbb S_i = \bigcup \bigcup_{i \mathop \in I} \mathbb S_i$
Proof
By the definition of set union, we have:
- $\ds \forall i \in I: \forall S \in \mathbb S_i: S \in \bigcup_{j \mathop \in I} \mathbb S_j$
Hence, by Set is Subset of Union: General Result:
- $\ds \forall i \in I: \forall S \in \mathbb S_i: S \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$
By Union is Smallest Superset: General Result, it follows that:
- $\ds \forall i \in I: \bigcup \mathbb S_i \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$
Therefore, by Union is Smallest Superset: Family of Sets, we conclude that:
- $\ds \bigcup_{i \mathop \in I} \bigcup \mathbb S_i \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$
By Set is Subset of Union General Result and then Set is Subset of Union of Family, we have:
- $\ds \forall i \in I: \forall S \in \mathbb S_i: S \subseteq \bigcup \mathbb S_i \subseteq \bigcup_{j \mathop \in I} \bigcup \mathbb S_j$
Because $\subseteq$ is transitive, we can apply the definition set union to conclude that:
- $\ds \forall S \in \bigcup_{i \mathop \in I} \mathbb S_i: S \subseteq \bigcup_{j \mathop \in I} \bigcup \mathbb S_j$
Hence, by Union is Smallest Superset: General Result:
- $\ds \bigcup \bigcup_{i \mathop \in I} \mathbb S_i \subseteq \bigcup_{j \mathop \in I} \bigcup \mathbb S_j$
By definition of set equality, the result follows.
$\blacksquare$