Union of Subsets is Subset/Family of Sets
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Theorem
Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Then for all sets $X$:
- $\ds \paren {\forall i \in I: S_i \subseteq X} \implies \bigcup_{i \mathop \in I} S_i \subseteq X$
where $\ds \bigcup_{i \mathop \in I} S_i$ is the union of $\family {S_i}$.
Proof
Suppose that $\forall i \in I: S_i \subseteq X$.
Consider any $\ds x \in \bigcup_{i \mathop \in I} S_i$.
By definition of set union:
- $\exists i \in I: x \in S_i$
But as $S_i \subseteq X$ it follows that $x \in X$.
Thus it follows that:
- $\ds \bigcup_{i \mathop \in I} S_i \subseteq X$
So:
- $\ds \paren {\forall i \in I: S_i \subseteq X} \implies \bigcup_{i \mathop \in I} S_i \subseteq X$
$\blacksquare$
Sources
- 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.4$: Sets of Sets: Exercise $1.4.4 \ \text{(i)}$