Union is Smallest Superset/General Result
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Theorem
Let $S$ and $T$ be sets.
Let $\powerset S$ denote the power set of $S$.
Let $\mathbb S$ be a subset of $\powerset S$.
Then:
- $\ds \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
Family of Sets
In the context of a family of sets, the result can be presented as follows:
Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Then for all sets $X$:
- $\ds \paren {\forall i \in I: S_i \subseteq X} \iff \bigcup_{i \mathop \in I} S_i \subseteq X$
where $\ds \bigcup_{i \mathop \in I} S_i$ is the union of $\family {S_i}$.
Proof
Let $\mathbb S \subseteq \powerset S$.
By Union of Subsets is Subset: Subset of Power Set:
- $\ds \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$
$\Box$
Now suppose that $\ds \bigcup \mathbb S \subseteq T$.
Consider any $X \in \mathbb S$ and take any $x \in X$.
From Set is Subset of Union: General Result we have that:
- $\ds X \subseteq \bigcup \mathbb S$
Thus:
- $\ds x \in \bigcup \mathbb S$
But:
- $\ds \bigcup \mathbb S \subseteq T$
So it follows that:
- $X \subseteq T$
So:
- $\ds \bigcup \mathbb S \subseteq T \implies \paren {\forall X \in \mathbb S: X \subseteq T}$
$\Box$
Hence:
- $\ds \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
$\blacksquare$
Also see
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.6 \ \text{(f)}$
- 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.): $\S 1.4$: Exercise $1.4.4 \ \text{(i)}$