Intersection is Largest 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 $X$:
- $\ds \paren {\forall i \in I: X \subseteq S_i} \iff X \subseteq \bigcap_{i \mathop \in I} S_i$
where $\ds \bigcap_{i \mathop \in I} S_i$ is the intersection of $\family {S_i}$.
Proof
Let $X \subseteq S_i$ for all $i \in I$.
Then from Set is Subset of Intersection of Supersets: General Result:
- $\ds X \subseteq \bigcap_{i \mathop \in I} S_i$
$\Box$
Now suppose that $\ds X \subseteq \bigcap_{i \mathop \in I} S_i$.
From Intersection is Subset: Family of Sets we have:
- $\ds \forall i \in I: \bigcap_{j \mathop \in I} S_j \subseteq S_i$
So from Subset Relation is Transitive, it follows that:
- $\ds \forall i \in I: X \subseteq \bigcap_{j \mathop \in I} S_j \subseteq S_i$
So it follows that $\forall i \in I: X \subseteq S_i$.
So:
- $\ds X \subseteq \bigcap_{i \mathop \in I} S_i \implies \paren {\forall i \in I: X \subseteq S_i}$
$\Box$
Hence:
- $\ds \paren {\forall i \in I: X \subseteq S_i} \iff X \subseteq \bigcap_{i \mathop \in I} S_i$
$\blacksquare$
Also see
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.1 \ (1)$
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $4 \ \text{(a)}$