Set Intersection Preserves Subsets/Families of Sets
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Theorem
Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$.
Let:
- $\forall \beta \in I: A_\beta \subseteq B_\beta$
Then:
- $\ds \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha$
Corollary 1
Let $I$ be an indexing set.
Let $\family {B_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of a set $S$.
Let $A$ be a set such that $A \subseteq B_\alpha$ for all $\alpha \in I$.
Then:
- $\ds A \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha$
Corollary 2
Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$.
Let:
- $\forall \beta \in I: A_\beta \subseteq B_\beta$
Then:
- $\ds \bigcap_{\alpha \mathop \in I} B_\alpha = \O \implies \bigcap_{\alpha \mathop \in I} A_\alpha = \O$
Proof
\(\ds x\) | \(\in\) | \(\ds \bigcap_{\alpha \mathop \in I} A_\alpha\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \alpha \in I: \, \) | \(\ds x\) | \(\in\) | \(\ds A_\alpha\) | Definition of Intersection of Family | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \alpha \in I: \, \) | \(\ds x\) | \(\in\) | \(\ds B_\alpha\) | Definition of Subset | |||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \bigcap_{\alpha \mathop \in I} B_\alpha\) | Definition of Intersection of Family |
By definition of subset:
- $\ds \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha$
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $1 \ \text{(e)}$