Set Intersection is Self-Distributive/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 $S$.
Then:
- $\ds \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cap B_\alpha} = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcap_{\alpha \mathop \in I} B_\alpha}$
where $\ds \bigcap_{\alpha \mathop \in I} A_\alpha$ denotes the intersection of $\family {A_\alpha}$.
Proof
\(\ds x\) | \(\in\) | \(\ds \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cap B_\alpha}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \alpha \in I: \, \) | \(\ds x\) | \(\in\) | \(\ds A_\alpha \cap B_\alpha\) | Definition of Intersection of Family | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \alpha \in I: \, \) | \(\ds x\) | \(\in\) | \(\ds A_\alpha\) | Definition of Set Intersection | |||||||||
\(\, \ds \land \, \) | \(\ds x\) | \(\in\) | \(\ds B_\alpha\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \bigcap_{\alpha \mathop \in I} A_\alpha\) | Definition of Intersection of Family | ||||||||||
\(\, \ds \land \, \) | \(\ds x\) | \(\in\) | \(\ds \bigcap_{\alpha \mathop \in I} B_\alpha\) | Definition of Intersection of Family | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcap_{\alpha \mathop \in I} B_\alpha}\) | Definition of Set Intersection |
Thus by definition of subset:
- $\ds \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cap B_\alpha} \subseteq \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcap_{\alpha \mathop \in I} B_\alpha}$
$\Box$
\(\ds x\) | \(\in\) | \(\ds \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcap_{\alpha \mathop \in I} B_\alpha}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \bigcap_{\alpha \mathop \in I} A_\alpha\) | Definition of Set Intersection | ||||||||||
\(\, \ds \land \, \) | \(\ds x\) | \(\in\) | \(\ds \bigcap_{\alpha \mathop \in I} B_\alpha\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \alpha \in I: \, \) | \(\ds x\) | \(\in\) | \(\ds A_\alpha\) | Definition of Intersection of Family | |||||||||
\(\, \ds \land \, \) | \(\ds x\) | \(\in\) | \(\ds B_\alpha\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \alpha \in I: \, \) | \(\ds x\) | \(\in\) | \(\ds A_\alpha \cap B_\alpha\) | Definition of Set Intersection | |||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cap B_\alpha}\) | Definition of Intersection of Family |
Thus by definition of subset:
- $\ds \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcap_{\alpha \mathop \in I} B_\alpha} \subseteq \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cap B_\alpha}$
$\Box$
By definition of set equality:
- $\ds \map {\bigcap_{\alpha \mathop \in I} } {A_\alpha \cap B_\alpha} = \paren {\bigcap_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcap_{\alpha \mathop \in I} B_\alpha}$
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $1 \ \text{(d)}$
- 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 $1.4.4 \ \text{(iv)}$