Intersection Distributes over Union/Family of Sets
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Theorem
Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ be a indexed family of subsets of a set $S$.
Let $B \subseteq S$.
Then:
- $\ds \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cap B} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B$
where $\ds \bigcup_{\alpha \mathop \in I} A_\alpha$ denotes the union of $\family {A_\alpha}_{\alpha \mathop \in I}$.
Corollary
Let $I$ and $J$ be indexing sets.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\beta}_{\beta \mathop \in J}$ be indexed families of subsets of a set $S$.
Then:
- $\ds \bigcup_{\tuple {\alpha, \beta} \mathop \in I \times J} \paren {A_\alpha \cap B_\beta} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap \paren {\bigcup_{\beta \mathop \in J} B_\beta}$
where $\ds \bigcup_{\alpha \mathop \in I} A_\alpha$ denotes the union of $\family {A_\alpha}_{\alpha \mathop \in I}$.
Proof
| \(\ds x\) | \(\in\) | \(\ds \bigcup_{\alpha \mathop \in I} \paren {A_\alpha \cap B}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists \alpha \in I: \, \) | \(\ds x\) | \(\in\) | \(\ds A_\alpha \cap B\) | Definition of Union of Family | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds A_\alpha\) | Definition of Set Intersection | ||||||||||
| \(\, \ds \land \, \) | \(\ds x\) | \(\in\) | \(\ds B\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \paren {\bigcup_{\alpha \mathop \in I} A_\alpha}\) | Set is Subset of Union of Family | ||||||||||
| \(\, \ds \land \, \) | \(\ds x\) | \(\in\) | \(\ds B\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B\) | Definition of Set Intersection |
By definition of subset:
- $\ds \bigcup_{\alpha \mathop \in I} \paren {A_\alpha \cap B} \subseteq \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B$
$\Box$
| \(\ds x\) | \(\in\) | \(\ds \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \paren {\bigcup_{\alpha \mathop \in I} A_\alpha}\) | Definition of Set Intersection | ||||||||||
| \(\, \ds \land \, \) | \(\ds x\) | \(\in\) | \(\ds B\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists \alpha \in I: \, \) | \(\ds x\) | \(\in\) | \(\ds A_\alpha\) | Definition of Union of Family | |||||||||
| \(\, \ds \land \, \) | \(\ds x\) | \(\in\) | \(\ds B\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \exists \alpha \in I: \, \) | \(\ds x\) | \(\in\) | \(\ds A_\alpha \cap B\) | Definition of Set Intersection | |||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \bigcup_{\alpha \mathop \in I} \paren {A_\alpha \cap B}\) | Set is Subset of Union of Family |
By definition of subset:
- $\ds \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B \subseteq \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cap B}$
$\Box$
By definition of set equality:
- $\ds\map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cap B} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B$
$\blacksquare$
Also see
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 9$: Families
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $1 \ \text{(f)}$
- 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 $1.4.4 \ \text{(v)}$