# Intersection Distributes over Union

## Theorem

$R \cap \paren {S \cup T} = \paren {R \cap S} \cup \paren {R \cap T}$

### Family of Sets

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:

$\displaystyle \map {\bigcup_{\alpha \mathop \in I} } {A_\alpha \cap B} = \paren {\bigcup_{\alpha \mathop \in I} A_\alpha} \cap B$

where $\displaystyle \bigcup_{\alpha \mathop \in I} A_\alpha$ denotes the union of $\family {A_\alpha}_{\alpha \mathop \in I}$.

### General Result

Let $S$ and $T$ be sets.

Let $\powerset T$ be the power set of $T$.

Let $\mathbb T$ be a subset of $\powerset T$.

Then:

$\displaystyle S \cap \bigcup \mathbb T = \bigcup_{X \mathop \in \mathbb T} \paren {S \cap X}$

## Proof

 $\displaystyle$  $\displaystyle x \in R \cap \paren {S \cup T}$ $\displaystyle$ $\leadstoandfrom$ $\displaystyle x \in R \land \paren {x \in S \lor x \in T}$ Definition of Set Union and Definition of Set Intersection $\displaystyle$ $\leadstoandfrom$ $\displaystyle \paren {x \in R \land x \in S} \lor \paren {x \in R \land x \in T}$ Conjunction is Left Distributive over Disjunction $\displaystyle$ $\leadstoandfrom$ $\displaystyle x \in \paren {R \cap S} \cup \paren {R \cap T}$ Definition of Set Union and Definition of Set Intersection

$\blacksquare$

## Demonstration by Venn Diagram  In the left hand diagram, $R$ is depicted in yellow and $S \cup T$ is depicted in blue.

Their intersection $R \cap \paren {S \cup T}$ where they overlap is depicted in green.

In the right hand diagram, $\paren {R \cap S}$ is depicted in yellow and $\paren {R \cap T}$ is depicted in blue.

Their intersection, where they overlap, is depicted in green.

Their union is the total shaded area: yellow, blue and green.

As can be seen by inspection, the areas are the same.

## Also known as

This result and Union Distributes over Intersection are together known as the Distributive Laws.

## Examples

### $3$ Arbitrarily Chosen Sets

Let:

 $\displaystyle A$ $=$ $\displaystyle \set {3, -i, 4, 2 + i, 5}$ $\displaystyle B$ $=$ $\displaystyle \set {-i, 0, -1, 2 + i}$ $\displaystyle C$ $=$ $\displaystyle \set {- \sqrt 2 i, \dfrac 1 2, 3}$

#### Intersection with Union

 $\displaystyle A \cap \paren {B \cup C}$ $=$ $\displaystyle \set {3, -i, 4, 2 + i, 5} \cap \set {-i, 0, -\sqrt 2 i, -1, 2 + i, \dfrac 1 2, 3}$ $\displaystyle$ $=$ $\displaystyle \set {3, -i, 2 + i}$

#### Union of Intersections

 $\displaystyle \paren {A \cap B} \cup \paren {A \cap C}$ $=$ $\displaystyle \set {-i, 2 + i} \cup \set 3$ $\displaystyle$ $=$ $\displaystyle \set {3, -i, 2 + i}$

Thus it is seen that:

$A \cap \paren {B \cup C} = \paren {A \cap B} \cup \paren {A \cap C}$

### Arbitrary Integer Sets

Let:

 $\displaystyle A$ $=$ $\displaystyle \set {2, 4, 6, 8, \dotsc}$ $\displaystyle B$ $=$ $\displaystyle \set {1, 3, 5, 7, \dotsc}$ $\displaystyle C$ $=$ $\displaystyle \set {1, 2, 3, 4}$

Then:

$\paren {A \cup B} \cap C = \set {1, 2, 3, 4} = \paren {A \cap C} \cup \paren {B \cap C}$