# Image of Intersection under One-to-Many Relation

## Contents

## Theorem

Let $S$ and $T$ be sets.

Let $\RR \subseteq S \times T$ be a relation.

Then:

- $\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$

if and only if $\RR$ is one-to-many.

### General Result

Let $S$ and $T$ be sets.

Let $\mathcal R \subseteq S \times T$ be a relation.

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

Then:

- $\displaystyle \forall \mathbb S \subseteq \powerset S: \mathcal R \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \mathcal R \sqbrk X$

if and only if $\mathcal R$ is one-to-many.

### Family of Sets

Let $S$ and $T$ be sets.

Let $\mathcal R \subseteq S \times T$ be a relation.

Then $\mathcal R$ is a one-to-many relation if and only if:

- $\displaystyle \mathcal R \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \mathcal R \sqbrk {S_i}$

where $\family {S_i}_{i \mathop \in I}$ is *any* family of subsets of $S$.

## Proof

### Sufficient Condition

Suppose that:

- $\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$

If $S$ is singleton, the result follows immediately as $\RR$ would *have* to be one-to-many.

So, assume $S$ is not singleton.

Suppose $\RR$ is specifically *not* one-to-many.

So:

- $\exists x, y \in S: \exists z \in T: \tuple {x, z} \in T, \tuple {y, z} \in T, x \ne y$

and of course $\set x \subseteq S, \set y \subseteq S$.

So:

- $z \in \RR \sqbrk {\set x}$
- $z \in \RR \sqbrk {\set y}$

and so by definition of intersection:

- $z \in \RR \sqbrk {\set x} \cap \RR \sqbrk {\set y}$

But:

- $\set x \cap \set y = \O$

Thus from Image of Empty Set is Empty Set:

- $\RR \sqbrk {\set x \cap \set y} = \O$

and so:

- $\RR \sqbrk {\set x \cap \set y} \ne \RR \sqbrk {\set x} \cap \RR \sqbrk {\set y}$

$\Box$

### Necessary Condition

Let $\RR$ be one-to-many.

From Image of Intersection under Relation, we already have:

- $\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$

So we just need to show:

- $\RR \sqbrk {S_1} \cap \RR \sqbrk {S_2} \subseteq \RR \sqbrk {S_1 \cap S_2}$

Let $t \in \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$.

Then:

\(\displaystyle t\) | \(\in\) | \(\displaystyle \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle t\) | \(\in\) | \(\displaystyle \RR \sqbrk {S_1} \land t \in \RR \sqbrk {S_2}\) | Definition of Set Intersection | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \exists s_1 \in S_1: \tuple {s_1, t}\) | \(\in\) | \(\displaystyle \RR \land \exists s_2 \in S_2: \tuple {s_2, t} \in \RR\) | Definition of Relation | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle s_1\) | \(=\) | \(\displaystyle s_2\) | Definition of One-to-Many Relation | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle s_1 = s_2\) | \(\in\) | \(\displaystyle S_1 \land s_1 = s_2 \in S_2\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle s_1 = s_2\) | \(\in\) | \(\displaystyle S_1 \cap S_2\) | Definition of Set Intersection | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map \RR {s_1} = \map \RR {s_2}\) | \(\in\) | \(\displaystyle \RR \sqbrk {S_1 \cap S_2}\) | Image of Element is Subset | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}\) | \(\subseteq\) | \(\displaystyle \RR \sqbrk {S_1 \cap S_2}\) | Definition of Subset |

So if $\RR$ is one-to-many, it follows that:

- $\RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$

$\Box$

Putting the results together, we see that:

- $\RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ if and only if $\RR$ is one-to-many.

$\blacksquare$