# Image of Intersection under One-to-Many Relation

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

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

Then:

$\ds \forall \mathbb S \subseteq \powerset S: \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$

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

### Family of Sets

Let $S$ and $T$ be sets.

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

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

$\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}$

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:

 $\ds t$ $\in$ $\ds \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ $\ds \leadsto \ \$ $\ds t$ $\in$ $\ds \RR \sqbrk {S_1}$ Definition of Set Intersection $\, \ds \land \,$ $\ds t$ $\in$ $\ds \RR \sqbrk {S_2}$ $\ds \leadsto \ \$ $\ds \exists s_1 \in S_1: \,$ $\ds \tuple {s_1, t}$ $\in$ $\ds \RR$ Definition of Relation $\ds \exists s_2 \in S_2: \,$ $\, \ds \land \,$ $\ds \tuple {s_2, t}$ $\in$ $\ds \RR$ $\ds \leadsto \ \$ $\ds s_1$ $=$ $\ds s_2$ Definition of One-to-Many Relation $\ds \leadsto \ \$ $\ds s_1 = s_2$ $\in$ $\ds S_1$ $\, \ds \land \,$ $\ds s_1 = s_2$ $\in$ $\ds S_2$ $\ds \leadsto \ \$ $\ds s_1 = s_2$ $\in$ $\ds S_1 \cap S_2$ Definition of Set Intersection $\ds \leadsto \ \$ $\ds \map \RR {s_1} = \map \RR {s_2}$ $\in$ $\ds \RR \sqbrk {S_1 \cap S_2}$ Image of Element is Subset $\ds \leadsto \ \$ $\ds \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ $\subseteq$ $\ds \RR \sqbrk {S_1 \cap S_2}$ 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.

$\blacksquare$