# Image of Intersection under One-to-Many Relation

## Theorem

Let $S$ and $T$ be sets.

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

Then:

$\forall S_1, S_2 \subseteq S: \mathcal R \left[{S_1 \cap S_2}\right] = \mathcal R \left[{S_1}\right] \cap \mathcal R \left[{S_2}\right]$

if and only if $\mathcal R$ 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:

$\forall S_1, S_2 \subseteq S: \mathcal R \left[{S_1 \cap S_2}\right] = \mathcal R \left[{S_1}\right] \cap \mathcal R \left[{S_2}\right]$

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

So, assume $S$ is not singleton.

Suppose $\mathcal R$ is specifically not one-to-many.

So:

$\exists x, y \in S: \exists z \in T: \left({x, z}\right) \in T, \left({y, z}\right) \in T, x \ne y$

and of course $\left\{{x}\right\} \subseteq S, \left\{{y}\right\} \subseteq S$.

So:

• $z \in \mathcal R \left[{\left\{{x}\right\}}\right]$
• $z \in \mathcal R \left[{\left\{{y}\right\}}\right]$

and so by definition of intersection:

$z \in \mathcal R \left[{\left\{{x}\right\}}\right] \cap \mathcal R \left[{\left\{{y}\right\}}\right]$

But:

$\left\{{x}\right\} \cap \left\{{y}\right\} = \varnothing$

Thus from Image of Empty Set is Empty Set:

$\mathcal R \left[{\left\{{x}\right\} \cap \left\{{y}\right\}}\right] = \varnothing$

and so:

$\mathcal R \left[{\left\{{x}\right\} \cap \left\{{y}\right\}}\right] \ne \mathcal R \left[{\left\{{x}\right\}}\right] \cap \mathcal R \left[{\left\{{y}\right\}}\right]$

$\Box$

### Necessary Condition

Let $\mathcal R$ be one-to-many.

From Image of Intersection under Relation, we already have:

$\mathcal R \left[{S_1 \cap S_2}\right] \subseteq \mathcal R \left[{S_1}\right] \cap \mathcal R \left[{S_2}\right]$

So we just need to show:

$\mathcal R \left[{S_1}\right] \cap \mathcal R \left[{S_2}\right] \subseteq \mathcal R \left[{S_1 \cap S_2}\right]$

Let $t \in \mathcal R \left[{S_1}\right] \cap \mathcal R \left[{S_2}\right]$.

Then:

 $\displaystyle t$ $\in$ $\displaystyle \mathcal R \left[{S_1}\right] \cap \mathcal R \left[{S_2}\right]$ $\displaystyle \implies \ \$ $\displaystyle t$ $\in$ $\displaystyle \mathcal R \left[{S_1}\right] \land t \in \mathcal R \left[{S_2}\right]$ Definition of Set Intersection $\displaystyle \implies \ \$ $\displaystyle \exists s_1 \in S_1: \left({s_1, t}\right)$ $\in$ $\displaystyle \mathcal R \land \exists s_2 \in S_2: \left({s_2, t}\right) \in \mathcal R$ Definition of Relation $\displaystyle \implies \ \$ $\displaystyle s_1$ $=$ $\displaystyle s_2$ Definition of One-to-Many Relation $\displaystyle \implies \ \$ $\displaystyle s_1 = s_2$ $\in$ $\displaystyle S_1 \land s_1 = s_2 \in S_2$ $\displaystyle \implies \ \$ $\displaystyle s_1 = s_2$ $\in$ $\displaystyle S_1 \cap S_2$ Definition of Set Intersection $\displaystyle \implies \ \$ $\displaystyle \mathcal R \left({s_1}\right) = \mathcal R \left({s_2}\right)$ $\in$ $\displaystyle \mathcal R \left[{S_1 \cap S_2}\right]$ Image of Element is Subset $\displaystyle \implies \ \$ $\displaystyle \mathcal R \left[{S_1}\right] \cap \mathcal R \left[{S_2}\right]$ $\subseteq$ $\displaystyle \mathcal R \left[{S_1 \cap S_2}\right]$ Definition of Subset

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

$\mathcal R \left[{S_1 \cap S_2}\right] = \mathcal R \left[{S_1}\right] \cap \mathcal R \left[{S_2}\right]$

$\Box$

Putting the results together, we see that:

$\mathcal R \left[{S_1 \cap S_2}\right] = \mathcal R \left[{S_1}\right] \cap \mathcal R \left[{S_2}\right]$ if and only if $\mathcal R$ is one-to-many.

$\blacksquare$