# Image of Intersection under One-to-Many Relation/Family of Sets

## Theorem

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:

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

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

Then by definition of $\family {S_i}_{i \mathop \in I}$:

- $\forall i, j \in I: \mathcal R \sqbrk {S_i \cap S_j} = \mathcal R \sqbrk {S_i} \cap \mathcal R \sqbrk {S_j}$

and the sufficient condition applies for Image of Intersection under One-to-Many Relation.

So $\mathcal R$ is one-to-many.

$\Box$

### Necessary Condition

Suppose $\mathcal R$ is one-to-many.

From Image of Intersection under Relation: Family of Sets, we already have:

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

so we just need to show:

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

Let:

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

Then:

\(\displaystyle t\) | \(\in\) | \(\displaystyle \bigcap_{i \mathop \in I} \mathcal R \sqbrk {S_i}\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \forall i \in I: t\) | \(\in\) | \(\displaystyle \mathcal R \sqbrk {S_i}\) | Definition of Set Intersection | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \forall i \in I: \exists x \in S_i: \tuple {x, t}\) | \(\in\) | \(\displaystyle \mathcal R\) | Definition of Relation | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle x\) | \(\in\) | \(\displaystyle \mathcal R \sqbrk {\bigcap_{i \mathop \in I} S_i}\) | $\mathcal R$ is one-to-many | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map {\mathcal R} x\) | \(\subseteq\) | \(\displaystyle \mathcal R \sqbrk {\bigcap_{i \mathop \in I} S_i}\) | Image of Element is Subset | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \bigcap_{i \mathop \in I} \mathcal R \sqbrk {S_i}\) | \(\subseteq\) | \(\displaystyle \mathcal R \sqbrk {\bigcap_{i \mathop \in I} S_i}\) | Definition of Subset |

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

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

$\Box$

Putting the results together:

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

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

$\blacksquare$