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

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Theorem

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:

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

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

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

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

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

So $\RR$ is one-to-many.

$\Box$


Necessary Condition

Suppose $\RR$ is one-to-many.


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

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

so we just need to show:

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


Let:

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

Then:

\(\ds t\) \(\in\) \(\ds \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}\)
\(\ds \leadsto \ \ \) \(\ds \forall i \in I: \, \) \(\ds t\) \(\in\) \(\ds \RR \sqbrk {S_i}\) Definition of Set Intersection
\(\ds \leadsto \ \ \) \(\ds \forall i \in I: \exists x \in S_i: \, \) \(\ds \tuple {x, t}\) \(\in\) \(\ds \RR\) Definition of Relation
\(\ds \leadsto \ \ \) \(\ds x\) \(\in\) \(\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i}\) $\RR$ is one-to-many
\(\ds \leadsto \ \ \) \(\ds \map \RR x\) \(\subseteq\) \(\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i}\) Image of Element is Subset
\(\ds \leadsto \ \ \) \(\ds \bigcap_{i \mathop \in I} \RR \sqbrk {S_i}\) \(\subseteq\) \(\ds \RR \sqbrk {\bigcap_{i \mathop \in I} S_i}\) Definition of Subset


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

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

$\Box$


Putting the results together:

$\RR$ is one-to-many 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 $S$.

$\blacksquare$

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