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

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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:

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


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:

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

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

$\blacksquare$