Image of Intersection under One-to-Many Relation/General Result

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Theorem

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.


Proof

Sufficient Condition

Suppose:

$\displaystyle \mathcal R \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \mathcal R \sqbrk X$

where $\mathbb S$ is any subset of $\powerset S$.

Then by definition of $\mathbb S$:

$\forall S_1, S_2 \in \mathbb S: \mathcal R \sqbrk {S_1 \cap S_2} = \mathcal R \sqbrk {S_1} \cap \mathcal R \sqbrk {S_2}$

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: General Result, we already have:

$\displaystyle \mathcal R \sqbrk {\bigcap \mathbb S} \subseteq \bigcap_{X \mathop \in \mathbb S} \mathcal R \sqbrk X$

so we just need to show:

$\displaystyle \forall \mathbb S \subseteq \powerset S: \bigcap_{X \mathop \in \mathbb S} \mathcal R \sqbrk X \subseteq \mathcal R \sqbrk {\bigcap \mathbb S}$


Let:

$\displaystyle t \in \bigcap_{X \mathop \in \mathbb S} \mathcal R \sqbrk X$

Then:

\(\displaystyle t\) \(\in\) \(\displaystyle \bigcap_{X \mathop \in \mathbb S} \mathcal R \sqbrk X\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \forall X \in \mathbb S: t\) \(\in\) \(\displaystyle \mathcal R \sqbrk X\) $\quad$ Definition of Set Intersection $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \forall X \in \mathbb S: \exists x \in X: \tuple {x, t}\) \(\in\) \(\displaystyle \mathcal R\) $\quad$ Definition of Relation $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(\in\) \(\displaystyle \bigcap \mathbb S\) $\quad$ Definition of One-to-Many Relation $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \mathcal R \sqbrk x\) \(\in\) \(\displaystyle \mathcal R \sqbrk {\bigcap \mathbb S}\) $\quad$ Image of Element is Subset $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \bigcap_{X \mathop \in \mathbb S} \mathcal R \sqbrk X\) \(\subseteq\) \(\displaystyle \mathcal R \sqbrk {\bigcap \mathbb S}\) $\quad$ Definition of Subset $\quad$


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

$\displaystyle \forall \mathbb S \subseteq \powerset S: \mathcal R \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \mathcal R \sqbrk X$

$\Box$


Putting the results together:

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

$\displaystyle \mathcal R \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \mathcal R \sqbrk X$

where $\mathbb S$ is any subset of $\powerset S$.

$\blacksquare$