Image of Intersection under One-to-Many Relation

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Theorem

Let $S$ and $T$ be sets.

Let $\RR \subseteq S \times T$ be a relation.


Then:

$\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$

if and only if $\RR$ 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:

$\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$

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

So, assume $S$ is not singleton.


Suppose $\RR$ is specifically not one-to-many.

So:

$\exists x, y \in S: \exists z \in T: \tuple {x, z} \in T, \tuple {y, z} \in T, x \ne y$

and of course $\set x \subseteq S, \set y \subseteq S$.


So:

$z \in \RR \sqbrk {\set x}$
$z \in \RR \sqbrk {\set y}$

and so by definition of intersection:

$z \in \RR \sqbrk {\set x} \cap \RR \sqbrk {\set y}$

But:

$\set x \cap \set y = \O$


Thus from Image of Empty Set is Empty Set:

$\RR \sqbrk {\set x \cap \set y} = \O$

and so:

$\RR \sqbrk {\set x \cap \set y} \ne \RR \sqbrk {\set x} \cap \RR \sqbrk {\set y}$

$\Box$


Necessary Condition

Let $\RR$ be one-to-many.


From Image of Intersection under Relation, we already have:

$\RR \sqbrk {S_1 \cap S_2} \subseteq \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$


So we just need to show:

$\RR \sqbrk {S_1} \cap \RR \sqbrk {S_2} \subseteq \RR \sqbrk {S_1 \cap S_2}$


Let $t \in \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$.

Then:

\(\displaystyle t\) \(\in\) \(\displaystyle \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle t\) \(\in\) \(\displaystyle \RR \sqbrk {S_1} \land t \in \RR \sqbrk {S_2}\) Definition of Set Intersection
\(\displaystyle \leadsto \ \ \) \(\displaystyle \exists s_1 \in S_1: \tuple {s_1, t}\) \(\in\) \(\displaystyle \RR \land \exists s_2 \in S_2: \tuple {s_2, t} \in \RR\) Definition of Relation
\(\displaystyle \leadsto \ \ \) \(\displaystyle s_1\) \(=\) \(\displaystyle s_2\) Definition of One-to-Many Relation
\(\displaystyle \leadsto \ \ \) \(\displaystyle s_1 = s_2\) \(\in\) \(\displaystyle S_1 \land s_1 = s_2 \in S_2\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle s_1 = s_2\) \(\in\) \(\displaystyle S_1 \cap S_2\) Definition of Set Intersection
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map \RR {s_1} = \map \RR {s_2}\) \(\in\) \(\displaystyle \RR \sqbrk {S_1 \cap S_2}\) Image of Element is Subset
\(\displaystyle \leadsto \ \ \) \(\displaystyle \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}\) \(\subseteq\) \(\displaystyle \RR \sqbrk {S_1 \cap S_2}\) Definition of Subset


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

$\RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$

$\Box$


Putting the results together, we see that:

$\RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ if and only if $\RR$ is one-to-many.

$\blacksquare$