Image of Intersection under One-to-Many Relation

Theorem
Let $S$ and $T$ be sets.

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

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

$\RR$ is one-to-many.

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

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:

So if $\RR$ is one-to-many, it follows that:
 * $\RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$

Putting the results together, we see that:
 * $\RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$ $\RR$ is one-to-many.