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:
- $\forall S_1, S_2 \subseteq S: \RR \sqbrk {S_1 \cap S_2} = \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}$
if and only if $\RR$ is one-to-many.
General Result
Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation.
Let $\powerset S$ be the power set of $S$.
Then:
- $\ds \forall \mathbb S \subseteq \powerset S: \RR \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} \RR \sqbrk X$
if and only if $\RR$ is one-to-many.
Family of Sets
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 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:
\(\ds t\) | \(\in\) | \(\ds \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds t\) | \(\in\) | \(\ds \RR \sqbrk {S_1}\) | Definition of Set Intersection | ||||||||||
\(\, \ds \land \, \) | \(\ds t\) | \(\in\) | \(\ds \RR \sqbrk {S_2}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists s_1 \in S_1: \, \) | \(\ds \tuple {s_1, t}\) | \(\in\) | \(\ds \RR\) | Definition of Relation | |||||||||
\(\ds \exists s_2 \in S_2: \, \) | \(\, \ds \land \, \) | \(\ds \tuple {s_2, t}\) | \(\in\) | \(\ds \RR\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds s_1\) | \(=\) | \(\ds s_2\) | Definition of One-to-Many Relation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds s_1 = s_2\) | \(\in\) | \(\ds S_1\) | |||||||||||
\(\, \ds \land \, \) | \(\ds s_1 = s_2\) | \(\in\) | \(\ds S_2\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds s_1 = s_2\) | \(\in\) | \(\ds S_1 \cap S_2\) | Definition of Set Intersection | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \RR {s_1} = \map \RR {s_2}\) | \(\in\) | \(\ds \RR \sqbrk {S_1 \cap S_2}\) | Image of Element is Subset | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \RR \sqbrk {S_1} \cap \RR \sqbrk {S_2}\) | \(\subseteq\) | \(\ds \RR \sqbrk {S_1 \cap S_2}\) | 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.
$\blacksquare$