Image of Intersection under One-to-Many Relation

Theorem
Let $$\mathcal R \subseteq S \times T$$ be a relation.

Let $$S_1$$ and $$S_2$$ be subsets of $$S$$.

Then:
 * $$\mathcal R \left({S_1 \cap S_2}\right) = \mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right)$$

iff $$\mathcal R$$ is one-to-many.

Generalized Result
Let $$\mathcal R \subseteq S \times T$$ be a relation.

Let $$S_i \subseteq S: i \in \N^*_n$$.

Then:
 * $$\mathcal R \left({\bigcap_{i = 1}^n S_i}\right) = \bigcap_{i = 1}^n \mathcal R \left({S_i}\right)$$

iff $$\mathcal R$$ is one-to-many.

Sufficient Condition
Suppose that:
 * $$\forall S_1, S_2 \subseteq S: \mathcal R \left({S_1 \cap S_2}\right) = \mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right)$$

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

So, assume $$S$$ is not singleton, and suppose $$\mathcal R$$ is specifically not one-to-many.

So:
 * $$\exists x, y \in S: \exists z \in T: \left({x, z}\right) \in T, \left({y, z}\right) \in T$$.

and of course $$\left\{{x}\right\} \subseteq S, \left\{{y}\right\} \subseteq S$$.

So:
 * $$z \in \mathcal R \left({\left\{{x}\right\}}\right)$$
 * $$z \in \mathcal R \left({\left\{{y}\right\}}\right)$$

and so $$z \in \mathcal R \left({\left\{{x}\right\}}\right) \cap \mathcal R \left({\left\{{y}\right\}}\right)$$

But $$\left\{{x}\right\} \cap \left\{{y}\right\} = \varnothing$$.

Thus:
 * $$\mathcal R \left({\left\{{x}\right\} \cap \left\{{y}\right\}}\right) = \varnothing$$

and so:
 * $$\mathcal R \left({\left\{{x}\right\} \cap \left\{{y}\right\}}\right) \ne \mathcal R \left({\left\{{x}\right\}}\right) \cap \mathcal R \left({\left\{{y}\right\}}\right)$$

Necessary Condition
Let $$\mathcal R$$ be one-to-many

From Image of Intersection, we already have:


 * $$\mathcal R \left({S_1 \cap S_2}\right) \subseteq \mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right)$$.

So we just need to show:


 * $$\mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right) \subseteq \mathcal R \left({S_1 \cap S_2}\right)$$.

Let $$t \in \mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right)$$.

Then:

$$ $$ $$ $$ $$ $$ $$ $$

So if $$\mathcal R$$ is one-to-many, it follows that:
 * $$\mathcal R \left({S_1 \cap S_2}\right) = \mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right)$$

Putting the results together, we see that:
 * $$\mathcal R \left({S_1 \cap S_2}\right) = \mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right)$$ iff $$\mathcal R$$ is one-to-many.

Generalized Proof
Proof by induction:

For all $$n \in \N^*, n \ge 2$$, let $$P \left({n}\right)$$ be the proposition:
 * $$\bigcap_{i = 1}^n \mathcal R \left({S_i}\right) = \mathcal R \left({\bigcap_{i = 1}^n S_i}\right)$$ iff $$\mathcal R$$ is one-to-many.

Basis for the Induction
$$P(2)$$ is the case:
 * $$\mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right) = \mathcal R \left({S_1 \cap S_2}\right)$$ iff $$\mathcal R$$ is one-to-many.

This has been proved above.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 2$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:


 * $$\bigcap_{i = 1}^k \mathcal R \left({S_i}\right) = \mathcal R \left({\bigcap_{i = 1}^k S_i}\right)$$ iff $$\mathcal R$$ is one-to-many.

Then we need to show:


 * $$\bigcap_{i = 1}^{k+1} \mathcal R \left({S_i}\right) = \mathcal R \left({\bigcap_{i = 1}^{k+1} S_i}\right)$$ iff $$\mathcal R$$ is one-to-many.

Induction Step
This is our induction step:

$$ $$ $$ $$

So $$P \left({k}\right) \implies P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $$\forall n \in \N, n \ge 2: \mathcal R \left({\bigcap_{i = 1}^n S_i}\right) = \bigcap_{i = 1}^n \mathcal R \left({S_i}\right)$$

iff $$\mathcal R$$ is one-to-many.