Image of Intersection under One-to-Many Relation

Let $$\mathcal{R} \subseteq S \times T$$ be a relation which is one-to-many.

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

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

Generalized Result
Let $$S_i \subseteq S: i \in \mathbb{N}^*_n$$.

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

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

Generalized Proof
From Image of Intersection, we already have:

$$\mathcal{R} \left({\bigcap_{i = 1}^n S_i}\right) \subseteq \bigcap_{i = 1}^n \mathcal{R} \left({S_i}\right)$$.

So we just need to show:

$$\bigcap_{i = 1}^n \mathcal{R} \left({S_i}\right) \subseteq \mathcal{R} \left({\bigcap_{i = 1}^n S_i}\right)$$.

Proof by induction:

For all $$n \in \mathbb{N}^*$$, let $$P \left({n}\right)$$ be the proposition $$\bigcap_{i = 1}^n \mathcal{R} \left({S_i}\right) \subseteq \mathcal{R} \left({\bigcap_{i = 1}^n S_i}\right)$$.


 * $$P(1)$$ is true, as this just says $$\mathcal{R} \left({S_1}\right) \subseteq \mathcal{R} \left({S_1}\right)$$.

Basis for the Induction

 * $$P(2)$$ is the case $$\mathcal{R} \left({S_1}\right) \cap \mathcal{R} \left({S_2}\right) \subseteq \mathcal{R} \left({S_1 \cap S_2}\right)$$, which 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:

$$\bigcap_{i = 1}^k \mathcal{R} \left({S_i}\right) \subseteq \mathcal{R} \left({\bigcap_{i = 1}^k S_i}\right)$$.

Then we need to show:

$$\bigcap_{i = 1}^{k+1} \mathcal{R} \left({S_i}\right) \subseteq \mathcal{R} \left({\bigcap_{i = 1}^{k+1} S_i}\right)$$.

Induction Step
This is our induction step:

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

Therefore $$\bigcap_{i = 1}^n \mathcal{R} \left({S_i}\right) \subseteq \mathcal{R} \left({\bigcap_{i = 1}^n S_i}\right)$$.

So $$\mathcal{R} \left({\bigcap_{i = 1}^n S_i}\right) = \bigcap_{i = 1}^n \mathcal{R} \left({S_i}\right)$$.