Image of Subset under Relation equals Union of Images of Elements

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

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

Let $X \subseteq S$ be a subset of $S$.

Then:
 * $\ds \RR \sqbrk X = \bigcup_{x \mathop \in X} \map \RR x$

where:
 * $\RR \sqbrk X$ is the image of the subset $X$ under $\RR$
 * $\map \RR x$ is the image of the element $x$ under $\RR$.

Proof
By definition:
 * $\RR \sqbrk X = \set {y \in T: \exists x \in X: \tuple {x, y} \in \RR}$
 * $\map \RR x = \set {y \in T:\tuple {x, y} \in \RR}$

First:

Then:

So:
 * $\ds \bigcup_{x \mathop \in X} \map \RR x \subseteq \RR \sqbrk X$

and:
 * $\ds \RR \sqbrk X \subseteq \bigcup_{x \mathop \in X} \map \RR x$

The result follows by definition of set equality.