Image of Subset under Relation is Subset of Image

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

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

Let $A, B \subseteq S$ such that $A \subseteq B$.

Then the image of $A$ is a subset of the image of $B$:
 * $A \subseteq B \implies \RR \sqbrk A \subseteq \RR \sqbrk B$

In the notation of direct image mappings, this can be written:
 * $A \subseteq B \implies \map {\RR^\to} A \subseteq \map {\RR^\to} B$

Corollary 1
The same applies to the preimage, as follows.

Corollary 2
The same applies for a mapping $f: S \to T$ and its inverse $f^{-1} \subseteq T \times S$, whether $f^{-1}$ is a mapping or not.

Corollary 3
And similarly:

Proof
Suppose $\RR \sqbrk A \nsubseteq \RR \sqbrk B$.

The result follows by the Rule of Transposition.