Image of Subset under Relation is Subset of Image

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

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

Then the image of $A$ is a subset of the image of $B$:
 * $A \subseteq B \implies \mathcal R \left({A}\right) \subseteq \mathcal R \left({B}\right)$

In the language of induced mappings, that would be written:
 * $A \subseteq B \implies f_{\mathcal R} \left({A}\right) \subseteq f_{\mathcal R} \left({B}\right)$

Corollary 1
The same applies to the preimage.

Let $C, D \subseteq T$.

Then:
 * $C \subseteq D \implies f_{\mathcal R^{-1}} \left({C}\right) \subseteq f_{\mathcal R^{-1}} \left({D}\right)$

where $\mathcal R^{-1}$ is the inverse of $\mathcal R$.

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.

Let $f: S \to T$ be a mapping.

Let:
 * $A, B \subseteq S$
 * $C, D \subseteq T$

Then:
 * $A \subseteq B \implies f \left({A}\right) \subseteq f \left({B}\right)$
 * $C \subseteq D \implies f^{-1} \left({C}\right) \subseteq f^{-1} \left({D}\right)$

Proof
Suppose $\mathcal R \left({A}\right) \not \subseteq \mathcal R \left({B}\right)$.

... and the result follows by the Rule of Transposition.

Proof of Corollary 1
As $\mathcal R^{-1}$ is itself a relation, by definition of inverse relation, the main result applies directly.

Proof of Corollary 2
As $f: S \to T$ is a mapping, it is also a relation, and thus:
 * $f \subseteq S \times T$

and so is its inverse:
 * $f^{-1} \subseteq T \times S$

Hence, as for Corollary 1, the main result applies directly.