# Image of Subset under Mapping is Subset of Image

## Corollary to Image of Subset under Relation is Subset of Image

Let $S$ and $T$ be sets.

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

$A \subseteq B \implies f \sqbrk A \subseteq f \sqbrk B$

This can be expressed in the language and notation of direct image mappings as:

$\forall A, B \in \powerset S: A \subseteq B \implies \map {f^\to} A \subseteq \map {f^\to} B$

## Proof

As $f: S \to T$ is a mapping, it is also a relation, and thus:

$f \subseteq S \times T$

The result follows directly from Image of Subset under Relation is Subset of Image.

$\blacksquare$