Image is Subset of Codomain

Theorem

Let $\RR = S \times T$ be a relation.

For all subsets $A$ of the domain of $\RR$, the image of $A$ is a subset of the codomain of $\RR$:

$\forall A \subseteq \Dom \RR: \RR \sqbrk A \subseteq T$

In the notation of direct image mappings, this can be written as:

$\forall A \in \powerset S: \map {\RR^\to} A \in \powerset T$

Let $\RR = S \times T$ be a relation.

The image of $\RR$ is a subset of the codomain of $\RR$:

$\Img \RR \subseteq T$

These results also hold for mappings:

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

For all subsets $A$ of the domain $S$, the image of $A$ is a subset of the codomain of $f$:

$\forall A \subseteq S: f \sqbrk A \subseteq T$

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

The image of $f$ is a subset of the codomain of $f$:

$\Img f \subseteq T$

$\ds A$

$\subseteq$

$\ds \Dom \RR$

$\ds \leadsto \ \ $

$\ds \RR \sqbrk A$

$\subseteq$

$\ds \Img \RR$

$\ds $

$\subseteq$

$\ds T$

$\blacksquare$