Image of Domain is Subset of Codomain
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Contents |
Theorem
Let $\mathcal R = S \times T$ be a relation.
The image of $\mathcal R$ is a subset of the codomain of $\mathcal R$:
- $\operatorname{Im} \left({\mathcal R}\right) \subseteq T$
Corollary
This also holds for mappings:
Let $f: S \to T$ be a mapping.
The image of $f$ is a subset of the codomain of $f$:
- $\operatorname{Im} \left({f}\right) \subseteq T$
Proof
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \operatorname{Im} \left({\mathcal R}\right)\) | \(=\) | \(\displaystyle \mathcal R \left({\operatorname{Dom} \left ({\mathcal R}\right)}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of image | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \operatorname{Dom} \left ({\mathcal R}\right)\) | \(\subseteq\) | \(\displaystyle \operatorname{Dom} \left ({\mathcal R}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Subset of Itself | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle \operatorname{Im} \left({\mathcal R}\right)\) | \(\subseteq\) | \(\displaystyle T\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Image is Subset of Codomain |
$\blacksquare$
Proof of Corollary
As a mapping is by definition also a relation, the result follows immediately.
$\blacksquare$
