Equivalence of Definitions of Image of Mapping
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Theorem
The following definitions of the concept of image of a mapping are equivalent:
Definition 1
The image of a mapping $f: S \to T$ is the set:
- $\Img f = \set {t \in T: \exists s \in S: \map f s = t}$
That is, it is the set of values taken by $f$.
Definition 2
The image of a mapping $f: S \to T$ is the set:
- $\Img f = f \sqbrk S$
where $f \sqbrk S$ is the image of $S$ under $f$.
Proof
Let $f: S \to T$ be a mapping.
Let:
- $Y = \set {t \in T: \exists s \in S: \map f s = t}$
Then by definition:
- $Y$ is the image of $f$ by definition 1.
But by definition of image of subset under mapping:
- $Y = f \sqbrk S$
Thus $Y$ is the image of $f$ by definition 2.
Hence the result.
$\blacksquare$