Image of Domain of Mapping is Image Set
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Theorem
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
The image of $S$ is the image set of $f$:
- $f \sqbrk S = \Img f$
Proof
By definition, a mapping is a relation.
Thus Image of Domain of Relation is Image Set applies.
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 8$: Functions
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Functions
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.10$: Functions: Remark $10.8 \ \text{(b)}$