Image of Domain of Mapping is Image Set

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 $10.8 \ \text{(b)}$
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