Image of Subset under Mapping is Subset of Image
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Corollary to Image of Subset under Relation is Subset of Image
Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping from $S$ to $T$.
Let $A, B \subseteq S$ such that $A \subseteq B$.
Then the image of $A$ is a subset of the image of $B$:
- $A \subseteq B \implies f \sqbrk A \subseteq f \sqbrk B$
This can be expressed in the language and notation of direct image mappings as:
- $\forall A, B \in \powerset S: A \subseteq B \implies \map {f^\to} A \subseteq \map {f^\to} B$
Proof
As $f: S \to T$ is a mapping, it is also a relation, and thus:
- $f \subseteq S \times T$
The result follows directly from Image of Subset under Relation is Subset of Image.
$\blacksquare$
Sources
- 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): $\S 2.2$: Homomorphisms: $\text{(ii)}$
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.10$: Functions: Exercise $5 \ \text{(c)}$
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1: \ \text{(i)}$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 21.3$: The image of a subset of the domain; surjections
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(e)}$
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): image