Image of Set Difference under Mapping/Corollary 3
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Theorem
Let $f: S \to T$ be a surjection.
Let $A \subseteq S$ be a subset of $S$.
Then:
- $T \setminus f \sqbrk A \subseteq f \sqbrk {S \setminus A}$
where $\setminus$ denotes set difference.
Proof
As $T$ is a surjection, $T = f \sqbrk S$.
Thus Image of Set Difference under Mapping can be applied:
- $f \sqbrk S \setminus f \sqbrk A \subseteq f \sqbrk {S \setminus A}$
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $4$: Mappings: Exercise $6$
- 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $1$: Pairs, Relations, and Functions: Exercise $7 \ \text {(c)}$