Image of Set Difference under Mapping/Corollary 1
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Theorem
Let $f: S \to T$ be a mapping.
Let $S_1 \subseteq S_2 \subseteq S$.
Then:
- $\relcomp {f \sqbrk {S_2} } {f \sqbrk {S_1} } \subseteq f \sqbrk {\relcomp {S_2} {S_1} }$
where $\complement$ (in this context) denotes relative complement.
Proof
From Image of Set Difference under Relation: Corollary 1 we have:
- $\relcomp {\RR \sqbrk {S_2} } {\RR \sqbrk {S_1} } \subseteq \RR \sqbrk {\relcomp {S_2} {S_1} }$
where $\RR \subseteq S \times T$ is a relation on $S \times T$.
As $f$, being a mapping, is also a relation, it follows directly that:
- $\relcomp {f \sqbrk {S_2} } {f \sqbrk {S_1} } \subseteq f \sqbrk {\relcomp {S_2} {S_1} }$
$\blacksquare$