Image of Set Difference under Mapping/Proof 1
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Theorem
Let $f: S \to T$ be a mapping.
The set difference of the images of two subsets of $S$ is a subset of the image of the set difference.
That is:
Let $S_1$ and $S_2$ be subsets of $S$.
Then:
- $f \sqbrk {S_1} \setminus f \sqbrk {S_2} \subseteq f \sqbrk {S_1 \setminus S_2}$
where $\setminus$ denotes set difference.
Proof
As $f$, being a mapping, is also a relation, we can apply Image of Set Difference under Relation:
- $\RR \sqbrk {S_1} \setminus \RR \sqbrk {S_2} \subseteq \RR \sqbrk {S_1 \setminus S_2}$
$\blacksquare$