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:

$\RR \sqbrk {S_1} \setminus \RR \sqbrk {S_2} \subseteq \RR \sqbrk {S_1 \setminus S_2}$

$\blacksquare$