Image of Set Difference under Mapping/Proof 2
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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
\(\ds y\) | \(\in\) | \(\ds f \sqbrk {S_1} \setminus f \sqbrk {S_2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists x \in {S_1}: x \notin {S_2}: \, \) | \(\ds \tuple {x, y}\) | \(\in\) | \(\ds f\) | Definition of Image of Subset under Mapping | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists x \in {S_1} \setminus {S_2}: \, \) | \(\ds \tuple {x, y}\) | \(\in\) | \(\ds f\) | Definition of Set Difference | |||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(\in\) | \(\ds f \sqbrk {S_1 \setminus S_2}\) | Definition of Image of Subset under Mapping |
$\blacksquare$