Image of Set Difference under Mapping/Proof 2

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

$\blacksquare$