Image of Set Difference under Mapping

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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.


Corollary 1

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.


Corollary 2

Let $f: S \to T$ be a mapping.

Let $X$ be a subset of $S$.


Then:

$\relcomp {\Img f} {f \sqbrk X} \subseteq f \sqbrk {\relcomp S X}$

where:

$\Img f$ denotes the image of $f$
$\complement_{\Img f}$ denotes the complement relative to $\Img f$.


This can be expressed in the language and notation of direct image mappings as:

$\forall X \in \powerset S: \relcomp {\Img f} {\map {f^\to} X} \subseteq \map {f^\to} {\relcomp S X}$

That is:

$\forall X \in \powerset S: \map {\paren {\complement_{\Img f} \circ f^\to} } X \subseteq \map {\paren {f^\to \circ \complement_S} } X$

where $\circ$ denotes composition of mappings.


Corollary 3

Let $f: S \to T$ be a surjection.

Let $A \subseteq S$ be a subset of $S$.

Then:

$T \setminus f \sqbrk A \subseteq f \sqbrk {S \setminus A}$

where $\setminus$ denotes set difference.


Proof 1

As $f$, being a mapping, is also a relation, we can apply Image of Set Difference under Relation:

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

$\blacksquare$


Proof 2

\(\ds y\) \(\in\) \(\ds f \sqbrk {S_1} \setminus f \sqbrk {S_2}\)
\(\ds \leadsto \ \ \) \(\ds \exists x \in {S_1}: x \notin {S_2}: \, \) \(\ds \tuple {x, y}\) \(\in\) \(\ds f\) 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

$\blacksquare$


Also see


Sources

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