Image of Set Difference under Relation

Theorem
Let $\mathcal R \subseteq S \times T$ be a relation. Let $A$ and $B$ be subsets of $S$.

Then:
 * $\mathcal R \left({A}\right) \setminus \mathcal R \left({B}\right) \subseteq \mathcal R \left({A \setminus B}\right)$

where $\setminus$ denotes set difference.

Corollary 1
In addition to the other conditions above:

Let $A \subseteq B$.

Then:
 * $\complement_{\mathcal R \left({B}\right)} \left({\mathcal R \left({A}\right)}\right) \subseteq \mathcal R \left({\complement_{B} \left({A}\right)}\right)$

where:
 * $\mathcal R \left({B}\right)$ denotes the image of $B$ under $\mathcal R$;
 * $\complement$ (in this context) denotes relative complement.

Corollary 2

 * $\complement_{\operatorname{Im} \left({\mathcal R}\right)} \left({\mathcal R \left({A}\right)}\right) \subseteq \mathcal R \left({\complement_S \left({A}\right)}\right)$

where $\operatorname{Im} \left({\mathcal R}\right)$ denotes the image of $\mathcal R$.

Proof of Corollary 1
We have that $A \subseteq B$.

Then by definition of relative complement:
 * $\complement_B \left({A}\right) = B \setminus A$
 * $\complement_{\mathcal R \left({B}\right)} \left({\mathcal R \left({A}\right)}\right) = \mathcal R \left({B}\right) \setminus \mathcal R \left({A}\right)$

Hence, when $A \subseteq B$:
 * $\complement_{\mathcal R \left({B}\right)} \left({\mathcal R \left({A}\right)}\right) \subseteq \mathcal R \left({\complement_B \left({A}\right)}\right)$

means exactly the same thing as:
 * $\mathcal R \left({B}\right) \setminus \mathcal R \left({A}\right) \subseteq \mathcal R \left({B \setminus A}\right)$

Proof of Corollary 2
By definition of the image of $\mathcal R$:
 * $\operatorname{Im} \left({\mathcal R}\right) = \mathcal R \left({S}\right)$

So, when $B = S$ in the corollary 1:
 * $\complement_{\operatorname{Im} \left({\mathcal R}\right)} \left({\mathcal R \left({A}\right)}\right) = \complement_{\mathcal R \left({S}\right)} \left({\mathcal R \left({A}\right)}\right)$

Hence:
 * $\complement_{\operatorname{Im} \left({\mathcal R}\right)} \left({\mathcal R \left({A}\right)}\right) \subseteq \mathcal R \left({\complement_S \left({A}\right)}\right)$

means exactly the same thing as:
 * $\complement_{\mathcal R \left({S}\right)} \left({\mathcal R \left({A}\right)}\right) \subseteq \mathcal R \left({\complement_S \left({A}\right)}\right)$

that is:
 * $\mathcal R \left({S}\right) \setminus \mathcal R \left({A}\right) \subseteq \mathcal R \left({S \setminus A}\right)$

Note
Note that equality does not hold in general.

See the note on Mapping Image of Set Difference for an example of a mapping (which is of course a relation) for which it does not.