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
$$ $$ $$

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.