$A⊆B$ then $R(A)⊆R(B)$

Question

Assume $R$ is a relation from $X$ to $Y$ , then Prove if $A⊆B⊆X$ then $R(A)⊆R(B)$.

my try:

$$n\ ∈\ R(A)$$$$=n\ ∈\left\{y\ ∈\ Y\ ∣∃x\left(\ x\ ∈\ A\right)∧\ xRy\right\}⇒n\ ∈\left\{y\ ∈\ Y\ ∣∃x\left(\ x\ ∈\ B\right)∧\ xRy\right\}$$$$=n\ ∈\ R(B)$$

hence if $A⊆B⊆X$ then $R(A)⊆R(B)$

Answer 1

You're right, but it's better style to not use so many symbols, but use words instead, and what's bad is that you reuse $A$ and $B$ as symbols: the relation $R$ is from $A$ to $B$ so $R \subseteq A \times B$. The property you mean is probably

$$\forall C,D \subseteq A: C \subseteq D \implies R[C] \subseteq R[D]$$

introducing new variables for subsets of $A$. We cannot have two different $A$'s in statements of theorems...

Suppose that $C \subseteq D (\subseteq A)$ and let $b \in R[C]$; this means by the definition (which you correctly stated) that there is some $a \in C$ with $aRb$. As $C \subseteq D$ we know that $a \in D$ too so the fact that $aRb$ then by definition implies that $b \in R[D]$ as well. This shows the inclusion $R[C] \subseteq R[D]$.

Motto: choose your symbols wisely and use fewer formulae.