If $A = \{0, 1, 2, 3\}$ and , then

Applying $y=x+1$ with both $x,y\in A=\{0,1,2,3\}$: $x=0,1,2$ give $y=1,2,3$, all in $A$, but $x=3$ gives $y=4\notin A$ and is excluded. So $R=\{(0,1),(1,2),(2,3)\}$.