Constructing the set $D$ that satisfies the equation $f(x)=x+1$

Question

Let, $f:A\rightarrow B$, where $ A, B ⊆ \mathbb R$ and $f(f(x))=f(x)+1$ holds $\forall x\in A$.

I want to construct the set $D$, such that $f(x)=x+1$ holds, $\forall x\in D$.

Substituting $f(x)\mapsto x$ we have,

$$f(x)=x+1$$

Then, let $B'⊆ B$ be an image set of $f$. Hence, we can write

$$f(x)=x+1$$

holds if and only if $$x\in D := A∩ B' .$$

Do I think correctly?

Answer 1

For any $x\in A$ and $y=f(x),$ in order that $1+f(x)=f(f(x))=f(y)$ we must have $f(x)=y\in dom(f)=A.$ So if $B'=\{f(x):x\in A\}$ then $B'\subseteq A.$ So $B'\cap A=B'.$ Hence $B'\subseteq D.$ But we may have $B'\ne D$.

Example 1. Let $A=[0,\infty)$ and $f(x)=x+1$ for all $x\in A.$ Then $D=A\ne [1,\infty)=B'.$

Example 2. Let $A=\{-2\}\cup [0,\infty).$ Let $f(-2)=1.$ Let $f(x)=x+1$ for $x\ge 0.$ Then $A\ne D =[0,\infty)\ne B'= [1,\infty).$