Mathematical Proofs

Question

Create an example of a function $f: \mathbb{R} \to \mathbb{R}$ such that $f(f(f(\mathbb{R}))) = f(f(\mathbb{R})) \neq f(\mathbb{R})$

Answer 1

Take any endofunction (i.e. a function from a finite set $F$ to itself) which does the job as, with $F=\{0,1,2\}$ $$ 0\to 1\to 2\to 2 \mbox{ a loop at number "two"} $$ Call it $f_0$. Now, in $\mathbb{R}\setminus F$ extend it by identity and create $f$ i.e. $$ f(x)=x \mbox{ if }x\notin F\mbox{ and }f(x)=f_0(x) \mbox{ if }x\in F. $$ (remark: $f$ is measurable), then you have a function such that $f\circ f\circ f=f\circ f$ and $$ f\circ f\circ f(\mathbb{R})=f\circ f(\mathbb{R})= \mathbb{R}\setminus \{0,1\} $$ whereas $f(\mathbb{R})=\mathbb{R}\setminus \{0\}$.

Late edit: It seems to me that this is NOT a research question and the MO should consider asking such a question within MSE.