Prove that if $~x,~y \in \mathbb R~$, $~f(x)=f(y)~$ and $~x=y~$.

Question

$~f:\mathbb R\to \mathbb R~$ is a function on $~\mathbb R~$, and $~\{f_z\}~$ is a sequence of real valued functions on $~\mathbb R~$ defined by $~f_1(x)=f(x), ~x \in \mathbb R ~$ and $~f_{z+1}=f[f_z(x)] ,~x ~\in \mathbb R ,~~~ z=1,2,3,4,\cdots~$

Assume there exists a positive integer $~z_0~$ such that $~f_{z_0}(x)=x \forall x \in \mathbb R ~$. Prove that the function is injective.

So I understand in order to prove this function is injective, I need to prove that if $~x,~y \in \mathbb R~$, $~f(x)=f(y)~$ and $~x=y~$.

If someone could please walk me through how to approach a problem like this that would be awesome and really appreciated, my prof doesn't do examples so figuring out the way to go about proving this seems really complicated.

Thank you in advance.

Answer 1

Suppose that $f$ is not injective. Then there are distinct real numbers $a$ and $b$ such that $a\neq b$ and $f(a)=f(b)$. But then $(\forall m\in\mathbb N):f_m(a)=f_m(b)$. So, since $a\neq b$, you can't have both $f_m(a)=a$ and $f_m(b)=b$.

Answer 2

Suppose $f(x)=f(y)$. Then $f_z(x)=f_z(y)\,,\forall z$. In particular, $x=f_{z_0}(x)=f_{z_0}(y)=y$. Thus $f$ is injective.