Let $\emptyset\neq X\supset A$, $f:A\rightarrow X$,
a point $x\in A$ is eventually periodic point of $f$, if :
$$\exists n_0\in\mathbb N \exists N\in\mathbb N: f^{n+N}(x)=f^n(x) \quad \forall n\ge n_0.$$
Note: $f^n(x) := \underbrace{f\circ f\circ f\circ \dots \circ f}_{ \text{$n$ times }}(x).$
I don't quite understand the following proof.
$$\text{Let } f:\mathbb R \rightarrow \mathbb R : f(x)= \lvert x-1\rvert,$$ $$f(0)=1,f(1)=0\implies 0 \text{ and } 1 \text{ are periodic points of $f$.}$$ $$f(-1)=2, f(-k)=k+1, f(k)=k-1 ,\forall k\ge 2,k\text{ integer.} \implies $$ $$\implies \text{every integer point is eventually periodic point of $f$ .}$$
I understand the part that $1$ and $-1$ are periodic points, but im a bit confused about the rest of the proof.
Consider the following diagram: $$ \begin{matrix} &&&& -1 && -2 && -3 && -4 & \cdots \\ &&&& \downarrow && \downarrow && \downarrow && \downarrow \\ 0 & \longleftrightarrow & 1 & \longleftarrow & 2 & \longleftarrow & 3 & \longleftarrow & 4 & \longleftarrow & 5 & \longleftarrow & \cdots \end{matrix} $$
If $f$ is repeatedly applied to an integer $k$, we eventually end reach $1$. From that point, the sequence $f^n(k)$ becomes periodic: 1, 0, 1, 0, 1, 0, ...
For a formal proof, because $1$ is eventually periodic and $f(k)=k-1$ for positive $k$, it follows by induction that all positive integers are eventually periodic.
From $f(-k) = k+1$ for positive $k$ it then also follows immediately that all negative integers are eventually periodic.