Let $\epsilon>0$ be given. prove that $Iso_{\epsilon} ( x,d) $ is not closed under the function composition operation and thus does not form a group.

Question

Let $f:(X,d)\longrightarrow (X,d)$ be a bijection on metric space $(X,d) $. For $\epsilon>0$ the bijection map $f$ is called $\epsilon$_self _isometry,whenever
$$\sup \underset{x,x\prime \in X} \mid d( x,x\prime) - d( g( x) ,g( x\prime ))\mid \leq \epsilon $$
It is clear that if $f:(X,d)\longrightarrow (X,d)$ is an isometry (that is $d(x,x\prime)=d(g(x), g(x\prime))$), then it is an $\epsilon$_self_isometry,for all $\epsilon>0$ .

Question.Let $\epsilon>0$ be given. prove that $Iso_{\epsilon} ( x,d) $ is not closed under the function composition operation and thus does not form a group.

Answer 1

Consider the map $\mathbb{(0,1)} \to \mathbb{R}$ defined by $x \mapsto x(1 + 2\epsilon/3)$. This map has distortion less than $\epsilon$. But applying it twice (on the naturally restricted domain) yields a map with distortion greater than $\epsilon$.