Let $(X,d)$ be a complete metric space, and suppose that $f:X \to X$ satisfies the condition: for each $\epsilon >0$, there exists $\delta > 0$ such that for all $x,y \in X$
$$ \epsilon \le d(x,y) < \epsilon+\delta \implies d(f(x),f(y)) < \epsilon.$$Clearly, this condition...