$\epsilon$-isometry of a compact metric space is $\epsilon$-surjective

Question

The question whether an isometric map $f : X \to X$ of a compact metric space is surjective has been asked (and answered positively) frequently.

Assume more generally that $\vert d(f(x),f(y)) - d(x,y)\vert \leq \epsilon$. Is it correct that $X = \overline{B_\epsilon f(X)}$?

I guess this is indeed true and assuming on the contrary one can construct some contradicting sequence in a clever way.

Answer 1

This is not true: an $\epsilon$-isometry can omit an arbitrarily large piece of the space.

A simple example: $X=\{k\epsilon/2: k=0,1,\dots,n\}$ with metric $d(a,b)=a+b$ (unless $a=b$, when it's zero). The triangle inequality is easy to verify.

Let $f(x)=\max(0,x-\epsilon/2)$. This is a map from $X$ into itself and $$d(x,x')-\epsilon\le d(f(x),f(x'))\le d(x,x')$$ On the other hand, $X\setminus f(X) = \{n\epsilon/2\}$, which is a point at distance $n\epsilon/2$ from $f(X)$. Here $n$ can be arbitrarily large.

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Original, more complicated example (but a connected one).

Consider the following "triangular comb" which consists of the horizontal segment from $(0,0)$ to $(0,1)$ and also vertical segments from $(k/n,0)$ to $(k/n,k/n)$, for $k=1,\dots, n$.

Equip it with the path metric: the distance between two points is how long you have to travel from one to the other, within the comb. This is a nice compact geodesic metric space $X$.

Define $f:X\to X$ by $f(x,y) =( (x-1/n)^+, (y-1/n)^+)$ where ${a}^+$ means $\max(a,0)$. Observe that

• $f$ is a nonexpansive map: $d(f(p),f(q))\le d(p,q)$.
• $f$ does not decrease pairwise distances by much: specifically, $d(f(p),f(q))\ge d(p,q)-\frac2n$.

Yet, the complement of $f(X)$ in $X$ contains the ball of radius $1$ centered at $(1,1)$. On the picture below, $X\setminus f(X)$ is in red.