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Let $(M, d)$ be a metric space. I define a translation of $M$ to be a function $f$ from $M$ to $M$ such that $d(x, f(x)) = d(y, f(y))$ for all $x$ and $y$ in $M$. My conjecture is that every translation on $M$ is an isometry under the same metric.

Can anyone prove this, or give me a counterexample?

Seirios
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user107952
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    Let $(M,d)$ be a discrete metric space with $d(x,y) = 1$ if $x\neq y$. Let $x_0,x_1$ be two distinct points of $M$. Let $f(x_0) = x_1$, and $f(x) = x_0$ for $x\neq x_0$. Then $d(x,f(x)) = 1$ for all $x$, but it is an isometry (if and) only if $M = {x_0,x_1}$. I think it would hold, however, for spaces sufficiently similar to $\mathbb{R}^n$ with the Euclidean distance. – Daniel Fischer Nov 20 '13 at 10:14
  • @DanielFischer This is elaborate enough for an answer, I think. – AlexR Nov 20 '13 at 10:16
  • @DanielFischer Thanks!! I wasn't expecting an answer this quickly. – user107952 Nov 20 '13 at 10:22

1 Answers1

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If we don't require continuity of $f$, we have a counterexample $f\colon \mathbb{R}\to\mathbb{R}$ with

$$f(x) = \begin{cases}x + 1 &, x \in \mathbb{Q}\\ x-1 &, x \notin\mathbb{Q}.\end{cases}$$

For continuous $f$, we get a counterexample by considering a discrete metric space $(M,d)$, with $d(x,y) = 1$ for $x\neq y$, that contains at least three points. Let $x_0,x_1$ be two distinct points, and

$$f(x) = \begin{cases}x_0 &, x \neq x_0\\ x_1 &, x = x_0. \end{cases}$$

Then $d(x,f(x)) = 1$ for all $x$, but $d(f(x),f(y)) = 0$ for all $x,y \in M\setminus \{x_0\}$.

Daniel Fischer
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