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$f: M\to M$ an isometry between metric spaces, is $f$ bijective?

$f$ obviously is injective. I proved bijection for $M=\mathbb{R}^n$. But I'm not sure if is true in general metric spaces.

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2 Answers 2

up vote 10 down vote accepted

Let $M = \mathbb{N}$ with the inherited metric from $\mathbb{R}$. Then the fuction $f(x) = x+1$ is an isometry, but is not surjective. As Yoni points out in the comments, one can use $(0,\infty)$ in place of $\mathbb{N}$ if, say, one wants a connected example.

Further, we can do a similar thing on any infinite space with discrete metric.

Incidentally, when $M$ is compact, any isometry must be surjective. See the corresponding question asked here.

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You can also take the positive ray $\Bbb{R}^+$, and the same function $f(x) = x+1$ is an injective isometry that is not surjective. –  Yoni Rozenshein Mar 25 '13 at 4:28

Jasons answers give a very nice example, but there are even examples when you have much more than only a metric. Just giving an example for Hilbert spaces. Taking $\ell^2$ with the norm $$\|(a_k)_{k\in \mathbb{N}}\|=\sum_{k=0}^\infty |a_k|^2$$ (which is induced from the skalarproduct) $$\langle (a_k)_{k \in \mathbb{N}}, (b_k)_{k\in \mathbb{N}}\rangle= \sum_{k=0}^\infty a_k b_k $$ and taking something like a switch function so that $$f(a_0,a_1,a_2,\dots)=(0,a_0,a_1,a_2,\dots)$$ This one is obviously an isometry, but obviously not surjective.

For Hilbertspaces there are no finite dimensional examples, as with a scalarproduct ones always get that a isometry is affine linear (I will consider the linear cases which doesn't make a big difference), and as every isometry is injective the kernel is trivial (here I say it is linear) and hence the isometry is in finite dimensional case a bijection.

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