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A metric space $(X,d)$ is said to be bounded if there is $r\in\mathbb{R}$ such that for all $x,y\in X$ we have $d(x,y) \leq r$.

A self-isometry is a map $\iota:X\to X$ such that for all $x,y\in X$ we have $d(x,y) = d(\iota(x), \iota(y))$. Let's say that a metric space $(X,d)$ has the SIS-property if all self-isometries are surjective.

Are all SIS-spaces compact? Or bounded? Or does the SIS-property relate to some other property of metric spaces?

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    $\begingroup$ $\mathbb{R}$ (with the usual metric) has this property (proof: once the image of $0$ and $1$ is fixed, the image of every other point is determined; looking at cases, all maps have the form $x \mapsto \pm x + c$). The same argument works for any $\mathbb{R}^n$. So SIS does not imply compact or bounded. You might guess "complete" next, but this is wrong too since the same argument also applies to $\mathbb{Q}$. I don't know how to answer your last question. $\endgroup$
    – Will Brian
    Commented Jul 29, 2015 at 11:42
  • $\begingroup$ Thanks! - The other way round, completeness doesn't imply SIS (mathoverflow.net/questions/212531/… ), possibly compactness does. My last question is fuzzy, therefore not very good, but I would be interested in an answer nevertheless. $\endgroup$
    – Dominic van der Zypen
    Commented Jul 29, 2015 at 11:48
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    $\begingroup$ Compactness does. Suppose $C$ is compact and $f:C\to C$ is a self-isometry which is not onto. Then there exists $x\in C\setminus f(C)$. Then by compactness of $f(C)$ we have $d(x,f(C))=d>0$. Then you see that for every $m\neq n$ we have $d(f^m(x),f^n(x))\geq d$. Thus the sequence $f^n(x)$ contains no converging subsequence contradicting the compactness of $C$. $\endgroup$
    – Adam Przeździecki
    Commented Jul 29, 2015 at 12:58
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    $\begingroup$ $\mathbf{R}\smallsetminus\mathbf{Q}$ is SIS, while $\mathbf{R}\smallsetminus\mathbf{Q}_+$ is not. $\endgroup$
    – YCor
    Commented Dec 14, 2018 at 0:25

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Suppose X is a metric space such that (i) the group of bijective self-isometries acts transitively on X, and (ii) every closed ball is compact. Then X is an SIS-space. (Note that this gives an alternate proof that R^n is an SIS-space; the group of translations acts transitively on R^n.) Proof: Let I be a self-isometry of X. Let a in X. Let b=I(a). By (i), there exists a bijective self-isometry of X which maps b to a. Then TI is a self-isometry of X with TI(a)=a. Hence TI gives us a self-isometry of the closed ball B(r,a) of radius r centered at a, which by Adam Przeździecki's argument is surjective onto B(r,a). Since this holds for all r, therefore TI is surjective, and hence I is surjective.

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