Timeline for Surjectivity of self-isometries as property of metric spaces

6 events

when toggle format	what		by	license	comment
Dec 14, 2018 at 0:25	comment	added	YCor		$\mathbf{R}\smallsetminus\mathbf{Q}$ is SIS, while $\mathbf{R}\smallsetminus\mathbf{Q}_+$ is not.
Dec 14, 2018 at 0:02	answer	added	Mike Krebs		timeline score: 3
Jul 29, 2015 at 12:58	comment	added	Adam Przeździecki		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 $C$.
Jul 29, 2015 at 11:48	comment	added	Dominic van der Zypen		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.
Jul 29, 2015 at 11:42	comment	added	Will Brian		$\mathbb{R}$ (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.
Jul 29, 2015 at 11:27	history	asked	Dominic van der Zypen	CC BY-SA 3.0