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Dec 8, 2016 at 19:14 comment added Taras Banakh Some related problems were studied by Piotr Niemiec arxiv.org/abs/1201.5675
Nov 25, 2016 at 13:36 history edited Jaikrishnan CC BY-SA 3.0
edited title
Nov 25, 2016 at 13:32 comment added Pietro Majer I would consider, for all $(x,y)\in X\times X$, the set $\mathcal{F}(x,y):=\{(f(x),f(y))\,:\, f\in\mathcal{F}\}$. If $\mathcal{F}$ is a group of isometries w.r.to some distance $d$, the sets $D(x,y):=\overline{ \mathcal{F}(x,y)\cup \mathcal{F}(y,x)}$ make a closed partition of $X\times X$ (which is a slightly stronger necessary condition than said above), and $d$ must be constant on each of them.
Nov 25, 2016 at 13:11 comment added Jaikrishnan @Pietro Thanks! I have edited the question.
Nov 25, 2016 at 13:10 history edited Jaikrishnan CC BY-SA 3.0
Modified question as per Pietro's comments.
Nov 25, 2016 at 12:40 comment added Pietro Majer Note that if $\mathcal{F}$ is a collection of isometries w.r.to some metric, then so is the group it generates. Moreover, the point-wise limit of a sequence of isometries is still an isometry: this in general is not true for just a limit of homeomorphism, so it is a necessary condition. Therefore it would be better to assume from the beginning that $\mathcal{F}$ is a group of homeomorphisms, closed by point-wise convergence.
Nov 25, 2016 at 11:32 history asked Jaikrishnan CC BY-SA 3.0