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Mar 7, 2017 at 18:07 comment added Misha @Seirios: See the answer.
Mar 7, 2017 at 18:07 answer added Misha timeline score: 3
Feb 18, 2017 at 13:44 comment added Seirios @Misha: Can you give some hints for the proof in the case the action is properly discontinuous? I find the result interesting because it implies that an action on some length space is metrically proper and cocompact iff it is properly discontinuous and cocompact. (I do not know if this still holds in other contexts without a properness assumption.)
Jun 25, 2013 at 3:02 review First posts
Jun 25, 2013 at 5:04
Jun 14, 2013 at 3:07 comment added Misha Helge: What Lee says is correct, such actions are usually called "metrically properly discontinuous". In your setting, you probably meant "properly discontinuous" (rather than just proper). Then the assertion is correct and is a nice exercise in point-set topology although not as easy as the one for "metrically proper" actions.
Jun 13, 2013 at 21:14 comment added Lee Mosher Being a length metric is irrelevant. On the other hand, you ought to assume the metric space is proper (closed balls are compact), particularly since you are assuming that the action is proper (the induced map $G \times X \to X \times X$ is a proper function). In that case what you ask for is true, and is a trivial consequence of the definitions.
Jun 13, 2013 at 18:12 comment added Helge does one need the assumption that $(X,d)$, is a metric space, where $d$ is a length metric ?
Jun 13, 2013 at 18:09 history asked Helge CC BY-SA 3.0