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Let $H$ be an open group such that, $H$ act continously an by isometrieson a metric space $(X,d)$ ($\forall h\in H$, the map $X\ni x\longmapsto h.x\in X$ is an isometry.). Recall that for $x_{0}\in X$ the orbit of $x_{0}$ is the following set $orb(x_{0})=\{h.x_{0}:\,\,h\in H\}$. We say that this orbit is bounded if it is boundedas a subset of the metric space $(X,d)$.

I want to show that if the action by isometries of the open group $H$ on the metric space $(X,d)$ have an boounded orbit, then every orbit is bounded.

thank for any help.

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What is an open group? – Qiaochu Yuan Jan 4 '13 at 12:05
up vote 1 down vote accepted

Suppose the $H$-orbit of $x_0$ is bounded. Thus, there are $y \in X$ and $r \gt 0$ such that $d(y,hx_0) \lt r$ for all $h \in H$. For every $x_1 \in X$ and every $h \in H$ we have from the triangle inequality and $d(hx_0,hx_1) = d(x_0,x_1)$ that $$d(y,hx_1) \leq d(y,hx_0) + d(hx_0,hx_1) \lt r + d(x_0,x_1),$$ so the orbit of $x_1$ is in the ball of radius $r + d(x_0,x_1)$ around $y$.

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This argument only uses that $H$ acts by isometries on $X$. It is not necessary to know what an open group is or that the action is continuous. – Martin Jan 4 '13 at 12:09

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