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Let $X$ be a metric space, and let $G$ be a discrete subgroup of $\mbox{Isom}(X)$ in the compact-open topology. Fix $x \in X$. If $X$ is a proper metric space, it's not hard to show using Arzela-Ascoli that $Gx$ is discrete. However, is there an easy example of a metric space that is not proper so that $Gx$ is not discrete for a discrete $G \subset \mbox{Isom}(X)$?

I thought about permutations of an orthonormal basis in $l^2(\mathbb{N})$, but no luck there.

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Let $X = \mathbb{R}^2$ with the following metric: $$d((x_1,y_1),(x_2,y_2)) = \begin{cases} |y_1-y_2| & \text{if $x_1=x_2$,} \\ |y_1|+|y_2|+|x_1-x_2| & \text{if $x_1\ne x_2$}. \end{cases}$$ I don't know if there is a name for this metric, it is the length of the shortest path if we only allow arbitrary vertical segments and horizontal segments along the $x$-axis. Now let $\mathbb{R}$ act on $X$ by horizontal translation, i.e., $t \circ (x,y) = (x+t,y)$. This is obviously an isometry for every $t$, by definition of the metric. Since the $x$-axis is an orbit, and the metric restricted to the $x$-axis coincides with the usual metric, the orbit of any point $(x,0)$ under this action is non-discrete. In order to see that the group action is discrete, let $s,t \in \mathbb{R}$ with $s\ne t$, and observe that $d(t\circ(0,1),s\circ(0,1)) = d((t,1),(s,1)) = 2+|t-s| >2$, so the orbit of $(0,1)$ has no accumulation point in $X$.

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Looks good! Thanks! –  Zach L. Nov 1 '12 at 4:31
    
That's a nice example. Engelking in General Topology, Example 4.1.6, calls this the river metric on $\mathbb{R}^2$, the idea being that $\{y = 0\}$ is a river in a country in which you travel either perpendicularly to the river or swim in the river. –  commenter Nov 1 '12 at 4:52

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