# Is the limit set of a group action always closed?

Let $G$ be a discrete group acting on a compact metric space $X$. A point $x\in X$ is called a limit point, if there is a base point $x_0\in X$ and an injective sequence $(x_k)_{k\in\mathbb{N}}$ in the orbit $G(x_0)$ such that $x_k\to x$ as $k$ tends to infinity. Let $L$ be the set of limit points.

Question: is $L$ always closed?

For a fixed base point $x$, the accumulation set $L_x$ of $G(x)$ is closed. The limit set $L$ can be defined as $$L=\bigcup_{x\in X}L_{x}.$$ This is a union of infinitely many closed sets, therefore not necessarily closed. I would like to see an example of limit set that is not closed, or an argument showing that $L$ is always closed. Further assumptions on $G$ and $X$ are welcome, as long as you feel comfortable.

For example: consider Kleinian group acting on the unit disk, then the limit set $L$ is on the boundary and closed. However, this example is quite special, because the orbit of any base point has the same accumulation set.

Another example: I believe that in the case where $G$ acts as a group of isometries, $L$ is closed. This example is wrong. Thanks to Yves.

• @YvesCornulier Thanks. Then, let's say that $X$ is a compact metric space. (the terms limit point / accumulation point / cluster point messed up ... ) – Hao Chen Nov 26 '13 at 23:07
• Here's an easy example: define, on the closed unit disc $X$ of the complex plane, $f(z)=u(|z|)z$ where $u(r)=e^{i\pi(1-r)}$. Clearly $f$ is a self-homeomorphism, defining an action of $\mathbf{Z}$ on $X$. Then $L$ is exactly the set of points in $X$ whose radius is irrational, and thus is not closed. – YCor Nov 26 '13 at 23:11
• @YvesCornulier Nice one. So the answer is no for general cases. Thanks! – Hao Chen Nov 26 '13 at 23:19
• Thinking twice, considering an irrational rotation of the disc $X$ gives rise to an even simpler example: here $L$ is the complement of $\{0\}$ and thus is not closed, and the action is, this time, isometric. – YCor Nov 27 '13 at 8:46