limit points of a discrete group acting on a metric space Let $\Gamma$ be a discrete group acting continuously on a metric space $X$.
Let $a_1,a_2\in X$, and suppose $a_2$ is a limit point of the orbit $\Gamma a_1$. Suppose $a_3$ is a limit point of the orbit $\Gamma a_2$. Must $a_3$ also be a limit point of $\Gamma a_1$?
I'm happy to assume that the metric on $X$ is nonarchimedean, and that the closure of any orbit of $\Gamma$ is compact.
 A: That $a_2$ is a limit point of the orbit $\Gamma a_1$ means there is a sequence $(g_n)_{n \in \mathbb{N}}$ in $\Gamma$ such that $g_n(a_1) \neq a_2$ for all $n$, and $g_n(a_1) \to a_2$. Similarly, we have a sequence $(h_m)_{m\in \mathbb{N}}$ with $h_m(a_2) \neq a_3$ and $h_m(a_2) \to a_3$.
Let $\varepsilon > 0$. There is an $m$ with $h_m(a_2) \in B_{\varepsilon}(a_3) \setminus \{a_3\}$ (there is even an $m_0$ such that this holds for all $m \geqslant m_0$, but we don't need that). Since $B_{\varepsilon}(a_3) \setminus \{a_3\}$ is open, there is a $\delta > 0$ such that $B_{\delta}(h_m(a_2)) \subset B_{\varepsilon}(a_3) \setminus \{a_3\}$. By continuity of $h_m$, there is an $\eta > 0$ with $h_m(B_{\eta}(a_2)) \subset B_{\delta}(h_m(a_2))$. Choose an $n$ such that $g_n(a_1) \in B_{\eta}(a_2)$. Then $h_m(g_n(a_1)) \in B_{\varepsilon}(a_3) \setminus \{a_3\}$, showing that
$$\Gamma a_1 \cap B_{\varepsilon}(a_3) \setminus \{a_3\} \neq \varnothing$$
for every $\varepsilon > 0$, i.e. $a_3$ is a limit point of $\Gamma a_1$.
