Let $G$ be group and let $X$ be a topological space on which $G$ acts continuously. Now let us consider the following two properties relative to the group action:

(a) For every compact subset $K\subseteq X$ we have that $|\lbrace g\in G:gK\cap K\neq\emptyset\rbrace|<\infty$

(b) For all sequence $\{g_n\}_{n\geq 1}$ of pairwise distinct elements of $G$ and every $x\in X$ the sequence $\{g_n x\}_{n\geq 1}$ has no limit point in $X$.

It is easy to see that $(a)\Rightarrow (b)$. What about the converse?

Under the following assumptions one may show that $(b)\Rightarrow (a)$:

(*) Assume that $X$ is a locally compact metric space where the distance function is denoted by $d$. Assume that there is an absolute constant $C>0$ such that for all $x\in X$ there exists a neighborhood $U_x$ of $x$ such that for all $g\in G$ and all $u,v\in U_x$ one has that $d(gu,gv) < C\cdot d(u,v)$.

For example if $G$ acts through isometries on $X$ on a locally compact space then $(a)$ is equivalent to $(b)$.

1. Is it possible to weaken Assumption $(*)$ ?