TITLE: If the action of a group $G$ on $\mathbb{R}$ is properly discontinuous then G is isomorph to $\mathbb{Z}$?
QUESTION [5 upvotes]: Let $G$ be a topological group, acts on a topological space $X$, such that the map $f: G \times X \rightarrow X:(g,x)\mapsto g*x$ is continuous.
We say that this action is $properly\;discontinuous$ if for every element $x\in X$, there exist an open neighbourhood $U_{x}$ for $x$ such that $gU_{x}\cap U_{x}\neq \phi$ and $g\in G$ implies that $g=1_{G}$ 
I am trying to show that if the action of a group $G$ on $\mathbb{R}$ is properly discontinuous then G is isomorphic to $\mathbb{Z}$.
best regards

REPLY [4 votes]: It suffices to show the orbit $G(0)$ of $0$ is a discrete set, and any $x\in\mathbb{R}$ is "sandwiched" by two representatives of $G(0)$. Then the space $X/G$ is homeomorphic to $\mathbb{S}^1$ (the action is continuous), and so $G\approx\pi_1(X/G)=\pi_1(S^1)=\mathbb{Z}$.
The discreteness is natural. Suppose $G(0)$ is bounded above. Let $s\equiv\sup G(0)$; since $G\neq\{e\}$ and acts properly discontinuous, there is some $s'=gs\neq s$ with $g\neq e$. If $s'<s$, then $g(s,\infty)=(s',\infty)$ contains no representative of $G(0)$, contradicting with the definition of $s$. If $s'>s$, then for an interval $I$ containing $s$ such that $gI\cap I=\emptyset$ (such interval exists due to $G$ acting properly discontinuous), all elements of $gI$ are larger than $s$ and so $gI$ contains no representative of $G(0)$. This is however a contradiction since any interval $I$ containing $s=\sup G(0)$ (and so $gI$) must contain some representative of $G(0)$. An analogous argument using $i\equiv\inf G(0)$ shows $G(0)$ is also not bounded below.