If the action of a group $G$ on $\mathbb{R}$ is properly discontinuous then G is isomorph to $\mathbb{Z}$? 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 $\mathbb{Z}$.
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 A: The orbit $\mathcal O=G0$ of $0\in\mathbb R$ is discrete. It follows that it is countable and, morerover, there is either a strictly increasing or decreasing bijection $f:\mathbb Z\to\mathcal O$ or $f:\mathbb N_0\to\mathcal O$. Let me suppose it is increasing; if not, we do something similar.
If $g\in G$, there is a $n_g\in\mathbb Z$ such that $g(f(0))=f(n_g)$. I claim that $$g(f(a))=f(a+n_g)\qquad\text{for all $a$ in $\mathbb N_0$ or $\mathbb Z$,}$$ depending on which case we are in. Indeed, there is an $m$ such that $g(f(1))=f(m)$. If we had $m>1+n_g$, then the image of the interval $[f(0),f(1)]$ under $g$ would contain the interval $[f(n_g),f(n_g+2)]$, and there would exist a point in $(f(0),f(1))$ whose image under $g$ is $f(n_g+1)$. This is absurd. Doing this in general gives my claim.
Now I can define a map $\phi:g\in G\mapsto n_g\in\mathbb Z$. It is a morphism of groups, and it must be injective.
A: 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.
