algebraic-topologygeneral-topologytopological-groups
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
These properties are not equivalent. Here's a counterexample: Let $X=\mathbb R^2\smallsetminus\{(0,0)\}$, and define an action of $\mathbb Z$ on $X$ by $n\cdot (x,y) = (2^n x, 2^{-n} y)$. This is properly discontinuous by your definition, but it's not a proper action. The subset $K \times K \subseteq X\times X$ is compact, where $K = \{(x,y): \max(|x|,|y|)=1\}$, but $\rho^{-1}(K\times K)$ contains the sequence $(n, (2^{-n},1))$, which has no convergent subsequence.
I think one reason for your confusion is that different authors give different definitions of "properly discontinuous." Topologists concerned primarily with actions that determine covering maps often give the definition you gave:
(i) Every $x \in X$ has a neighborhood $U$ such that $gU \cap U \neq \emptyset$ implies $g = e$.
This is necessary and sufficient for the quotient map $X\to X/G$ to be a covering map. However, in order for the action to be proper (and thus for the quotient space to be Hausdorff), an additional condition is needed:
(ii) If $x,x'\in X$ are not in the same $G$-orbit, then there exist neighborhoods $U$ of $x$ and $U'$ of $x'$ such that $gU\cap U' = \emptyset$ for all $g\in G$.
When $X$ is a locally compact Hausdorff space and $G$ is a discrete group acting freely on $X$, the action is proper if and only if both conditions (i) and (ii) are satisfied. Differential geometers, who are typically concerned with forming quotient spaces that are manifolds, are more apt to define "properly discontinuous" to mean both (i) and (ii) are satisfied.
Because of this ambiguity (and because the term "properly discontinuous" leads to oxymoronic phrases such as "a continuous properly discontinuous action"), Allen Hatcher in his Algebraic Topology coined the term covering space action for an action satisfying condition (i). I've adopted that terminology, and I use free and proper action for an action satisfying (i) and (ii) (at least for locally compact Hausdorff spaces). I sincerely hope the term properly discontinuous will eventually die out.
You can find more about these issues in the second editions of my books Introduction to Topological Manifolds (Chapter 12) and Introduction to Smooth Manifolds (Chapter 21).
Exactly the same group action (but named little differently) is in the book "Differential Geometry"on page 68 written by Marcel Berger and Bernard Gostiaux.
Additonaly on the page 70 there is theorem of yours. It is done well with details.
Back to your questions.
Best Answer
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.