I have to study the following G-Action $$ \begin{cases}\mathbb{Z}^2 \times \mathbb{R}^2 & \longrightarrow \mathbb{R}^2 \\ (m,n)\times (x,y) & \longmapsto (x+m,y+n) \end{cases} $$ That is, as the title says, the Group $G=\mathbb{Z}^2$ acting on $\mathbb{R}^2$ by translation.
I managed to show the following property of the $G$-Action
i) $\forall x \in X ,\exists U \ni x$ neighborhood : $gU \cap U \neq \emptyset \implies g=e$
I also have to verify the property
ii) If $x,x' \in X$ are not in the same $G$-orbit, then $\exists U \ni x, U' \ni x'$ neighborhoods : $gU \cap U' =\emptyset , \ \forall g \in G$
I am stuck with showing ii)
My approach: Let $x=(x_1,x_2), x'=(x_1',x_2') \in \mathbb{R}^2$ such that $x \neq g x'$ where $g=(g_1,g_2) \in \mathbb{Z}^2$. That is $x,x' \in \mathbb{R}^2$ are not in the same $G$-Orbit.
Now let $\epsilon >0$ and choose $U:=B_\epsilon (x)$ and $U':=B_\epsilon (x')$ the open balls around $x,x'$
My goal is to show that they are disjoint for all elements of $g \in G$.
The only possible way I can think of is producing a contradiction by assuming that there exists an element lets call it $g'=(g_1',g_2') \in G$ such that $g'U \cap U' \neq \emptyset$ $$ \implies \exists z \in g'U \cap U' \implies z=(u_1',u_2')=(g_1'+u_1, g_2' + u_2) \\ \implies u':=(u_1',u_2')=g'u$$
Where $u_1,u_2 \in U, \ u_1',u_2' \in U$
Therefore the elements $u=(u_1,u_2)$ and $u'=(u_1',u_2')$ are in the same $G$-Orbit. I am not sure if this helps me in any regard, I fail to see a contradiction because my backbone on $G$-actions isn't strong enough yet.
How can I continue?
If all of it is wrong, how can I show ii)?
Edit: We call property ii) that the action is separable
You must to see what's happening on the fundamental domain what is the unit square.