to show a quotient space of group action is not Hausdorff Consider the group $\mathbb{Z}$ acting on $\mathbb{R}^2-\{(0,0)\}$ by $n\cdot(x,y)=(2^nx,2^{-n}y)$. How do we know the quotient space $\mathbb{R}^2-\{(0,0)\}/\mathbb{Z}$ is not Hausdorff?
 A: When asking questions like yours on this website, it's a good idea to tell us what you've tried already, so that answers can be tailored to fit your needs.  Since I don't know where you're stuck, here are some hints to guide you.
HINT 1: What does it mean for $(\mathbb{R}^2 \setminus \{(0,0)\}) / \mathbb{Z}$ to be Hausdorff or not Hausdorff?

 $(\mathbb{R}^2 \setminus \{(0,0)\}) / \mathbb{Z}$ is Hausdorff if for all $x, y \in \mathbb{R}^2 \setminus \{(0,0)\}$ such that $\mathbb{Z} \cdot x \neq \mathbb{Z} \cdot y$, there exists open neighborhoods $U \ni x$, $V \ni y$ in $\mathbb{R}^2 \setminus \{(0,0)\}$ such that $\mathbb{Z} \cdot U \cap \mathbb{Z} \cdot V = \emptyset$.
 Therefore, to show that $(\mathbb{R}^2 \setminus \{(0,0)\}) / \mathbb{Z}$ is not Hausdorff, you need to exhibit two points $x$ and $y$ from different orbits such that for every pair of open sets $U \ni x$, $V \ni y$ containing them, the orbits of $U$ and $V$ intersect.

HINT 2: How do you show a violation of Hausdorffness for $(\mathbb{R}^2 \setminus \{(0,0)\}) / \mathbb{Z}$?

 Take $x = (1, 0)$ and $y = (0, 1)$.  If you take a look at their orbits, it appears that they accumulate at the origin.  So while the orbits of $x$ and $y$ are different, it seems that open neighborhoods around these orbits will have to overlap close to the origin.  Can you show that this is the case for every pair of neighborhood containing $x$ and $y$?

If you're still stuck, here's another hint.

 Let $U$ be an open neighborhood of $x = (1,0)$ and $V$ an open neighborhood of $y = (0,1)$.  By shrinking these sets, we may assume without loss of generality that $U = (1 - \epsilon, 1 + \epsilon) \times (-\epsilon, \epsilon)$ and $V = (-\epsilon, \epsilon) \times (1 - \epsilon, 1 + \epsilon)$ for some $0 < \epsilon < 1$.  The group $\mathbb{Z}$ acts on these sets by \begin{align*} m \cdot U &= (2^m (1 - \epsilon), 2^m (1 + \epsilon)) \times (-2^{-m} \epsilon, 2^{-m} \epsilon) \\ n \cdot V &= (-2^n \epsilon, 2^n \epsilon) \times (2^{-n} (1 - \epsilon), 2^{-n} (1 + \epsilon)), \end{align*} for $m, n \in \mathbb{Z}$.  Can you find $m, n$ (depending on $\epsilon$) such that $m \cdot U$ and $n \cdot V$ intersect?

Here is one last hint:

 Take $m = -n$, and $n > \frac{1}{2} \log_2(\frac{1-\epsilon}{\epsilon})$ in the previous hint.

