How to show that this space is not a manifold? Define $X = (\mathbb{R}^2 \backslash \{0\})/\sim$ with the quotient topology induced by the projection $\pi: \mathbb{R}^2\backslash \{0\} \rightarrow X$ where the equivalence relation $\sim$ on $\mathbb{R}^2 \backslash \{0\}$ is given by $(x,y) \sim (\lambda^{-1} x, \lambda y)$ for some $\lambda \in \mathbb{R}\backslash\{0\}$.
How would you show that X is not a manifold? 
 A: Hagen von Eitzen is correct, $X$ is not Hausdorff because $[(0,1)]$ and $[(1,0)]$ cannot be seperated by disjoint open sets in $X$.
Here is a proof.  Consider the sequences $x_n, y_n$ of points in $\mathbb{R}^2\setminus 0$ given by $x_n = \left(\frac{1}{n},1\right)$ and $y_n = \left(1, \frac{1}{n}\right)$.
We clearly have $x:=\lim_{n\rightarrow \infty} x_n = (0,1)$ and $y:=\lim_{n\rightarrow \infty} y_n = (1,0)$.
Now, since $\pi$ is continuous, we see that $\pi(x_n)$ converges to $\pi(x)$.  Likewise, $\pi(y_n)$ converges to $\pi(y)$.
But note the following two points:


*

*$\pi(x)\neq \pi(y)$.  This is because if $(0,1) = (\lambda^{-1} 1, \lambda 0)$ for some $\lambda \in \mathbb{R}\setminus 0$, then the second coordinate gives $1 = 0$, an obvious contradiction.

*$\pi(x_n) = \pi(y_n)$ for all $n$.  To see this, set $\lambda = n$.  Then $[y_n] = [(1,1/n)] = [ (\lambda^{-1}, \lambda/n)] = [(1/n, 1)] = [x_n]$.
Point 1 says the two sequences $\pi(x_n)$ and $\pi(y_n)$ have two different limits and Point 2 says the two sequences $\pi(x_n)$ and $\pi(y_n)$ are really the same sequence.  So, the one sequence $\pi(x_n) = \pi(y_n)$ has two distinct limits.  This cannot occur in a Hausdorff space.
