A 'weird' topology: Hausdorff, quotients. Let $X$ be obtained by taking two disjoint copies of the interval $[0,2]$ (with the Euclidean topology) and gluing each $t$ in the first copy with the corresponding $t$ in the second copy, for all $t \in [0,2]$ different from the middle point. Explicitely, one may take the space 
$$ Y=[0,2]\times \{0\} \cup [0,2] \times \{1\} \subset \mathbb{R}^2$$
with the topology induced from the Euclidean topology, and $X$ is the space obtained from $Y$ by gluing $(t,0)$ to $(t,1)$ for all $t\in [0,2], t\neq 1$. We endow $X$ with the quotient topology. 


*

*Is $X$ Hausdorff?

*Show that $X$ can also be obtained as a quotient of the circle $S^1$. 


I find this a very weird topology and hard to understand it. Also, I don't know how to proof some space is Hausdorff, I can't give a perfect proof of it. Can somebody help me? 
 A: Roughly, the space $X$ is like an interval $[0, 2]$ but at $1$ there is a small infinitesimal "bump", and if you zoom in you'll see two pieces. Thus it's called the "line with two origins".
$X$ is not Hausdorff. Take image of the points $1 \times \{0\}$ and $1 \times \{1\}$ by the quotient map $q : Y \to X$. Any neighborhood of the two points will intersect.
$X$ can be obtained from the quotient of $S^1$ as follows. Define an equivalence relation $\sim$ on $S^1 \subset \Bbb R^2$ as follows: $(x, y) \sim (x, -y)$ for all points on $S^1$ such that $(x, y) \neq (0, 1)$. Essentially, all we did was to identify the two points of the intervals in $Y$ beforehand to get a circle and then did the rest of the identifications.
A: if you shift by $-1$ and consider any sequence in $Y$ which decreases monotonically to zero. then the image of this sequence in $X$ converges to two distinct points, so $X$ cannot be Hausdorff.
call the two exceptional points $A$ and $B$. let  $N$ be an open neighbourhood of $A$ not containing $B$. then $N \cup \{B\}\setminus \{A\}$ is an open neighbourhood of $B$ not containing $A$. since the sequence does not contain either $A$ or $B$ the criteria for convergence in $X$ to $A$ and $B$ are identical to the criterion for convergence to zero in $Y$
