X is not Hausdorff and for every point in X we have an open neighborhood that is homeomorphic to $\mathbb{R}$ We let $\mathbb{R}\coprod\mathbb{R}$ the disjoint union of two copies of $\mathbb{R}$ and write $(i,x)$ with $i \in \{1,2\}$ and $x \in \mathbb{R}$ for its elements so that the first index distinguishes the two copies of $\mathbb{R}$.
We view $\mathbb{R}\coprod\mathbb{R}$ as a topological space with the topology of the disjoint union induced by the standard topology on $\mathbb{R}$.
Let $\sim$ be the equivalence relation on $\mathbb{R}\coprod\mathbb{R}$ generated by requiring $(1,x) \sim (2,x)$ whenever $x \in \mathbb{R}\setminus\{0\}$. We let $X = (\mathbb{R}\coprod\mathbb{R})/\sim$ be the quotient space.
(i) Show that X is not a Hausdorff space.
(ii) Show that every point in X admits an open neighborhood that is homeomorphic to $\mathbb{R}$.
(i) I think that I have to show this by contradiction, but I don't know how. I have difficulties with understanding how this space X looks like. I already know that $\mathbb{R}\coprod\mathbb{R}$ = $\{(i, x_{i}) \mid i \in \{1,2\}, x_{i} \in \mathbb{R}\}$. And all points that satisfy the equivalence relation will be squeezed to one point. So how do two points from X look like and how can I derive then that X can not be a Hausdorff space?
(ii) I think I need a map that is a homeomorphism/embedding $f: X \rightarrow \mathbb{R}$. But I don't know how to construct such a map and how to continue then.
 A: Call $0_{0}$ and $0_{1}$ the two origins. Because by definition of the quotient space topology, the open neighbourhoods of $0_{i} \in(\mathbb{R} \sqcup \mathbb{R}) / \sim$ are precisely those that contain subsets of the form
$$
(-\epsilon, \epsilon)_{i}:=(-\epsilon, 0) \cup\left\{0_{i}\right\} \cup(0, \epsilon) .
$$
But this means that the "two origins" $0_{0}$ and $0_{1}$ may not be separated by neighbourhoods, since the intersection of $(-\epsilon, \epsilon)_{0}$ with $(-\epsilon, \epsilon)_{1}$ is always non-empty:
$$
(-\epsilon, \epsilon)_{0} \cap(-\epsilon, \epsilon)_{1}=(-\epsilon, 0) \cup(0, \epsilon) .
$$
For the second part just let me sketch the main idea. For every point that is not one of the two origins then just pick an open neighbourhood and map in the same neighbourhood of the real line. For the two origins the procedure is the same, just pick the interval formed by the "upper" and "lower" lines.
You may want to check line with two origins as this example. It is a nice example of a locally Euclidean space (every point admits a neighbourhood homeomorphic to the real line) that fails to be Hausdorff. Hence, when one defines a topological manifold, the requirement to be Hausdorff is not redundant.
A: Hint: What are neighbourhoods of $(1,0)$ and $(2,0)$?
