TITLE: Quotient space of $\mathbb{R}^2$
QUESTION [8 upvotes]: This is an example of from the book Topology of Janich. In the picture $X = \mathbb{R}^2$ with standard topology, and the lines represent the equivalence classes, which are closed $1$ dimensional manifolds. The example is aimed to show: Even if the equivalence classes are closed subsets of the Hausdorff space $\mathbb{R}^2$, the quotient space $X/\sim$ could still be non-Hausdorff. It says that one quotient space in the picture is Hausdorff while the other is not. However I am having hard time finding out which is. Could somebody help me please.

REPLY [2 votes]: The picture on the right gives a non-Hausdorff quotient space. Specifically, the two vertical lines $l_1$ and $l_2$ next to the cap-shaped lines do not have disjoint open neighbourhood. (from then on, I call cap-shaped lines "cap" lines)
We start at a point $x_1$ on one line $l_1$ of the two lines and see what its neighbourhood contains. Draw an open ball about that point. This ball intersects a certain "cap" line, say $c_1$. Note also that when this ball intersect $c_1$, then this ball also intersects all the "cap" lines above $c_1$. So here, our attempt to construct an arbitrary open neighbourhood $U_1$ (an open set in quotient space) of $l_1$ shows that this neighbourhood contains $c_1$ and all "cap" lines above $c_1$.
By the same argument applied to the other line $l_2$, an open neighbourhood $U_2$ (an open set in the quotient space) of $l_2$ shall contain a certain "cap" line $c_2$ and all the cap lines above $c_2$.
This is how $U_1\cap U_2$ is non-empty: there are "cap" lines which are simultaneously above $c_1$ and $c_2$, and such lines are elements in $U_1\cap U_2$, hence making $U_1\cap U_2$ non-empty.