Let $X=\Bbb R \times \{-1,1\}$ with the standard topology. Let $\sim$ be the equivalence relation $(x,j)\sim (y,j)$ if $x=y \neq 0$, for $x,y \in \Bbb R$ and $j,k \in \{-1,1\}$. Prove that the quotient space $X/\sim$ with the quotient topology is not Hausdorff.
I found a nice explanation here The Line with two origins However the explanation is more about the intuition , they do it "1-dimensionally" i.e working with a single line every time, and there are not many details of the computations. The intuition is clear though, I would like some help with the details.
I am trying to prove it working with the two lines in the original space, but I have a problem at the end.
Take $p=(0,1), q=(0,-1)$ in $X/\sim$ and $U_p$ and $U_q$ open nbhds respectively of $p $ and $q$. I want to show that the intersection is always non-empty by showing that the intersection of the pre-images via the projection map $\pi: X\to X/\sim$ is non-empty. .
Now a basis for the topology of $X$ ,given that its is a product space, is given by the product of open subsets of $\Bbb R$ and $\{-1,1\}$, i.e. $(-\epsilon, \epsilon)\times\{j\}$, where $j\in \{-1,1\}$. This is because the open subsets of a discrete space are the points themselves
By the definition of quotient topology, since $U_p$ and $U_q $are open in $X/\sim$, $\pi^{-1}(U_p)$ and $\pi^{-1}(U_q)$ are open in $X$.By definition of openness there exists open nbhds contained in each of these 2 sets, which using the basis of the product space must be $(-\epsilon, \epsilon)\times\{j\}$ and $(-\epsilon', \epsilon')\times\{j'\}$. Since $p=(0,1)$ is in the first one and $q=(0,-1)$ is in the second one, this forces $j=1, j'=-1$. But now I am in trouble because $(-\epsilon, \epsilon)\times\{1\}$ and $(-\epsilon', \epsilon')\times\{-1\}$ are disjoint (all because of the second components), while the idea was to see that they intersect
What am I doing wrong and how do I fix it?
