Why is $\mathbb{R}/(-1,1)$ not Hausdorff?

Question

My guess would be that $S:=\mathbb{R}/(-1,1)$ is homeomorphic to $S':=(\leftarrow,-1]\cup\{0\}\cup[1,\rightarrow)$ and to somehow show that $S'$ contains points which are non-separable by neighborhoods. Another guess is that I should either consider the points 0 and 1 or 1 and -1. I'm not really sure if this hunch is in any way correct or what I should do next (if it is).

Answer 1

Recall that a Topological space $X$ is Hausdorff if for any $x,y\in X$ there exist open sets $U,V$ such that $x\in U$, $y\in V$, and $U\cap V=\varnothing$.

Considering $\mathbb{R}$ with the usual topology, any open neighborhood of $1$ contains a point in $(-1,1)$. Thus, with the quotient topology $\mathbb{R}/(-1,1)$, any open neighborhood of $[1]$ (the equivalence class containing $1$, i.e. $[1]=\{1\}$) contains $[0]$ (the equivalence class containing $0$, i.e. $[0]=(-1,1)$). That is, $[0]$ and $[1]$ cannot be separated by open sets, and thus $\mathbb{R}/(-1,1)$ is not Hausdorff.

Answer 2

You can see this in another way. Consider $$\pi :\Bbb R\rightarrow \Bbb R/(-1;1)$$ you can prove that the space $\Bbb R/(-1;1)$ isn't $T1$. Indeed consider the equivalence class of the $[0]$ and $\pi ^{-1} ([0])=(-1;1)$ which is open but not closed in $\Bbb R$