Let $Y=X/ \sim$ be the topology quotient. Determine if $Y$ is a Hausdorff space. Let $X=\mathbb{R}^n$ be the standard topology. Define in $\mathbb{R}^n$ the equivalence relation $\sim$ as follows: if $x,y\in \mathbb{R}^n$, then $x\sim y$ if and only if $\|x\|=\|y\|$ or $\|x\|\cdot\|y\|=1$. Let $Y=X/ \sim$ be the topology quotient. Determine if $Y$ is a Hausdorff space.
I think $Y$ is Hausdorff, I have tried to prove it as follows: Let $[x],[y]\in Y, [x]\neq[y]$ then $x\neq y$ with which since $\mathbb{R}^n$ with the usual topology is Hausdorff we have that there are neighborhoods $U$ of $x$ and $V$ of $y$ such that $U\cap V= \phi $, but here I am stuck, what I want is to prove that $[x]\in p(U), [y]\in p(V), p(U)\cap p(V)=\phi$, where $p:X\rightarrow X/ \sim, x\mapsto [x]$, but I do not know how to do it or if this is possible. Could anyone help me please? Is it okay what I've done? Thank you very much.
 A: You have to be careful when taking $U$ open in $X$ and then trying to show that $p(U)$ is open in $Y$, since this may not be true. Recall that a subset $W\subseteq Y$ is open in the quotient topology defined on $Y$ if and only if the pre-image $p^{-1}(W)=\{x\in X:[x]\in W \}$ is an open subset of $X=\mathbb{R}^n$. Since the quotient map $p$ is surjective, we know that $p(p^{-1}(W))=W$. Therefore your plan to find disjoint open sets $U,V\subseteq X$ containing $x,y$ respectively and then use $p(U),p(V)$ will only work if $U$ and $V$ are pre-images of open subsets of $Y$. This means that if $x\in U$, then we require $x'\in U$ for all $x'\in[x]$. We therefore seek $U\subseteq X$ satisfying $p^{-1}(p(U))=U$.
Let's first examine what $[x]$ looks like as a subset of $X$. Clearly $[0]=\{0\}$, so consider $x\neq 0$. Given $r>0$, let $S_{r}=\{a\in X:||a||=r\}$ denote the surface of the $(n-1)$-sphere of radius $r$ centred at the origin. We deduce from the definition of $\sim$ that $[x]=S_{||x||}\cup S_{||x||^{-1}}$. Therefore, just using the fact that $X$ is Hausdorff is not quite enough; we need to separate $(n-1)$-spheres and not just points.
Suppose we have $R>0$. Given $\delta>0$, let $A(R,\delta)=\{x\in X:R-\delta<||x||<R+\delta\}$ and let $\hat{A}(R,\delta)=\{x\in X:R-\delta<||x||^{-1}<R+\delta\}$. It is easy to check that these sets $A$ and $\hat{A}$ are open subsets of $X$. Now define $U(R,\delta)=A(R,\delta)\cup\hat{A}(R,\delta)$. Observe that $U(R,\delta)$ is open and, for all $u\in U(R,\delta)$, we have $[u]\subseteq U(R,\delta)$. From this we deduce that $p^{-1}(p(U(R,\delta)))=U(R,\delta).$ Hence, if $x\in X$ satisfies $||x||=R$ or $||x||=R^{-1}$, then $p(U(R,\delta))$ is an open neighbourhood of $[x]$ in $Y$.
Now suppose we $x,y\in X\setminus\{0\}$ with $[x]\neq[y]$. We can then find distinct $R_{1},R_{2}\geqslant 1$ such that $[x]\in p(U(R_{1},\delta_{1}))$ and $[y]\in p(U(R_{2},\delta_{2}))$ for any $\delta_{i}>0$. By taking the $\delta_{i}$ to be sufficiently small, we can ensure that $p(U(R_{1},\delta_{1}))$ and $p(U(R_{2},\delta_{2}))$ are disjoint, and so these are the desired open sets in $Y$ which separate $[x]$ and $[y]$.
The only remaining case is to look at open neighbourhoods of $[0]$. One can take $$Z(\varepsilon)=\{x\in X: ||x||<\varepsilon \,\, \mbox{or} \,\, ||x||>\varepsilon^{-1}\}.$$ As with the $U$ sets, it is easy to show that $p(Z(\varepsilon))$ is an open neighbourhood of $[0]$ in $Y$. For $[x]\in Y\setminus\{[0]\}$, we can then take $\delta,\varepsilon>0$ sufficiently small so that $p(U(||x||,\delta))$ and $p(Z(\varepsilon))$ are disjoint open neighbourhoods in $Y$ of $[x]$ and $[0]$ respectively.
