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.
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.