Consider $\mathbb{R}^3/E_i$ with $E_1 = \{(x,y,z) \in \mathbb{R}^2 \vert x^2 + y^2 + z^2 \leq 1\}$ and $E_2 = \{(x,y,z) \in \mathbb{R}^2 \vert x^2 + y^2 + z^2 < 1\}$. I want to study the quotient topology and say for which $E_i$ it is Hausdorff.
This is the theorem I want to apply:
If $(X,\tau)$ is Hausdorff, the quotient is Hausdorff iff $\forall p,q \in X$ not equivalent, there exist two saturated neighbourhood $U_p$ and $U_q$, such that $U_p \cap U_q = \emptyset$.
Let's start with $E_1$. I pick a point on the surface of the sphere and a point outside the sphere. To me it seems that I can take a small enough neighbourhood such that there is no intersection and so it is Hausdorff.
On the other hand, in $E_2$ I cannot pick a point on the surface. So I pick one "inside" the sphere and one outside. To meet this seems hausforff again, but I think my way of thinking is misleading me.
Any tips?
In the case of $E_1$, pick two points $p,q\in\mathbb{R}^3$. If they are both outside the unitspher, of course that there are neighbourhoods that satisfy the conditions of the theorem: just take two open balls that dont intersect also such that neithr of them intersects the unit sphere.
Now, if $p\in E_1$ and $q\not\in E_1$, let $r=\|q\|$. Consider the open ball $B_{\frac{1+r}2}(0)$, which is a neighborhood of $p$, and $B_{r+\frac{r-1}2}(q)$, which is a neighborhood of $q$. Their intersection is empty.
Therefore, $\mathbb{R}^3/E_1$ is Hausdorff.
But $\mathbb{R}^3/E_2$ isn't. Take any two points of the surface of the unit sphere and any two saturated neigbourhoods $U_p$ and $U_q$ of $p$ and $q$ respectively. In $\mathbb{R}^3$, both of them will have points of $E_2$. Therefore, and since they are saturated, $U_p\cap U_q\neq\emptyset$.