This question is from Section 5.1 ($T_0 - T_2$ spaces) of Wayne Patty's topology and I am struck on it. So, I am asking for help here.
Let $I =[0,1]$ and define $x \sim y$ provided that $x-y$ is rational. Then $\sim$ is an equivalence relation on $I$. Let $p: I \to I/\sim$ be the natural map. Prove that $I/\sim$ is not Hausdorff and $p$ is open.
Attempt : any open set in $I/\sim$ will look like $\{r_i+ i'\}$ where $r_i$ are set of rationals and $i'$ is any irrational number.
Open : Now, I think natural map $p$ will map any rational element to itself but I am not sure where it will map any irrational element.
Hausdorff : $I/\sim$ will contain equivalence classes of irrational numbers and 1 equivalence class will contain only 1 irrational number and sum of all rational numbers to it. So, I think open sets in $I/\sim$ will only be of the form of equivalence classes of each irrational number. Also for proving it is not Hausdorff, I took two elements $1+\sqrt 3$ and $2+\sqrt 3$ and as the only open sets are the equivalence class of $\sqrt 3$ which is not disjoint. So, It can't be Hausdorff.
Hope my argument for Hausdorff is fine ? How should I approach proving $p$ to be not open?
Kindly tell.
$q=\Bbb Q \cap [0,1]$ is one of the classes of $[0,1]{/}\sim$. $\{q\}$ is not closed in $[0,1]{/}\sim$ as the rationals, i.e. $p^{-1}[\{q\}]$, aren't closed in $[0,1]$ (def. of quotient topology). So $[0,1]{/}\sim$ is not $T_1$, let alone $T_2$.
If $O$ is open and non-empty try to show that $p^{-1}[p[O]] = [0,1]$ (so every $x \in O$ is equivalent to some $y \in O$) and so $p[O]$ is open by the definition of the quotient topology again.