Consider the following relation on $\mathbb{R}$ : $x\sim y$ if and only if for some $n\in \mathbb{Z}$, $x= 2^n y$.
Clearly this is an equivalence relation; let $\pi$ denote the canonical surjection onto the quotient, which we see as a quotient space.
I have an exercise that tells me that for this topology, $\{\pi(0)\}$ is dense in $\mathbb{R}/{\sim}$; however I can't seem to understand why the following isn't true :
let $x> 0$, and consider $\epsilon$ such that $0\notin ]x-\epsilon, x+\epsilon[$. Then $U = \displaystyle\bigcup_{n\in \mathbb{Z}} ]2^n x-2^n\epsilon, 2^n x+2^n\epsilon[$. Clearly $U$ is open, it doesn't contain $0$.
Moreover, if $y\in U$ and $y\sim y'$, say $y= 2^m y'$, then if $y= 2^n x - 2^n\epsilon + 2^nz$ where $0<z<2\epsilon$, we have $y'= 2^{n+m}x - 2^{n+m}\epsilon + 2^{n+m}z$, so $y' \in U$: $U$ is saturated wrt to $\sim$ and so $\pi^{-1}(\pi(U)) = U$, so that $\pi(U)$ is open, and $\pi(0)\notin \pi(U)$, $\pi(x)\in \pi(U)$. Therefore $\{\pi(0)\}$ is not dense in $\mathbb{R}/{\sim}$.
Where did I do something wrong ?
You are right, $\{\pi(0)\}$ is not dense. In fact, it's closed, since $\pi^{-1}(\{\pi(0)\}) = \{0\}$ is closed in $\mathbb{R}$. But the only open set containing $\pi(0)$ is $\mathbb{R}/{\sim}$. The writer of the exercise seems to have confused these two things.