Let's define equivalence relation on group $\mathbb{R}$ by setting $x\sim y$ iff $x-y\in\mathbb{Z}$ by all $x,y\in\mathbb{R}$.
What is the quotient topology?
For example if we set $x=\sqrt{2}$ and $y=\sqrt{2} + 2$, then $x-y\in\mathbb{Z}$.
So this looks similar to the case where you replace $\mathbb{Z}$ with $\mathbb{Q}$, but $\mathbb{Z}$ is not dense in $\mathbb{R}$. In that case the quotient topology is indiscrete.
Could someone please open up this a bit for me?
Hint: Consider the map $$\operatorname{cis}:\Bbb R\to \Bbb R^2\\\operatorname{cis}(x)=(\cos(2\pi x),\,\sin(2\pi x))$$ and show that the induced map $\overline{\operatorname{cis}}:\Bbb R/\sim\,\to \Bbb R^2$ on the quotient space is a homeomorphism onto the image. What's the image $\operatorname{cis}(\Bbb R)$ ?
Consider $a \in [0,1]$. Then for all $n \in \mathbb{Z}$, we have that $a\sim a+n$. Think of this as a circle with radius $1$ rolling along the integers, identifying each of these numbers. It is a way to "wrap" the real line around a circle a lot of times, so that the map is surjective, but highly non-injective.
Depending on the context of your question, if you take the interval $[0,1]$ you are restricting the function to the interval, but identifying $0 \sim 1$, so that this interval with the quotient topology is indeed homeomorphic to a circle [if you're feeling technical, $S^1$].
