$(0,\infty)/\sim$ homeomorphic to $S^1$ where $x\sim y$ iff $x/y=2^n$ for some $n\in\mathbb{Z}$

38 Views Asked by At

I'm studying for an exam and came across the following problem:

Consider the relation on $(0,\infty)$ given by $x\sim y$ if there is an integer $n$ such that $x/y=2^n$. Prove that $\sim$ is an equivalence relation and show that the quotient space $(0,\infty)/\sim$ is homeomorphic to $S^1$.

My thoughts: so I think the E.R. part is straightforward. Note $x/x=2^0$ so $\sim$ is reflexive; if $x/y=2^n$ then $y/x=2^{-n}$ so $\sim$ is symmetric; and if $x/y=2^n$ and $y/z=2^m$ then $x/z=(x/y)/(z/y)=2^n/2^{-m}=2^{m+n}$ so $\sim$ is transitive.

I'm stuck on figuring out why the quotient space is $S^1$. This can probably be visualized, I'm just having a hard time imagining it. Or maybe the way to go is an explicit map? Is the ray $(0,\infty)$ getting wrapped around on itself to make the circle? Any advice/hints will be greatly appreciated.