This is the last question on a homework set from my Introduction to Algebraic Topology class.
Consider the action of $\mathbb{Z}$ on $\mathbb{R}^m\setminus\{0\}$ given by $n.x=2^nx$. We have to show that the quotient space $\mathbb{R}^m\setminus\{0\}\,/\,\mathbb{Z}$ is homeomorphic to $\mathbb{S}^{m-1}\times\mathbb{S}^1$.
I have approached this question by sketching the situation in two and three dimensions. Fix a direction in the plane/space and consider all the orbits lying on this line. If we pick a representative of some orbit on this line, we get the equivalence class of this element by sending all points which are in the same orbit to this point; my teacher tried to help me here by suggesting that we are kind of "rolling over" all the points from the same orbit to one point, and if we do this for all the orbits, we will end up with $\mathbb{S}^1$; then, doing this in all directions in the plan/space we will end up with the Cartesian product $\mathbb{S}^{1}\times\mathbb{S}^1$/$\mathbb{S}^2\times\mathbb{S}^1$. At first this was clear, but not so much anymore. It's clear however that the $\mathbb{S}^{m-1}$ should come from the "looking in all directions" part, and that $\mathbb{S}^1$ should come from considering the orbits in one direction, but how is what my question comes down to:
Is it true that each equivalence class along one direction represents one point on $\mathbb{S}^1$, and that $[x]$ and $[-x]$ are antipodal points?
Any help in viewing this is greatly appreciated. I did my best to phrase the question as clear as possible, but let me know if anything is unclear or should be changed.