I have the following question:
Let $M$ be the quotient space $\mathbb{R}^3\backslash\{0\}$ obtained by identifying $(x,y,z)$ with $(2^mx,2^my,2^mz)$ for any $m \in \mathbb{Z}$. Is $M$ homeomorphic to $S^2 \times S^1$? Prove your assertion.
I am not exactly sure. The two topological properties I was thinking about was compactness. It seems like the quotient is compact, but I maybe not.
Any help would be greatly appreciated.
If you see this as a quotient by the action of $\mathbb{Z}$ on $\mathbb{R}^3 \setminus \{0\}$, you can prove that the set $D=\{ 1 \le x^2+y^2+z^2 \le 2 \}$ is a fundamental domain. Then the quotient is homeomorphic to $D$ quotiented by the same action. But $D$ can also be seen as $S^2 \times [1, 2]$ and the quotient simply associates $S^2 \times \{1\}$ with $S^2 \times \{2\}$, so the original quotient is in fact homeomorphic to $S^2 \times [1, 2]/\{1, 2\}$ which is indeed equivalent to $S^2 \times S^1$.