Let $M$ be the quotient space of $\mathbb{R}^3\setminus\{0\}$ obtained by identifying the points $(x,y,z)$ with $(2^mx,2^my,2^mz)$ for any integer $m$.
My attempt:
First we claim that $\mathbb{R}^3\setminus \{0\}$ is homeomorphic to $\mathbb{S}^2\times \mathbb{R}_{>0}$. Define $$f:\mathbb{R}^3\setminus \{0\}\to \mathbb{S}^2\times \mathbb{R}_{>0}$$ by $$x\mapsto \left(\frac{x}{|x|},|x|\right)$$ and $$f^{-1}:\mathbb{S}^2\times\mathbb{R}_{>0}\to \mathbb{R}^3\setminus \{0\}$$ by $$(c,x)\mapsto cx.$$ We want to find the quotient space of $\mathbb{R}_{>0}$ with the equivalence relation $x\sim 2^mx$. Define a map $$\varphi:\mathbb{R}_{>0}\to [1,2]$$ by $x\mapsto \frac{x}{2^m}$. This is well-defined since for each real number $x>0$, there exists $m$ such that $$2^m\leq x\leq 2^{m+1}.$$ It can be shown that $\varphi$ is a quotient map and homemorphism. Thus $$\mathbb{R}_{>0}\cong [1,2].$$ By identifying $1\sim 2$ we have $$[1,2]\cong \mathbb{S}^1.$$ Therefore $$\mathbb{S}^2\times \mathbb{R}_{>0}/\sim$$ is homeomorphic to $\mathbb{S}^2\times \mathbb{S}^1$ and $M\cong \mathbb{S}^2\times\mathbb{S}^1$.
Questions:
(a) Is my attempt correct?
(b) It is possible that I am being too technical here, but do I need to first say that $[1,2]$ is homeomorphic to $[0,1]$ before saying $[1,2]\cong \mathbb{S}^1$?