Decomposition of open sets in $\mathbb{R^d}$ I am trying to prove the following problem. It's an exercise in Stein's Real Analysis text book. 
Problem: Suppose $\mathbb{R^d}-\{0\}$ is represented as $\mathbb{R_+}\times S^{d-1}$ with $\mathbb{R_+}=\{0<r<\infty\}$ and $S^{d-1}=\{x\in \mathbb{R^d},|x|=1\}$ is the unit sphere. Then show that every open set in $\mathbb{R^d}\backslash{\{0\}}$ can be written as a countable union of open rectangles of this products.
Stein gives the following hint in his book
Hint: Considere the countable collection of rectangle of the form $$\{r_j<r<r_k^{'}\}\times\{\gamma\in S^{d-1}:|\gamma-\gamma_l|<1/n\}$$ Here $\gamma_j$ and $\gamma_k^{'}$ range over all positive rationals and $\{\gamma_l\}$ is a countable dense se of $S^{d-1}$
Attempt: Notice that any open set in $\mathbb{R^d}$ can be expressed as a countable union of almost disjoint rectangles. So it suffices to show the claim holds for open rectangles. Then I got stuck. So we have an open rectangle in the space how do we transform it to what we want..
 A: The idea is similar to that of balls. But now you are given some sectors 
$$ S_{j, k,n} = \{|r_j-r|< \frac{1}{n}\} \times \{\gamma \in \mathbb S^{d-1}: |\gamma - \gamma_k| \leq \frac{1}{n}\}$$
We can do that for any open set. Let $V$ be any open set in $\mathbb R^d \setminus \{0\}$. Let $(a_i)$ be a dense subset of $V$. So it suffices to find, for each $i$, a sector $S_{j, k, n}$ so that $a_i \in S_{j, k,n} \subset V$. 
Now for each $a_i$, there is $d_i>0$ so that the open ball centered at $a_i$ with radius $d_i$ are in $V$: $B_{d_i}(a_i) \subset V$. 
The polar coordinate of $a_i$ is given by $(|a_i|, \frac{a_i}{|a_i|})$. Let $n$ be large so that 
$$(*)\ \ \ \ \ \frac{1}{n} < \frac{d_i}{2(|a_i|+3)}.$$ 
Now find $r_j$ and $\gamma_k$ so that 
$$\big| r_j- |a_i| \big| < \frac{1}{n}, \ \ \ |\gamma_k - \frac{a_i}{|a_i|} | < \frac{1}{n}.$$ 
Thus $a_i \in S_{j, k, n}$. Now we show that $S_{j, k, n} \subset B_{d_i} (a_i)$. Pick $x\in S_{k, j, n}$. Let $y\in \mathbb R^d\setminus \{0\}$ so that $y = (|x|, \frac{a_i}{|a_i|})$ when written in polar coordinate. Then by triangle inequality 
$$|x - a_i| \leq |x-y | + |y-a_i| \leq |x| \bigg( \bigg| \frac{x}{|x|} - \frac{a_i}{|a_i|}\bigg|\bigg) + \big| |x| - |a_i|\big|$$
As $|x| \leq ||x| -r_j| + r_j \leq \frac{1}{n} + |a_i| + \frac{1}{n} = |a_i|+\frac{2}{n}$, so 
$$ |x| \bigg( \bigg| \frac{x}{|x|} - \frac{a_i}{|a_i|}\bigg|\bigg) \leq |x| \bigg( \bigg| \frac{x}{|x|} - \gamma_k\bigg| + \bigg| \gamma_k - \frac{a_i}{|a_i|}\bigg|\bigg)\leq (|a_i| + \frac{2}{n}) \frac{2}{n} \leq (|a_i|+2)\frac{2}{n}$$
Also, 
$$\big| |x| - |a_i|\big|\leq \big| |x| - r_j \big| + \big|r_j - |a_i|\big| \leq \frac{2}{n}. $$
Thus 
$$|x-a_i| \leq (|a_i|+3) \frac{2}{n} < d_i. $$
Hence $x\in B_{d_i}(a_i)$. So $S_{j, k, n} \subset V$ and so we are done. 
