Durrett 1.1.4: A sigma field $F$ is said to be countably generated if there is a countable collection $C \subset F$ so that $\sigma(C)=F$. Show that $R^d$ is countably generated where $R$ represents Borel Sets.
Answer:
I need to show that $C \subset R^d$ so that $\sigma(C)=R^d$. I know, from online that a family of cubes is a countable set that generates $R^d$. I don't really understand why this is the case. This would mean that the smallest sigma algebra containing a family or collection of cubes is equal to the borel sets, $R^d$, which includes open sets, closed sets, intersections of closed sets, intersections of open sets, unions of open sets, and unions of closed sets of $\mathbb{R^d}$.
I guess, I am having trouble visualizing since we are working with d dimensions. In $\mathbb{R^d}$, are the largest possible open/closed sets just cubes? Perhaps, some visual intuition would really help as I am having trouble visually/intuitively understanding what is happening.
Alternatively: I was thinking could I just take:$ \ [q, \infty)$ x $ \ [q, \infty)$ x ...$ \ [q, \infty)$ d times, where q is a rational number and this would generate $R^d$ as I know that $[q,\infty)$ where q is a rational number generates $R$. Any help would be much appreciated.