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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.

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    $\begingroup$ Hint: $\mathbb{R}^d$ is a separable topological space, so it's ANII, thus there is a countable family $(U_n)_{n\in\mathbb{N}}$ of open sets such that for every open set $U$ we can write $U=U_{n_1}\cup U_{n_2}\cup\cdots$. Let $C=\{U_n : n\in\mathbb{N}\}$. $\endgroup$ – Albert Apr 22 '18 at 23:42
  • $\begingroup$ So does my example:$ \ [q, \infty)$ x $ \ [q, \infty)$ x ...$ \ [q, \infty)$ d times, where q is a rational number and this would generate $R^d$ , not work? $\endgroup$ – rain Apr 22 '18 at 23:43
  • $\begingroup$ I think it doesn't work. But instead try to consider the family of sets of the form $[q_1,\infty)\times \cdots\times [q_d,\infty)$ for $(q_1,...,q_d)\in \mathbb{Q}^d$, which is also countable. $\endgroup$ – Albert Apr 22 '18 at 23:46
  • $\begingroup$ Thanks. So is (q1, .....q_d) a vector or d-tuple where each component is a rational number (just to clarify) ? $\endgroup$ – rain Apr 22 '18 at 23:50
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    $\begingroup$ That's right :) $\endgroup$ – Albert Apr 22 '18 at 23:54

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