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The Borel set can be generated using half open intervals $\prod\limits_{i=1}^m (a_i, b_i]$ in $\mathbb{R}^m$ using the $\sigma$-algebra operations, here the product is the catesian product.

But why is it the same as the set generated by open balls in $\mathbb{R^m}$ ?

The $m=1$ case is easy to see by using some countable intersections/unions involving $1/n$, but I'm not sure what to do for other dimensions.

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  • $\begingroup$ Hint: $(A\times B)\cap (C\times D)=(A\cap C)\times(B\cap D)$. $\endgroup$ – Math1000 Mar 12 '15 at 12:11
  • $\begingroup$ @Math1000 Hmm I don't really get that, don't I need to do something about the "circular" nature of balls? I don't have this problem with $m=1$. $\endgroup$ – simonzack Mar 12 '15 at 12:13
  • $\begingroup$ Well, I recall facts like that being useful when trying to prove various theorems about product topologies. That was just the quick idea I had off the top of my head before I left for work; don't know how helpful it would be. $\endgroup$ – Math1000 Mar 12 '15 at 13:04
  • $\begingroup$ Maybe I'm wrong, but I think you can say that two topologies are equivalent if given one open set in each, you can find two open sets in the other which contain and are contained in the first. Does this help? $\endgroup$ – Alfred Yerger Mar 12 '15 at 13:32
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Any open set is a countable union of open balls (or open boxes).

Combined with the fact that $\prod_{i=1}^m (a_i, b_i)$ also generates the Borel set, we have the result.

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