Show this set generates the Borel subset of the reals?

Question

This question is in a past paper I'm doing:

Let $\mathcal{I}$ be the collection of all intervals in $\mathbb{R}$ not containing zero. Show that $\mathcal{I}$ generates the $\sigma$-algebra of Borel subsets of $\mathbb{R}$.

I've been trying to write an arbitrary interval $(a,b)$ as some union or intersection of intervals in $\mathcal{I}$ but haven't had any luck.

Answer 1

If $I$ is an interval and $0 \in I$ then $I^{c}=(I^{c}\cap (-\infty,0))\cup (I^{c}\cap (0,\infty)) $. Can you verify that $I^{c}\cap (-\infty,0)$ and $I^{c}\cap (0,\infty)$ are intervals not containing $0$?