# The definition of Borel sigma algebra

In the text of Probability Essentials by J.Jacod & P.Protter, a theorem:

The Borel $\sigma$- algebra of $R$ is generated by intervals of the form $(-\infty,a ]$, where $a \in Q$.

As far as I've known the Borel sigma algebra is generated by all open subsets of $R$, which surely contains sets like (x, y), here x is irrational. So my question is how can 'a', in theorem, a rational generate such set (x, y).

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$(x,y) = \displaystyle\bigcup_{\substack{a,b\in\mathbb{Q} \\ x<a<b<y}} (a,b)$.
@DylanZhu Filling in more details: $(a,b'] = (-\infty,b'] \cap (-\infty,a']^c$, and $(a,b) = \cup_{b'\in\mathbb{Q}, b'<b} (a,b']$. Let me know if you have any specific questions. – Slade Oct 13 '13 at 19:04
The rationals are dense. For any $(x,x+\epsilon)$, pick $n > 1/\epsilon$, then some rational $t/n$ must land in there. – Slade Oct 13 '13 at 20:09