Borel Sigma Algebra generated by Open Intervals So I know that the Borel $\sigma$-algebra of $\mathbb{R}$ is the $\sigma$-algebra generated by open sets. I have been able to prove that this Borel $\sigma$-algebra is also generated by the family of open intervals of the form $(a,b),  a,b \in \mathbb{R}$ Now I want to show that the family of open intervals $(a,\infty)$ also generate the Borel $\sigma$-algebra. So what I have done is first show that 
$(a,b) = \bigcup_{q \in \mathbb{Q}, q<b}  ((a, \infty)-(q,\infty))$. This shows  that every open interval is a countable union of intervals of the form $(a,\infty)$ and thus $\sigma$-algebra generated by $(a,b)$ is contained in the $\sigma$-algebra generated by $(a, \infty)$ But how would I show the other way around, i.e. that every interval $(a, \infty)$ is in the Borel $\sigma$- algebra?
(Can we just say that every interval $(a, \infty)$ is infact an open interval in itself and thus belongs to  the $\sigma$-algebra generated by $(a,b)$?)
 A: Denote $\mathcal{B}=\sigma\left(\tau\right)=\sigma\left(\left\{ \left(a,b\right)\mid a,b\in\mathbb{R}\right\} \right)$ (this equality was found out by you allready)
where $\tau$ denotes the topology and $\mathcal{B}_{1}=\sigma\left(\left\{ \left(a,\infty\right)\mid a\in\mathbb{R}\right\} \right)$. 
Then $\left\{ \left(a,\infty\right)\mid a\in\mathbb{R}\right\} \subset\tau$
so that $\mathcal{B}_{1}\subseteq\mathcal{B}$. 
To prove the converse
inclusion it is enough to show that $\left\{ \left(a,b\right)\mid a,b\in\mathbb{R}\right\} \subset\mathcal{B}_{1}$. 
We have $\left(a,c\right]=\left(a,\infty\right)-\left(c,\infty\right)\in\mathcal{B}_{1}$
for each $c$, and consequently $\left(a,b\right)=\bigcup_{n\in\mathbb{N}}\left(a,b-\frac{1}{n}\right]\in\mathcal{B}_{1}$.
A: Hint: $$(a,\infty) = \bigcup_{n \in \mathbb{N}_0} (a+n,a+n+2).$$
And yes: If you know that the Borel-$\sigma$ algebra contains all open sets, then $(a,\infty) \in \mathcal{B}(\mathbb{R})$ follows also from the fact that $(a,\infty)$ is an open set.
