Show that the Borel $\sigma$-algebra $\beta$ is the smallest $\sigma$-algebra $A$ that contains all intervals of the form $[a,b)$ I think I should show that $\beta$ and $A$ contain both open intervals and intervals of the form $[a,b)$, but I have no idea how to proceed with either of those.
 A: Half-open intervals $\mathcal{A}=\{[a,b):a<b\}$ are Borel sets. Thus, $\sigma(\mathcal{A})\subset \mathcal{B}_{\mathbb{R}}$. On the other hand, any open set in $\mathbb{R}$ can be approximated by a countable union of the intervals from $\mathcal{A}$ which implies that $\mathcal{B}_{\mathbb{R}}\subset\sigma(\mathcal{A})$.
A: Let $\mathcal{I}$ be the collection of all open intervals, and $\mathcal{A}$ be the collection of all half open-intervals $[a,b)$. We shall prove $\mathcal{B}_\mathbb{R} = \sigma(\mathcal{I})$ and $\sigma(\mathcal{I}) = \sigma(\mathcal{A})$.
Open intervals are open sets so $\mathcal{I} \subseteq \mathcal{B}_\mathbb{R} \implies \sigma(\mathcal{I}) \subseteq \mathcal{B}_\mathbb{R}$.
Take an open set $U \subseteq \mathbb{R}$. For $x \in U$ there exists an open interval $I_x \subset U$ such that $x \in U$. Take $I_x$ to be the largest such interval. We have $U = \bigcup_{x\in U} I_x$. The intervals $I_x$ are disjoint, otherwise they wouldn't be the largest. Since a union of open intervals can be disjoint only if it is at most countable (since each of them contains a distinct rational number), there exists countably many intervals $I_{x_{n}}$ such that $U = \bigcup_{n=1}^\infty I_{x_{n}}$. Thus, $U \in \sigma(\mathcal{I})$.
Hence, $\mathcal{B}_\mathbb{R} = \sigma(\mathcal{I})$.
Now, take an open interval $(a, b) \in \mathcal{I}$ and notice
$$ (a, b) = \bigcup_{n=1}^\infty \underbrace{\left[a +\frac{1}{n}, b\right)}_{\in \mathcal{A}} \in \sigma(\mathcal{A})$$
so $\mathcal{I} \subseteq \sigma(\mathcal{A})\implies \sigma(\mathcal{I}) \subseteq \sigma(\mathcal{A})$.
Similarly, take $[a,b) \in \mathcal{A}$ and notice:
$$ [a, b) = \bigcap_{n=1}^\infty \underbrace{\left(a -\frac{1}{n}, b\right)}_{\in \mathcal{I}} \in \sigma(\mathcal{I})$$
so $\mathcal{A} \subseteq \sigma(\mathcal{I})\implies \sigma(\mathcal{A}) \subseteq \sigma(\mathcal{I})$.
Hence, $\sigma(\mathcal{I}) = \sigma(\mathcal{A})$.
Finally, we can conclude $\mathcal{B}_\mathbb{R} = \sigma(\mathcal{A})$.
