TITLE: If $\mathcal{B}$ is a base of a topology space $\left(X,\tau\right)$, then the Borel $\sigma$-algebra is generated by $\mathcal{B}$?
QUESTION [14 upvotes]: Let $\left(X,\tau\right)$ a topology space and   $\mathcal{B}$  a base of the topology,  my question is:
The Borel $\sigma$-algebra is generated by $\mathcal{B}$ ?

REPLY [7 votes]: As Cameron Buie pointed out, the claim is not true in general.

However, it is true if $\tau$ is second-countable and $\mathcal B$ is a countable topological basis (in particular, it is true in separable metric spaces). To see this, suppose that $U\in\mathcal \tau$ is any open set. Then, there exists a subcollection $\mathcal B_{U}\subseteq\mathcal B$, necessarily countable, such that $$U=\bigcup_{B\in\mathcal B_U}B.$$ This union is countable, so that $U\in\sigma(\mathcal B)$. Since $U$ is an arbitrary open set, it follows that $\tau\subseteq\sigma(\mathcal B)$, and, in turn, $\sigma(\tau)\subseteq\sigma(\mathcal B)$.
The other direction $\sigma(\mathcal B)\subseteq\sigma(\tau)$ is easy, given that $\mathcal B\subseteq\tau$. It follows that the Borel $\sigma$-algebra $\sigma(\tau)$ equals the $\sigma$-algebra generated by the topological basis $\mathcal B$.

In fact, if $\tau$ is second-countable, then any topological basis $\mathcal C$, countable or not, will generate the Borel $\sigma$-algebra. This is because then $\mathcal C$ contains a countable subcollection $\mathcal B\subseteq\mathcal C$ that is still a basis (see here for details). As shown above, $\sigma(\tau)=\sigma(\mathcal B)\subseteq\sigma(\mathcal C)$, and the other direction $\sigma(\mathcal C)\subseteq\sigma(\tau)$ is again easy.