# If $\mathcal{B}$ is a base of a topology space $\left(X,\tau\right)$, then the Borel $\sigma$-algebra is generated by $\mathcal{B}$?

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 $\mathcal{B}$ ?

• That's not a complete question, "The Borel $\sigma$-algebra is generated by $\mathcal{B}$?" Aug 30, 2015 at 22:02

Not necessarily. For example, consider the discrete topology on $\Bbb R,$ which has $$\mathcal B=\bigl\{\{x\}:x\in\Bbb R\bigr\}$$ as a base. The $\sigma$-algebra generated by $\mathcal B$ is the set of all subsets $A$ of $\Bbb R$ such that either (1) $A$ is finite or countably infinite, or (2) $\Bbb R\setminus A$ is finite or countably infinite. However, the appropriate Borel $\sigma$-algebra is the power set of $\Bbb R,$ so for example, $[0,1]$ is an element of the Borel $\sigma$-algebra that is not an element of the $\sigma$-algebra generated by $\mathcal B$.

To some extent, it depends on the topology in question--for example, given any second-countable topology and any base for that topology, the $\sigma$-algebra generated by the base is the Borel $\sigma$-algebra (and I suspect that only second-countable topologies necessarily have this property, unless we assume the Axiom of Choice). On the other hand, it also depends on what base you choose, since every topology has itself as a base.

• Not only second-countable topologies,If the base B is the topology T, for example. Even if we take a base B of smallest possible cardinal, and T is not second -countable ,it can be that every member of T is a union of countably many members of B. Example ;The order topology on cardinal ordinal K,when K is a strong limit and cf(K) is countable. (See my Q on the "grasp" of a space.) Aug 30, 2015 at 23:18
• True, we can always (as I mentioned) simply use the topology, itself, as a base. And it's true that second-countability is at least as strong as the requirement that for any base, every open set is the union of countably-many elements of the base. I wasn't sure if it was strictly stronger, though. Your example shows that it clearly is, if we assume Choice (though I suspect it need not be if we don't). Thanks! Aug 30, 2015 at 23:34
• By the way, with regard to your question on "grasp," is my comment there on target, or am I missing something? Aug 30, 2015 at 23:45
• To Cameron Blue : I made a comment there on your comment there. Aug 31, 2015 at 1:35
• Second-countability is not necessary. If $X$ is hereditary Lindelöf (e.g. second-countable, or separable metrizable) then the Borel $\sigma$-algebra $\sigma(\tau)$ is generated by any subbase of $\tau$ [Fremlin, "Measure Theory" 4A3D].
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.