# Topological space and Borel-$\sigma$-Algebra

I have a very little (and maybe banal) question concerning a topological space and it's Borel-$\sigma$-algebra. If I have a topological space $(X,\tau)$, then one says that $\mathcal{B}(X)$ is the smallest $\sigma$-algebra which contains the open sets of $X$.

Is it the same to say: $\mathcal{B}(X)$ is the smallest $\sigma$-algebra containing the topology $\tau$?

To my opinion: Yes, because the sets in $\tau$ are just the sets called "open sets".

With regards

• Yes, it's the same. – Daniel Fischer Aug 1 '13 at 12:34
• Oh, very quick answer. Thank you! Then this little "denotation-problem" is disposed of once and for all. :-) – math12 Aug 1 '13 at 12:36

Yes, it's the same. The open sets of $X$ are exactly the sets in the topology of $X$.
• One question - there are many different topologies of $X$, and only one topology of these contains all open sets. So not all topologies on $X$ contain Borel sets, right? I guess it would be best to say that $B(R)$ is a set containing all topologies of $X$. – user4205580 Dec 23 '15 at 16:03
• @user4205580 "Open sets" is contextual. Once you have fixed a topology $\mathcal{T}$ on $X$, its elements are by definition the "open sets of $X$," or more precisely, "the open sets of the topological space $(X,\mathcal{T})$." – Neal Dec 24 '15 at 13:04
• So $\tau$ defines what open sets are in the set $X$? – user4205580 Dec 25 '15 at 11:16
• @user4205580 Yes, the definition of "open set" is "element of $\mathcal{T}$." – Neal Dec 25 '15 at 12:37