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

  • 4
    $\begingroup$ Yes, it's the same. $\endgroup$ Commented Aug 1, 2013 at 12:34
  • $\begingroup$ Oh, very quick answer. Thank you! Then this little "denotation-problem" is disposed of once and for all. :-) $\endgroup$
    – user34632
    Commented Aug 1, 2013 at 12:36

1 Answer 1


Yes, it's the same. The open sets of $X$ are exactly the sets in the topology of $X$.

  • $\begingroup$ 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 $X$. $\endgroup$ Commented Dec 23, 2015 at 16:03
  • $\begingroup$ @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 $(X,\mathcal{T})$." $\endgroup$
    – Neal
    Commented Dec 24, 2015 at 13:04
  • $\begingroup$ So $\tau$ defines what open sets are in the set $X$? $\endgroup$ Commented Dec 25, 2015 at 11:16
  • $\begingroup$ @user4205580 Yes, the definition of "open set" is "element of $\mathcal{T}$." $\endgroup$
    – Neal
    Commented Dec 25, 2015 at 12:37

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