borel set on the usual topology in real space is a topology itself or not

Question

let B borel set on the usual topology in the real space R^n

then B contains the entire set and empty set B is closed under the finite intersection B is closed under the countable union

I am doubt if B is the topology in R^n by definition

Topology needs to be closed under the arbitrary union

Plz let me know if B is a topology and if not is there any counterexample where B is not closed under the arbitrary union

Answer 1

As you say, topologies need to be closed under arbitrary unions. If $X$ is a Hausdorff space, one-point sets are closed, and so are Borel sets. Were the Borel sets to be a topology, arbitrary unions of one-point sets would be Borel; all subsets must be Borel. Of course, for the usual topology on $\Bbb R$, not all subsets are Borel.

Answer 2

All open sets (i.e. $\mathcal{T}$) are Borel sets, but the Borel sets $\mathcal{B}$ are defined to be the smallest $\sigma$-algebra that contains $\mathcal{T}$. And $\sigma$-algebras are closed under countable unions, and complements, (and countable intersections, as follows from these two) so we also have all closed sets in $\mathcal{B}$, not just the open sets, and the countable intersection of open sets ($G_\delta$'s) (the irrational are such a set in the reals) and countable unions of closed sets ($F_\sigma$'s) (all countable subsets of the reals, none of which is open) as well. This shows that $\mathcal{B}$ is quite a bit larger than $\mathcal{T}$ in terms of subsets, in that $\mathcal{B}\setminus \mathcal{T}$ can be quite big. This is the case for most "nice" metric spaces, e.g., like $\mathbb{R}$.

It is rare that $\mathcal{T} = \mathcal{B}$ for a space $X$. E.g. this happens for a discrete space $X$ where both collections equal $\mathcal{P}(X)$, all subsets of $X$.