# Topological space and $\sigma$-algebra

Let $$(X, \tau)$$ be a topological space and $$\mathcal{B}(X)$$ be the smallest $$\sigma$$-algebra generated by the topology $$\tau$$.

I know that the topology $$\tau$$ is closed under arbitrary (possibly uncountable) unions, while the $$\sigma$$-algebra $$\mathcal{B}(X)$$ is only closed under countable unions. I'm a little confused about the following: how can $$\mathcal{B}(X)$$ contain $$\tau$$, but is only closed under countable unions? Can someone shed some lights on it (maybe provide a small example for the the usual topology in $$\mathbb{R}$$ vs. $$\mathcal{B}(\mathbb{R})$$?)

• If $E$ is a non-Borel set, then singleton sets in $E$ are all Borel sets but their union is not. Commented Oct 14, 2023 at 7:19

$$\mathcal B(X)$$ being closed under countable unions doesn't mean it is forbidden to contain an uncountable union. It simply means that uncountable unions are not, in general, required to be in $$\mathcal B(X)$$. Some uncountable unions may happen to still lie in the $$\sigma$$-algebra anyway, which is fine.
geetha290krm's comment gives an example of an uncountable union of Borel sets that is not Borel - namely, take any non-Borel set to begin with, and consider the singletons that make up that set. So this indicates that uncountable unions are not always $$\mathcal B(X)$$. Note that this example supposes singletons are Borel sets, which follows whenever, for example, the underlying topology $$\tau$$ is $$T_1$$, meaning singletons are closed, and presumes the existence of a non-Borel set, which is the case in $$\mathbb R$$ for example.
But of course, an open set in $$\mathbb R$$, for example, is an uncountable union of singletons, which does happen to still be a Borel set.