How can every continuous function be measurable if topologies can contain an uncountable number of open sets?

Question

If I have a topological space $(Y, \tau)$ with topology $\tau$ which contains an uncountable union of sets in $\tau$, and a measurable space $(X, M)$ with $\sigma-$algebra $M$ can by definition contain only countably many unions of measurable sets, do there exist continuous function $f: X \to Y$ which are not measurable? Also, how is it possible that I could define a Borel $\sigma-$algebra on $(Y, \tau)$? A particular theorem states "Let $(Y, \tau)$ be a topological space. There exists a smallest $\sigma-$algebra $B$ in $Y$ such that every open set in $Y$ belongs to $B$." and yet here this appears impossible since $\sigma-$algebras contain at most countably many sets. Where's my mistake?

Answer 1

$\sigma$-algebras can contain much more than just a countable collection of sets - even for a countable basis, the collection of all countable unions and intersections is uncountable. For example, take the basis to be $\mathbb{N}$, then $\mathcal{P}\left(\mathbb{N}\right)\subset\sigma\left(\mathbb{N}\right)$, and $\left|\mathcal{P}\left(\mathbb{N}\right)\right|=\left|\mathbb{R}\right|$. If your basis is the open sets of a general topological space, the corresponding $\sigma$-algebra can be even bigger, and mostly bigger than the cardinality of the basis.