It is stated that the the following set equality easily comes from continuity of paths of Brownian motion $B_t$, but I can't seem to make sense of it -
$$\{(\omega,t)\in \Omega\times (0,\infty) : B_t(\omega)\in (a,b)\} = \bigcap_{n=1}^{\infty} \bigcup_{t\in\mathbb{Q}_{+}} \{\omega\in\Omega:B_t(\omega)\in (a,b)\}\times (t-\frac{1}{n},t+\frac{1}{n})$$
Let's call RHS $S$. So $S$ is a countable intersection of sets depending on $n$ which is in turn a countable union of some sets. Am I correct in interpreting that, $(x,y)$ belongs to $S$, if and only if for every $n\ge 1$ there is some positive rational $t$ such that $s\in (t-1/n,t+1/n)$ and for this $s$, $B_s(x)\in (a,b)$?
But then isn't the set $\{\omega\in\Omega:B_t(\omega)\in (a,b)\}$ basically all the $\omega$ such that sample paths generated by them is completely contained in $(a,b)$? Doesn't that make things too special?
I think I'm generally confused by what is meant by $S$. It would be great if you could explain in detail what this means