Almost sure equality between two sets involving a brownian motion

Question

In the stochastic processes course I am taking, it is stated that if $B$ is a Brownian motion, then up to a set of zero probability, we have $$\bigcup_{1 \leq t \leq s} B^{-1}_t(0) = \bigcap_{n = 1}^\infty \bigcup_{t \in [1, s]\cap \mathbb{Q}} B_t^{-1}\left(-\frac{1}{n}, \frac{1}{n}\right)$$ I can see that since $B_t$ is almost surely continuous, the left hand side is included in the right hand side. For the other side, I can see that given $$\omega \in \bigcap_{n = 1}^\infty \bigcup_{t \in [1, s]\cap \mathbb{Q}} B_t^{-1}\left(-\frac{1}{n}, \frac{1}{n}\right),$$ for all $n \geq 1$, there exists $t \in [1, s]$ such that $B_t(\omega) \in (-1/n, 1/n)$, but I can't how to use this to deduce that there exists $t \in [1,s]$ such that $B_t(\omega) = 0$. In particular, couldn't $B_t(\omega)$ be decreasing but never $0$? Is anyone able to help me with this?

Answer 1

Easy consequence of compactness of $[1,s]$. If we have points $t_n \in [1,s]\cap \mathbb Q$ such that $|B_{t_n}|<\frac 1 n$ for all $n$, then there is a subsequence of $t_n$ conveging to some point $t$ of $[1,s]$ and we get $B_t=0$ in the limit.