Zeros of Brownian motion

Question

I wanted to find out why for a Brownian motion $(B_t)_t$ almost surely for every $s\geq 0$ there exist $u,t\geq s$ s.t. $$B_u<0<B_t.$$ I know that BM has to have zeros on every intervall $[0, \epsilon].$ But as $B_s$ varies in $\omega$, I am confused.

Answer 1

If $W_t$ is a Brownian motion, then $B_t = t W_{1/t}$ with $B_0 = 0$ is also a Brownian motion. The switching of signs of $W_t$ on the interval $[s, \infty)$ corresponds to the switching of signs of $B_t$ on the interval $(0, \tfrac{1}{s}]$.