Almost surely, for all $s \ge 0$, there exist $t$, $u \ge s$ with $B_t < 0 < B_u$?

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Let $(B_t)_{t \ge 0}$ be a Brownian motion starting from $0$. Then, do we have that, almost surely, for all $s \ge 0$, there exist $t$, $u \ge s$ with $B_t < 0 < B_u$?