Let $L=\sup\{t\in[0,1]:B_t=0\}$. How does one show that $L<1$ a.s.? I know that that $B_t$ has infinitely many zeros in [0,1], (in fact, in $(0,\epsilon)$). I also know that the probability of $B_1=0$ is zero, so what is left to show, is that there are is no sequence of zeros which gets arbitrarily close to 1. But I'm not sure how to do this, or whether this is the right approach.
Many thanks in advance!
Let $f:[0,1]\to \mathbb{R}$ be a continuous function. Consider $E(f):=\{x\in [0,1]:f(x)=0\}$ and suppose $\sup E(f)=1$. Then, there exists a sequence $(x_n)_{n \in \mathbb{N}}\subseteq E(f)$ s.t. $x_n\uparrow 1$. Since $f$ is continuous, $f(x_n)\to f(1)$ and since $f(x_n)=0,\,\forall n$ then $f(1)=0$.
Now recall $t\mapsto B_t(\omega)$ are continuous functions. We have: $$P(\sup E(B)=1)\leq P(B_1=0)=0$$ and we conclude.