Prove $\mathbb P\{\sup_{t\geq 0}W_t=+\infty \}=1$ where $(W_t)$ is a Brownian motion.

Question

Let $(W_t)_t$ a Brownian motion. How can I prove that $$\mathbb P(\sup_{t\geq 0}W_t=+\infty )=1\ \ ?$$

Attempts

Set $Z=\sup_{t\geq 0}W_t$. I know that $Z\sim cZ$ for all $c>0$, and thus $$\mathbb P(Z=\infty) =\mathbb P\left(\bigcap_{n\in N^*}(Z\geq n)\right)=\lim_{n\to \infty }\mathbb P(Z\geq n)=P(Z>0)$$ because for all $c>0$, $$\mathbb P(Z>n)=\mathbb P(Z>\frac{n}{c})\underset{c\to 0}{\to} \mathbb P(Z>0).$$

Question

Since $W_0=0$, we have that $\mathbb P(Z\geq 0)=1$, and thus $\mathbb P(Z=0)+\mathbb P(Z=\infty )=1$. But, how can I prove that $\mathbb P(Z=0)=0$ ? I know that it mean that $W_t\leq 0$ for all $t$, but I don't see how to prove that this is $0$.

Answer 1

Let $Z_t:=\sup_{0\le s\le t}B_s$. By the reflection principle, $Z_t\overset{d}{=}|B_t|$, so that $\mathsf{P}(Z\le 0)\le \mathsf{P}(Z_1\le 0)=0$.