supremum of brownian motion almost surely > 0 Is it true that
\begin{equation}
M_t = \sup_{s \leq t} B_s > 0 \ \ \text{a.s.}
\end{equation}
for all $t>0$? I remember reading this somewhere, but intuitively, can't the Brownian motion B stay below 0 for some time with probability $>0$?

One way to argue is using Blumenthal's zero one law. Define $A_n=\{B_{1/n}>\frac{1}{\sqrt{n}}\}$, and set $B=\{B_{1/n}>\frac{1}{\sqrt{n}} ~\text{i.o.}\} $
Then,
\begin{align*}
\mathbb{P}(B) &=\mathbb{P}(\limsup_n A_n)\\
&\geq \limsup_n\mathbb{P}( A_n)\\
&= \limsup_n\mathbb{P}(B_{1/n}>\frac{1}{\sqrt{n}})\\
&=\limsup_n\mathbb{P}(N(0,1)>1)\\
&=\mathbb{P}(N(0,1)>1)=M>0
\end{align*}
By Blumenthal's zero one law $\mathbb{P}(B)= 1 ~\text{or}~0$. So $\mathbb{P}(B)=1$. 
