Show Brownian Motion paths increasing has probability $\neq$ 1

Question

Essentially, we are to show then that $$ P(\{\omega:t\to B_t(\omega) \uparrow \})<1$$

How can I start with this? Supposedly the proof is just one line, I'm just not too sure what to apply here. Each independent increment must be increasing and therefore each $B_t − B_{t+1}$ > 0, which has probability $lim_{t \to \infty} \frac{1}{2^t} = 0$?

Answer 1

$P(B_{n+1}-B_n \geq 0 \, \forall n) =\lim_N P(B_{n+1}-B_n \geq 0 \, \text {for} \, 1\leq n \leq N)=\lim_N [P(B_1-B_0 \geq 0)]^{N}=\lim_N \frac 1 {2^{N}} =0$.

I have used the following:

$B_{n+1}-B_n, n=1,2...$ are i.i.d. with $N(0,1)$ distribution and $P(X\geq 0)=\frac1 2$ if $X \sim N(0,1)$.