If $0\leq s<t$, then the $W(s)$ and $W(t)-W(s)$ are independent.
How can we deduce this from the definition of Brownian motion?
Specifically, I know one time step (i.e. $[t,t+1]$) increments are independent, but we do not know how many time steps are apart from $s$ to $t$.
My intuition is that you can write $W(s)$ and $W(t)-W(s)$ as telescoping sum with how many one time-unit increments as possible and the product would always yield the cross-product of one-time step increment cross-products, so in effect, they are independent?
Independent increments means $\{W_{t_{i+1}} -W_{t_i}: 1\leq ,m\}$ is an independent collection whenever $t_1<t_2<...<t_n$. There is no restriction on the valaues of the numbers $t_{i+1}-t_i$ in this definition. Taking $t_1=0, t_2=s$ and $t_3=t$ we get the result.