# Brownian Motion proof: difference converging to 0 almost surely

I am reading a proof where it is assumed that $$\lim_{n \to \infty} \sup_{0 where $$t_n(.)$$ is some sequence of functions. Furthermore, we know that $$\lim_{s \to 0^{+}}s^{-1/2+\epsilon}W(s)=0 \text{ a.s.}\hspace{35mm} (2)$$ for $$W(s)$$ a brownian motion. The author then wants to prove that $$\lim_{n \to \infty} \sup_{0

The author then writes something along the lines of:

Take a sequence $$s_n \to s_0 \geq 0, n \to \infty$$". Then by (1) also $$t_n(s_n) \to s_0$$. For $$s_0>0$$ by continuity of Brownian motion (3) is true. For $$s_0=0$$ we use (2)

I have trouble understanding this proof, why are we looking at a sequence $$s_n$$? I suppose the mentioned continuity of the Brownian motion here is in fact uniform continuity because we are looking at $$\{s: 0 only.