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I am reading a proof where it is assumed that $$ \lim_{n \to \infty} \sup_{0<s\leq s_0}\left| \frac{t_n(s)}{s}-1 \right|=0 , \hspace{30mm} (1)$$ 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 $$\lim_{n \to \infty} \sup_{0<s\leq s_0} \left| \frac{W(t_n(s))-W(s)}{t_n(s)^{1/2-\epsilon}} \right|=0 \text{ a.s.} \hspace{10mm} (3)$$

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 $\{s: 0<s\leq s_0\}$ only.

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