I came across the following. For any fixed $n$, let $\{X_{n}(s) \}_{s\geq0}$ be a stochastic process and let $\{B_n(s) \}_{s\geq0}$ be a Brownian motion. We wish to study the behaviour of $\{X_{n}(s) \}_{s\geq0}$ as $n \to \infty$. The result I am looking at more or less says that we can define a sequence (over $n$) of Brownian motions $\{B_n(s) \}_{s\geq0}$ such that for a normalization of $X_{n}(s)$, $X_{n}(s)^*$, it holds that $$|X_{n}(s)^*-B_{n}(s)| \overset{p}{\to}0, \hspace{15mm} (n \to \infty)$$ uniformly for $s$ in some closed interval.
My question is, what is the use (or necessity even) of looking at a sequence of Brownian motions here rather than a single Brownian motion? Any Brownian motion has the same distributional properties and they only differ by possibly having different paths for any $\omega\in \Omega$ with $\Omega$ being the set of the probability space $(\Omega, \mathcal{F}, P)$ on which the sequences are defined. Couldn't we be done with a single Brownian motion since we are concerned with behaviour as $n \to \infty$ anyway?
I have added a Link to the theorem from which the theorem I am looking at is derived, note that $Q(t)$ is the quantile function and $X_{k,n}$ is the k-th order statistic of a random sample of size $n$.
