The Glivenko-Cantelli Theorem states that if $F$ is a distribution function, $X_1,\dots,X_n \sim F$, and $\hat{F}_n$ is the empirical distribution function, then $$\sup_{x \in \mathbb{R}} \lvert \hat{F}_n(x) - F(x) \rvert \xrightarrow{a.s.} 0 . \tag{1}$$
How does this differ from simply stating the following? $$\hat{F}_n(x) \xrightarrow{a.s.} F(x) \tag{2}$$
Using the definition of convergence almost surely, from (1): $$\begin{align*} \mathbb{P} \left( \lim_{n\rightarrow\infty} \left\lvert \sup_{x \in \mathbb{R}} \lvert \hat{F}_n(x) - F(x) \rvert \right\rvert = 0 \right) &= 1 \\ \mathbb{P} \left( \lim_{n\rightarrow\infty} \sup_{x \in \mathbb{R}} \lvert \hat{F}_n(x) - F(x) \rvert = 0 \right) &= 1 \tag{1a} \end{align*}$$
Using the definition of convergence almost surely, from (2): $$\begin{align*} \mathbb{P} \left( \lim_{n\rightarrow\infty} \left\lvert \hat{F}_n(x) - F(x) \right\rvert = 0 \right) = 1 \tag{2a} \end{align*}$$
To me, it seems that (1a) and (2a) are equivalent statements because of the least upper bound, and thus (1) and (2) are equivalent statements. But I have a feeling that I'm missing a subtle difference since otherwise I would think the Theorem would just be stated the simpler way (Equation (2)).