prove $\lim_{n\to\infty}\sup\{|f(x) - f_n(x) | : x \in S \} =0$

I understand how the theorem works but how would you prove that a sequence $f_n$ of functions on set $S \subset \mathbb{R}$ converges uniformly iff

$$\lim_{n\to\infty} \sup\{|f(x) - f_n(x) | : x \in S \}=0$$

It's actually trivial if you write everything down.

• For the direct implication:

Let $\epsilon > 0$

Let $N \in \mathbb N$ such that $n \geq N \Rightarrow \forall x \in S, |f_n(x)-f(x)|\leq \epsilon$

Then (by the very definition of $\sup$), for any $n \geq N$ we have that $\sup\{|f(x) - f_n(x) | : x \in S \} \leq \epsilon$

Equivalently said, $\lim_{n\to\infty} \sup\{|f(x) - f_n(x) | : x \in S \}=0$

• For the converse,

Let $\epsilon >0$

Let $N \in \mathbb N$ such that $n \geq N \Rightarrow \sup\{|f(x) - f_n(x) | : x \in S \} \leq \epsilon$.

Then for any $n \geq N$ and any $x \in S$, $|f(x) - f_n(x) | \leq \sup\{|f(x) - f_n(x) | : x \in S \} \leq \epsilon$ , which is the definition of uniform convergence.