proving a wiki statement about uniform convergence and supremum This wiki page says that a sequence $f_n$ is uniformally convergenct to  $f: S \to \mathbb{R}$ if and only if $||f_n - f||_{\sup} \to 0$. I am trying to prove this statement.
For the forward implication, we have that $f_n \to f$ uniformaly, so we have that $|f_n - f| < \epsilon$. Now, it seems as though the obvious option here is to simply take the supremum of this to get the result if we choose $N$ which gives us $|f_n - f| < \epsilon$. How would I do this step by step, I mean - it seems wrong to just take the supremum of everything as this doesn't use the fact that $f_n$ is uniformaly convergent to $f$
I have yet to try the backwards implication.

It's just another way to state the definition :
This is your base definition
$$\forall \epsilon > 0, \exists N, \forall n > N, \forall x, \quad |f_n(x)-f(x)|< \epsilon$$
This is what means $\|f_n-f\|_{\sup} \to 0$ :
$$\forall \epsilon > 0, \exists N, \forall n > N, \quad  \|f_n-f\|_{\sup} < \epsilon$$
And now replace $\|f_n-f\|_{\sup}$ by it's definition :
$$\forall \epsilon > 0, \exists N, \forall n > N, \quad  \sup_{x} |f_n(x)-f(x)| < \epsilon$$
And it's the same thing that
$$\forall \epsilon > 0, \exists N, \forall n > N, \forall x, \quad |f_n(x)-f(x)| < \epsilon$$
