1. ## Proof of supremum

Prove that if f is a bounded function on a nonempty set S, then |sup f| is less than or equal to sup|f|.

2. Obviously

$f(x)\leq|f(x)|$ for all x, so it holds for supremum:

$\sup_{x} f(x)\leq \sup_x |f(x)|$

Then we have

$- \sup_x |f(x)| \leq \sup_{x} f(x) \leq \sup_x |f(x)|$

hence

$|\sup_{x} f(x) | \leq \sup_x |f(x)|$