Supremum of Function is less than Supremum of Greater Function

Theorem
Let $f$ and $g$ be real functions.

Let $S$ be a subset of $\operatorname{Dom} \left({f}\right) \cap \operatorname{Dom} \left({g}\right)$.

Let $f \left({x}\right) \le g \left({x}\right)$ for every $x \in S$.

Let $\displaystyle \sup_{x \mathop \in S} g \left({x}\right)$ exist.

Then $\displaystyle \sup_{x \mathop \in S} f \left({x}\right)$ exists and:


 * $\displaystyle \sup_{x \mathop \in S} f \left({x}\right) \le \sup_{x \mathop \in S} g \left({x}\right)$.

Proof
We have:

Supremum of Sum equals Sum of Suprema also gives that $\sup f$ and $\sup \left({g - f}\right)$ exist.

We have: