Supremum of Absolute Value of Difference equals Supremum of Difference

Theorem
Let $S$ be a non-empty real set.

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

Then $\displaystyle \sup_{x, y \mathop \in S} \left\lvert{x - y}\right\rvert$ exists and:


 * $\displaystyle \sup_{x, y \mathop \in S} \left\lvert{x - y}\right\rvert = \sup_{x, y \mathop \in S} \left({x - y}\right)$.

Proof
Consider the set $\left\{{x - y: x, y \in S, x - y \le 0}\right\}$.

There is a number $x'$ in $S$ as $S$ is non-empty.

Therefore, $0 \in \left\{{x - y: x, y \in S, x - y \le 0}\right\}$ as $x = y = x'$ implies that $x - y = 0$, $x, y \in S$, and $x - y \le 0$.

Also, $0$ is an upper bound for $\left\{{x - y: x, y \in S, x - y \le 0}\right\}$ by definition.

Accordingly:


 * $\displaystyle \sup_{x, y \mathop \in S, x − y \mathop \le 0} \left({x - y}\right) = 0$

Consider the set $\left\{{x - y: x, y \in S, x - y \ge 0}\right\}$.

There is a number $x'$ in $S$ as $S$ is non-empty.

Therefore, $0 \in \left\{{x - y: x, y \in S, x - y \ge 0}\right\}$ as $x = y = x'$ implies that $x - y = 0$, $x, y \in S$, $x - y \ge 0$.

Accordingly:


 * $\displaystyle \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \left({x - y}\right) \ge 0$